Non-Rationalised Science NCERT Notes and Solutions (Class 6th to 10th)
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Non-Rationalised Science NCERT Notes and Solutions (Class 11th)
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Non-Rationalised Science NCERT Notes and Solutions (Class 12th)
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Class 12th (Physics) Chapters
1. Electric Charges And Fields	2. Electrostatic Potential And Capacitance	3. Current Electricity
4. Moving Charges And Magnetism	5. Magnetism And Matter	6. Electromagnetic Induction
7. Alternating Current	8. Electromagnetic Waves	9. Ray Optics And Optical Instruments
10. Wave Optics	11. Dual Nature Of Radiation And Matter	12. Atoms
13. Nuclei	14. Semiconductor Electronics: Materials, Devices And Simple Circuits

Chapter 1 Electric Charges And Fields

Introduction

Many everyday phenomena like seeing sparks when taking off synthetic clothes in dry weather, hearing a crackle, or experiencing a small electric shock when touching a car door handle are due to electric charges accumulating on insulating surfaces through rubbing and then discharging through the body. This build-up of electric charges is often referred to as **static electricity**.

The study of the forces, electric fields, and electric potentials that arise from electric charges that are stationary (not moving or changing with time) is known as **Electrostatics**.

Electric Charge

Historically, the property of attracting light objects after rubbing was first observed by Thales of Miletus with amber and wool around 600 BC. The term 'electricity' originates from the Greek word 'elektron', meaning amber.

Experiments involving rubbing various materials (like glass rods with silk, plastic rods with cat's fur) and observing their interactions with each other or with light objects (like paper bits or pith balls) revealed key properties:

Two glass rods rubbed with silk repel each other.
Two plastic rods rubbed with fur repel each other.
A glass rod rubbed with silk attracts a plastic rod rubbed with fur.
Similarly, materials used for rubbing also interact: silk repels silk, fur repels fur, but silk attracts plastic, and fur attracts glass.

These observations led to the conclusion that rubbing results in **electrification**, where bodies acquire an electric charge. There are only **two kinds** of electric charge. It was found that:

Like charges repel each other.
Unlike charges attract each other.

The property that differentiates these two kinds of charge is called the **polarity of charge**.

When two objects are rubbed and become electrified, one acquires one kind of charge, and the other acquires the second kind. If these charged bodies are brought back into contact, they lose their electrification and no longer attract or repel other objects. This indicates that the effects of the two kinds of charge neutralise or nullify each other.

The American scientist Benjamin Franklin named these two types of charges as **positive** and **negative**. By convention, the charge acquired by a glass rod rubbed with silk is called **positive**, and the charge acquired by a plastic rod rubbed with fur is called **negative**. An object possessing an electric charge is said to be **electrified** or **charged**. An object with no net charge is **electrically neutral**.

Historically, electricity and magnetism were considered separate phenomena until Oersted discovered in 1820 that electric current deflects a compass needle, showing a connection. This led to the development of **electromagnetism**, which describes the interdependence of electric and magnetic fields. Many fundamental forces in nature, including those holding matter together and controlling biological processes, are electromagnetic in origin.

A simple device used to detect the presence and nature of electric charge on a body is the **gold-leaf electroscope**. It consists of a metal rod with two thin gold leaves at the bottom. When a charged object touches the metal knob at the top, charge is transferred to the leaves, causing them to diverge (repel each other) due to like charges. The degree of divergence indicates the amount of charge.

Gold leaf electroscope and simple home-made electroscope.

To understand why bodies acquire charge, we consider their atomic structure. Matter is made of atoms, which are normally electrically neutral because the positive charge of protons in the nucleus is balanced by the negative charge of electrons orbiting the nucleus. Charging a neutral body involves adding or removing electrons. In solids, some electrons, being less tightly bound, can be transferred. A body becomes **positively charged** by losing electrons (deficit of negative charge) and **negatively charged** by gaining electrons (excess of negative charge).

When two bodies are rubbed, electrons are transferred from one to the other. The number of electrons transferred is a very small fraction of the total electrons in the body. Only the loosely bound electrons are transferred. Thus, charging by rubbing involves charge transfer, not creation or destruction of charge.

Conductors And Insulators

Materials differ in their ability to allow electric charge to pass through them.

Conductors: Substances that allow electricity to pass through them easily. They contain mobile electric charges (usually electrons in metals) that are relatively free to move throughout the material. Examples: Metals (copper, aluminium), human and animal bodies, Earth.
Insulators (Dielectrics): Substances that offer very high resistance to the passage of electricity. Charges placed on an insulator tend to stay at the location where they were placed. Examples: Glass, porcelain, plastic, nylon, wood, rubber.
Semiconductors: A third category with electrical conductivity intermediate between conductors and insulators.

When charge is transferred to a conductor, it quickly distributes itself over the entire surface. When charge is placed on an insulator, it remains localised.

This property explains why a metal spoon held in hand doesn't electrify significantly when rubbed, as charges would immediately leak through the body to the ground (since both are conductors). A plastic comb, being an insulator, retains charge when rubbed against dry hair.

Bringing a charged body into contact with the Earth is called **grounding** or **earthing**. Excess charge on the body flows to the Earth, which is a vast conductor. This process provides electrical safety in buildings. Household electrical wiring includes a live wire, a neutral wire, and an earth wire connected to a buried metal plate. Metallic appliance bodies are connected to the earth wire. If a live wire touches the metallic body, charge flows safely to the Earth, preventing electric shock.

Charging By Induction

Charging a body by direct contact with a charged object is one method. Another method is **charging by induction**, where a charged object is brought near a neutral conductor without touching it, causing a separation of charges in the conductor. This separated charge can then be fixed by grounding.

Steps for charging a metal sphere positively by induction using a negatively charged rod:

(i) Place a neutral metal sphere on an insulating stand.

(ii) Bring a negatively charged rod close to the sphere (without touching). Free electrons in the sphere are repelled and accumulate on the far side, leaving the near side positively charged. This charge separation occurs almost instantly until equilibrium is reached.

(iii) Connect the sphere to the ground with a conducting wire. The repelled electrons flow from the sphere to the ground. The positive charges on the near side remain held by the attraction of the negative rod.

(iv) Disconnect the sphere from the ground. The positive charge remains on the near end, still attracted by the rod.

(v) Remove the charged rod. The positive charge redistributes uniformly over the sphere's surface. The sphere is now positively charged.

Steps showing charging a metal sphere positively by induction using a negatively charged rod.

In this process, the initial charged object (the rod) does not lose any charge, unlike charging by contact. Charging a sphere negatively by induction follows similar steps, but electrons flow from the ground to the sphere when a positively charged rod is used.

Charging by induction also explains why charged objects attract uncharged light objects (like paper bits). The charged object induces opposite charges on the near surface of the neutral object and like charges on the farther surface. The attractive force between the charged object and the induced opposite charge on the near surface is stronger than the repulsive force between the charged object and the induced like charge on the farther surface (because force decreases with distance). This net attractive force pulls the light object towards the charged object.

Example 1.1. How can you charge a metal sphere positively without touching it?

Answer:

To charge a metal sphere positively without touching it, the process of charging by induction is used, typically involving a negatively charged object.

1. Place the neutral metal sphere on an insulating stand (to prevent charges from flowing to the ground prematurely).

2. Bring a negatively charged rod close to the sphere, but do not allow it to touch the sphere. The free electrons in the sphere will be repelled by the negative rod and move to the side of the sphere farthest from the rod. This leaves a net positive charge on the side of the sphere nearest to the rod.

3. While the negatively charged rod is held near the sphere, connect the sphere to the ground using a conducting wire. The accumulated negative charges on the far side of the sphere will flow from the sphere to the ground, driven by the repulsion from the rod. The positive charges on the near side remain attracted to the negative rod and stay on the sphere.

4. Disconnect the wire connecting the sphere to the ground. The positive charge is now trapped on the sphere, held in place by the attraction of the nearby negative rod.

5. Remove the negatively charged rod. The positive charge, no longer held by the rod, will redistribute uniformly over the surface of the metal sphere. The sphere is now positively charged.

This method allows the sphere to gain a positive charge without any direct transfer of charge from the negatively charged rod, meaning the rod retains its original charge.

Basic Properties Of Electric Charge

Electric charge, besides having two types (positive and negative) whose effects can cancel, possesses other fundamental properties.

When charged bodies are very small compared to the distance between them, they are treated as **point charges**, with all charge concentrated at a single point.

Additivity Of Charges

Electric charge is a **scalar** quantity, like mass. The total electric charge of a system containing multiple point charges is the simple algebraic sum of all the individual charges. If a system has charges $q_1, q_2, q_3, ..., q_n$, the total charge $Q_{total} = q_1 + q_2 + q_3 + ... + q_n$. Proper signs (positive or negative) must be used when adding charges.

Example: A system with charges +1, +2, -3, +4, and -5 units has a total charge of $(+1) + (+2) + (-3) + (+4) + (-5) = -1$ unit.

Unlike mass, which is always positive, charge can be positive or negative.

Charge Is Conserved

In an isolated system, the total electric charge remains constant over time. Charges can be transferred between bodies (as in charging by rubbing), or charged particles can be created or destroyed (e.g., a neutron decaying into a proton and an electron, where the created positive and negative charges are equal, maintaining zero net charge). However, the net charge of the isolated system is always conserved.

This principle of **conservation of charge** is a fundamental experimental observation.

Quantisation Of Charge

Experimental evidence shows that all free electric charges are integral multiples of a fundamental, basic unit of charge, denoted by $e$. This means any observable charge $q$ on a body is always given by:

$\mathbf{q = ne}$

where $n$ is an integer (positive, negative, or zero). This basic unit $e$ is the magnitude of the charge carried by an electron or a proton. By convention, the charge on an electron is $-e$, and the charge on a proton is $+e$.

This property is called **quantisation of charge**. It was first suggested by Faraday's laws of electrolysis and definitively demonstrated by Millikan's oil drop experiment in 1912.

The SI unit of electric charge is the **coulomb (C)**. It is defined based on the unit of electric current (ampere). One coulomb is the charge transferred in 1 second by a current of 1 ampere.

The value of the basic unit of charge in SI units is approximately:

$\mathbf{e = 1.602192 \times 10^{-19} \text{ C}}$

This means about $6 \times 10^{18}$ electrons are needed to make up a charge of $-1$ C. A coulomb is a very large unit for typical electrostatic phenomena.

Any observable charge is the difference between the number of protons ($n_2$) and electrons ($n_1$) multiplied by $e$: $q = (n_2 - n_1)e$. Since $n_1$ and $n_2$ are integers, the net charge is always an integer multiple of $e$. Charge can only change in discrete steps of $e$.

At the **macroscopic level**, where we deal with charges of the order of $\mu$C, the number of elementary charges ($n$) is enormous (e.g., $1 \mu\text{C} \approx 6 \times 10^{12} e$). The step size $e$ is so small compared to the total charge that the quantisation is not noticeable. Charge appears to be continuous. This is analogous to how a dotted line appears continuous from a distance.

However, at the **microscopic level**, where charges are small multiples of $e$ (e.g., a few tens or hundreds of $e$), the discrete nature of charge is significant, and quantisation cannot be ignored.

Example 1.2. If $10^9$ electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body?

Answer:

Number of electrons moving per second = $10^9$.

Charge of one electron, $e = 1.6 \times 10^{-19}$ C.

Charge transferred per second = (Number of electrons per second) $\times$ (Charge per electron)

Charge per second $= 10^9 \times 1.6 \times 10^{-19}\text{ C} = 1.6 \times 10^{-10}\text{ C/s}$.

We want to accumulate a total charge of 1 C.

Time required = (Total charge desired) / (Charge transferred per second)

Time $= \frac{1\text{ C}}{1.6 \times 10^{-10}\text{ C/s}} = 6.25 \times 10^9\text{ s}$.

To convert seconds to years, we divide by the number of seconds in a year ($365 \text{ days/year} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} \approx 3.15 \times 10^7\text{ s/year}$).

Time in years $= \frac{6.25 \times 10^9\text{ s}}{3.15 \times 10^7\text{ s/year}} \approx 198.4 \text{ years}$.

It would take approximately 200 years to accumulate a charge of 1 C at this rate. This demonstrates that 1 Coulomb is a very large unit of charge in everyday contexts.

Example 1.3. How much positive and negative charge is there in a cup of water?

Answer:

Let's assume the mass of a cup of water is 250 g.

The molecular formula of water is $\textsf{H}_2\text{O}$. The molecular mass of water is $2 \times \text{Atomic Mass (H)} + 1 \times \text{Atomic Mass (O)} = 2 \times 1.008 + 16.00 \approx 18.016 \text{ g/mol}$. We can approximate it as 18 g/mol.

One mole of water contains Avogadro's number of molecules, $N_A \approx 6.022 \times 10^{23}$ molecules.

Number of moles in 250 g of water $= \frac{\text{Mass}}{\text{Molecular Mass}} = \frac{250 \text{ g}}{18 \text{ g/mol}} \approx 13.89 \text{ mol}$.

Total number of water molecules in the cup $= \text{Number of moles} \times N_A \approx 13.89 \times 6.022 \times 10^{23} \text{ molecules} \approx 8.36 \times 10^{24} \text{ molecules}$.

Each water molecule ($\textsf{H}_2\text{O}$) consists of 2 hydrogen atoms and 1 oxygen atom.

Each H atom has 1 proton and 1 electron. Total protons/electrons from 2 H atoms = 2.
Each O atom has 8 protons and 8 electrons. Total protons/electrons from 1 O atom = 8.

So, each water molecule contains $2+8=10$ protons and $2+8=10$ electrons. This means a water molecule is electrically neutral.

The magnitude of positive charge in one molecule = 10 protons $\times e = 10e$.

The magnitude of negative charge in one molecule = 10 electrons $\times e = 10e$.

Total positive charge in the cup = (Total number of water molecules) $\times$ (Positive charge per molecule)

$Q_{positive} = (8.36 \times 10^{24}) \times (10 \times 1.602 \times 10^{-19} \text{ C})$

$Q_{positive} \approx 8.36 \times 10^{25} \times 1.602 \times 10^{-19} \text{ C}$

$Q_{positive} \approx 1.34 \times 10^7 \text{ C}$.

Total negative charge in the cup = (Total number of water molecules) $\times$ (Negative charge per molecule)

$Q_{negative} = (8.36 \times 10^{24}) \times (10 \times -1.602 \times 10^{-19} \text{ C})$

$Q_{negative} \approx -1.34 \times 10^7 \text{ C}$.

The total positive charge is about $1.34 \times 10^7$ C, and the total negative charge is about $-1.34 \times 10^7$ C. The net charge of the cup of water is zero, but it contains enormous amounts of positive and negative charges that exactly balance each other.

Coulomb’S Law

Coulomb's law provides a quantitative description of the electrostatic force between two point charges. It states that the magnitude of the force ($F$) between two point charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum is directly proportional to the product of the magnitudes of the charges ($|q_1 q_2|$) and inversely proportional to the square of the distance between them ($r^2$). The force acts along the straight line joining the two charges.

Mathematically, the magnitude of the force is given by:

$\mathbf{F = k \frac{|q_1 q_2|}{r^2}}$

where $k$ is a constant of proportionality.

Coulomb used a **torsion balance** to measure the force between charged spheres, allowing him to deduce the inverse square relationship by varying the distance and the direct proportionality to the product of charges by using spheres with known charge fractions (obtained by contact with identical uncharged spheres, based on conservation and additivity of charge).

In the International System (SI) of Units, the constant $k$ is chosen such that $k \approx 9 \times 10^9 \text{ N m}^2/\text{C}^2$. This choice defines the unit of charge as the **coulomb (C)** (as defined earlier based on the ampere). With this value of $k$, if $q_1 = q_2 = 1 \text{ C}$ and $r = 1 \text{ m}$, the force is $F = 9 \times 10^9$ N. Thus, 1 Coulomb is the charge that repels an equal charge at 1 m distance in vacuum with a force of $9 \times 10^9$ N.

For convenience, the constant $k$ is often expressed as $k = \frac{1}{4\pi\epsilon_0}$, where $\epsilon_0$ is called the **permittivity of free space**. Coulomb's law in this form is:

$\mathbf{F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2}}$

The value of $\epsilon_0$ in SI units is $\epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$.

Since force is a vector quantity, Coulomb's law can be written in vector form. Let $\vec{r}_1$ and $\vec{r}_2$ be the position vectors of charges $q_1$ and $q_2$. Let $\vec{r}_{21} = \vec{r}_2 - \vec{r}_1$ be the vector from $q_1$ to $q_2$, and $\hat{r}_{21} = \vec{r}_{21}/|\vec{r}_{21}|$ be the unit vector in that direction. The force on $q_2$ due to $q_1$, denoted by $\vec{F}_{21}$, is:

$\mathbf{\vec{F}_{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{21}^2} \hat{r}_{21}}$

where $r_{21} = |\vec{r}_{21}|$.

Remarks on the vector form:

This equation is valid for any sign of $q_1$ and $q_2$.
- If $q_1 q_2 > 0$ (like charges), $\vec{F}_{21}$ is in the direction of $\hat{r}_{21}$, representing repulsion.
- If $q_1 q_2 < 0$ (unlike charges), $\vec{F}_{21}$ is in the direction of $-\hat{r}_{21}$ (which is $\hat{r}_{12}$), representing attraction.
The force on $q_1$ due to $q_2$, $\vec{F}_{12}$, is given by interchanging 1 and 2: $\vec{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_2 q_1}{r_{12}^2} \hat{r}_{12}$. Since $\hat{r}_{12} = -\hat{r}_{21}$ and $r_{12}=r_{21}$, this gives $\vec{F}_{12} = -\vec{F}_{21}$, consistent with **Newton's third law** of motion.
Equation (1.3) describes the force between charges in vacuum. The presence of matter modifies the force (discussed in the next chapter).

Example 1.4. Coulomb’s law for electrostatic force between two point charges and Newton’s law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively.

(a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons. ($\textsf{m}_p = 1.67 \times 10^{–27}$ kg, $\textsf{m}_e = 9.11 \times 10^{–31}$ kg, $\textsf{G} = 6.67 \times 10^{–11} \textsf{ N m}^2/\textsf{kg}^2$, $e = 1.6 \times 10^{–19}$ C, $\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9} \textsf{ N m}^2/\textsf{C}^2$)

(b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Å (= $10^{-10}$ m) apart?

Answer:

(a) Comparison of forces:

(i) For an electron ($q_e = -e$, $m_e$) and a proton ($q_p = +e$, $m_p$) separated by distance $r$.

Electric force magnitude: $F_e = \frac{1}{4\pi\epsilon_0} \frac{|(-e)(+e)|}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$. It is attractive.

Gravitational force magnitude: $F_g = G \frac{m_e m_p}{r^2}$. It is attractive.

Ratio of magnitudes: $\frac{F_e}{F_g} = \frac{\frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}}{G \frac{m_e m_p}{r^2}} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{G m_e m_p}$. The $r^2$ terms cancel, showing the ratio is independent of distance.

Substituting values: $\frac{F_e}{F_g} = (9 \times 10^9 \text{ N m}^2/\text{C}^2) \times \frac{(1.6 \times 10^{-19}\text{ C})^2}{(6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (9.11 \times 10^{-31}\text{ kg}) \times (1.67 \times 10^{-27}\text{ kg})}$

$\frac{F_e}{F_g} \approx 9 \times 10^9 \times \frac{2.56 \times 10^{-38}}{6.67 \times 10^{-11} \times 1.52 \times 10^{-57}} = 9 \times 10^9 \times \frac{2.56 \times 10^{-38}}{10.14 \times 10^{-68}}$

$\frac{F_e}{F_g} \approx 9 \times 10^9 \times 0.25 \times 10^{30} \approx 2.25 \times 10^{39}$. The text gives a value of $2.4 \times 10^{39}$, which is close.

Conclusion: The electric force between an electron and a proton is about $10^{39}$ times stronger than the gravitational force between them.

(ii) For two protons ($q_p = +e$, $m_p$) separated by distance $r$.

Electric force magnitude: $F_e = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$. It is repulsive.

Gravitational force magnitude: $F_g = G \frac{m_p m_p}{r^2} = G \frac{m_p^2}{r^2}$. It is attractive.

Ratio of magnitudes: $\frac{F_e}{F_g} = \frac{\frac{1}{4\pi\epsilon_0} e^2}{G m_p^2}$.

$\frac{F_e}{F_g} \approx 9 \times 10^9 \times \frac{2.56 \times 10^{-38}}{6.67 \times 10^{-11} \times 2.79 \times 10^{-54}} = 9 \times 10^9 \times \frac{2.56 \times 10^{-38}}{18.61 \times 10^{-65}}$

$\frac{F_e}{F_g} \approx 9 \times 10^9 \times 0.137 \times 10^{27} \approx 1.23 \times 10^{36}$. The text gives $1.3 \times 10^{36}$, which is close.

Conclusion: Electric forces are enormously stronger than gravitational forces, indicating that gravity's dominance on large scales is due to the cancellation of positive and negative electric charges.

(b) Accelerations:

The distance between the electron and proton is $r = 1$ Å $= 10^{-10}$ m.

The magnitude of the electric force between them is:

$|F| = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} = (9 \times 10^9 \text{ N m}^2/\text{C}^2) \times \frac{(1.6 \times 10^{-19}\text{ C})^2}{(10^{-10}\text{ m})^2}$

$|F| = 9 \times 10^9 \times \frac{2.56 \times 10^{-38}}{10^{-20}}\text{ N} = 9 \times 10^9 \times 2.56 \times 10^{-18}\text{ N} = 23.04 \times 10^{-9}\text{ N} \approx 2.3 \times 10^{-8}\text{ N}$.

Acceleration of the electron ($a_e$) using Newton's second law ($F = m_e a_e$):

$a_e = \frac{|F|}{m_e} = \frac{2.3 \times 10^{-8}\text{ N}}{9.11 \times 10^{-31}\text{ kg}} \approx 0.252 \times 10^{23}\text{ m/s}^2 \approx 2.5 \times 10^{22}\text{ m/s}^2$.

This acceleration is vastly greater than the acceleration due to gravity ($g \approx 9.8 \text{ m/s}^2$).

Acceleration of the proton ($a_p$) using Newton's second law ($F = m_p a_p$):

$a_p = \frac{|F|}{m_p} = \frac{2.3 \times 10^{-8}\text{ N}}{1.67 \times 10^{-27}\text{ kg}} \approx 1.377 \times 10^{19}\text{ m/s}^2 \approx 1.4 \times 10^{19}\text{ m/s}^2$.

Although the proton's acceleration is smaller than the electron's (due to its larger mass), it is still extremely large compared to $g$. The effect of gravity on the motion of these charged particles at this distance is negligible.

Example 1.5. A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in Fig. 1.7(a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Fig. 1.7(b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in Fig. 1.7(c). What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centres.

Diagram for Example 1.5 showing initial charges and separation, then touching with uncharged spheres, then new separation.

Answer:

Let the initial charge on sphere A be $q_A$ and on sphere B be $q_B$. The initial distance between their centers is $r_1 = 10$ cm $= 0.1$ m.

The initial force of repulsion between A and B is $F_1 = \frac{1}{4\pi\epsilon_0} \frac{q_A q_B}{r_1^2}$. The force is repulsive, implying $q_A$ and $q_B$ have the same sign (both positive or both negative).

In Fig. 1.7(b), sphere A is touched by an identical but uncharged sphere C. Since A and C are identical conductors, the charge on A ($q_A$) is shared equally between A and C. After contact, sphere A has a new charge $q_A' = q_A/2$, and sphere C has charge $q_C = q_A/2$. C is then removed.

Similarly, sphere B is touched by an identical but uncharged sphere D. The charge on B ($q_B$) is shared equally between B and D. After contact, sphere B has a new charge $q_B' = q_B/2$, and sphere D has charge $q_D = q_B/2$. D is then removed.

Now, in Fig. 1.7(c), spheres A (with charge $q_A' = q_A/2$) and B (with charge $q_B' = q_B/2$) are brought to a new distance $r_2 = 5.0$ cm $= 0.05$ m between their centers. The force of repulsion between A and B is now $F_2$.

$F_2 = \frac{1}{4\pi\epsilon_0} \frac{q_A' q_B'}{r_2^2}$

Substitute the new charges and the new distance:

$F_2 = \frac{1}{4\pi\epsilon_0} \frac{(q_A/2) (q_B/2)}{(r_1/2)^2}$ (since $r_2 = r_1/2$)

$F_2 = \frac{1}{4\pi\epsilon_0} \frac{q_A q_B / 4}{r_1^2 / 4}$

$F_2 = \frac{1}{4\pi\epsilon_0} \frac{q_A q_B}{r_1^2} \times \frac{1/4}{1/4}$

$F_2 = \frac{1}{4\pi\epsilon_0} \frac{q_A q_B}{r_1^2}$

This is the same expression as the initial force $F_1$.

The expected repulsion of A (due to B) is **the same** as the initial repulsion.

Forces Between Multiple Charges

When a charge is subjected to forces from multiple other charges, the total force on it is determined by the **principle of superposition**. This principle states that the force on any charge due to a number of other charges is the **vector sum** of all the forces on that charge due to each of the other charges, calculated individually. The presence of other charges does not affect the force between any specific pair of charges.

Consider a system of $n$ stationary charges $q_1, q_2, ..., q_n$ at positions $\vec{r}_1, \vec{r}_2, ..., \vec{r}_n$. The total force $\vec{F}_1$ on charge $q_1$ due to all other charges $q_2, q_3, ..., q_n$ is the vector sum of the individual forces $\vec{F}_{12}$ (force on $q_1$ due to $q_2$), $\vec{F}_{13}$ (force on $q_1$ due to $q_3$), ..., $\vec{F}_{1n}$ (force on $q_1$ due to $q_n$).

$\mathbf{\vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} + ... + \vec{F}_{1n}}$

Each individual force $\vec{F}_{1i}$ is calculated using Coulomb's law for the pair of charges $q_1$ and $q_i$, ignoring all other charges:

$\vec{F}_{1i} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_i}{r_{1i}^2} \hat{r}_{1i}$

where $r_{1i}$ is the distance between $q_1$ and $q_i$, and $\hat{r}_{1i}$ is the unit vector from $q_i$ to $q_1$.

So, the total force on $q_1$ is:

$\mathbf{\vec{F}_1 = \sum_{i=2}^{n} \vec{F}_{1i} = \frac{1}{4\pi\epsilon_0} \sum_{i=2}^{n} \frac{q_1 q_i}{r_{1i}^2} \hat{r}_{1i}}$

The vector addition is performed using standard vector addition rules (like the parallelogram law).

All of electrostatics, including the calculation of electric fields for complex charge distributions, is built upon Coulomb's law and the principle of superposition.

Example 1.6. Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.9?

Diagram for Example 1.6 showing three charges at vertices of an equilateral triangle and a charge Q at the centroid.

Answer:

Let the vertices of the equilateral triangle be A, B, and C, and the centroid be O. Charges $q_1=q_2=q_3=q$ are placed at A, B, and C respectively. A charge Q is placed at O. Let the side length of the triangle be $l$.

The distance from the centroid O to each vertex (A, B, C) is the same. In an equilateral triangle with side $l$, this distance is $r_0 = \frac{l}{\sqrt{3}}$. So, $\textsf{AO} = \textsf{BO} = \textsf{CO} = r_0 = \frac{l}{\sqrt{3}}$.

The force on charge Q at O due to charge $q_1$ at A is $\vec{F}_{Q1}$. Since Q and $q_1$ have the same sign, this force is repulsive and directed along the line OA, away from A, i.e., along OA itself.

The magnitude of this force is $F_{Q1} = \frac{1}{4\pi\epsilon_0} \frac{|Qq_1|}{r_0^2} = \frac{1}{4\pi\epsilon_0} \frac{Qq}{(l/\sqrt{3})^2} = \frac{1}{4\pi\epsilon_0} \frac{3Qq}{l^2}$.

The force on charge Q at O due to charge $q_2$ at B is $\vec{F}_{Q2}$. This force is repulsive and directed along the line OB, away from B, i.e., along BO itself.

The magnitude of this force is $F_{Q2} = \frac{1}{4\pi\epsilon_0} \frac{|Qq_2|}{r_0^2} = \frac{1}{4\pi\epsilon_0} \frac{Qq}{(l/\sqrt{3})^2} = \frac{1}{4\pi\epsilon_0} \frac{3Qq}{l^2}$.

The force on charge Q at O due to charge $q_3$ at C is $\vec{F}_{Q3}$. This force is repulsive and directed along the line OC, away from C, i.e., along CO itself.

The magnitude of this force is $F_{Q3} = \frac{1}{4\pi\epsilon_0} \frac{|Qq_3|}{r_0^2} = \frac{1}{4\pi\epsilon_0} \frac{Qq}{(l/\sqrt{3})^2} = \frac{1}{4\pi\epsilon_0} \frac{3Qq}{l^2}$.

All three forces have the same magnitude. They are directed from O towards A, B, and C, respectively.

The total force on Q is the vector sum $\vec{F}_Q = \vec{F}_{Q1} + \vec{F}_{Q2} + \vec{F}_{Q3}$.

The vectors $\vec{F}_{Q1}, \vec{F}_{Q2}, \vec{F}_{Q3}$ are directed from O towards the vertices A, B, C. The angle between any two of these vectors is 120°.

The resultant of the forces $\vec{F}_{Q2}$ and $\vec{F}_{Q3}$ (acting towards B and C) can be found. By the parallelogram law, the magnitude of their resultant $F_{23}$ is $F_{23} = \sqrt{F_{Q2}^2 + F_{Q3}^2 + 2 F_{Q2} F_{Q3} \cos(120^\circ)}$. Since $F_{Q2} = F_{Q3} = F_m = \frac{1}{4\pi\epsilon_0} \frac{3Qq}{l^2}$, $F_{23} = \sqrt{F_m^2 + F_m^2 + 2 F_m F_m (-1/2)} = \sqrt{2F_m^2 - F_m^2} = \sqrt{F_m^2} = F_m$.

The direction of this resultant $\vec{F}_{23}$ is along the angle bisector of $\angle\text{BOC}$, which points from O towards A (opposite to $\vec{F}_{Q1}$).

So, the resultant of $\vec{F}_{Q2} + \vec{F}_{Q3}$ is a vector of magnitude $F_m$ directed along OA. The force $\vec{F}_{Q1}$ is also of magnitude $F_m$ and directed along OA.

The total force $\vec{F}_Q$ is the sum of $\vec{F}_{Q1}$ and $\vec{F}_{23}$. Since $\vec{F}_{Q1}$ is directed along OA and $\vec{F}_{23}$ is directed along OA, and they have the same magnitude $F_m$, the total force is $F_Q = F_m + F_m = 2 F_m$. Wait, the diagram shows the forces on Q at O acting *away* from A, B, C because Q and q have the same sign. So $\vec{F}_{Q1}$ is along OA, $\vec{F}_{Q2}$ is along OB, $\vec{F}_{Q3}$ is along OC. The sum of three equal vectors from the center to the vertices of an equilateral triangle is zero by symmetry. Yes, the forces are directed *radially outwards* from the centroid towards the vertices, so $\vec{F}_{Q1}$ is along OA, $\vec{F}_{Q2}$ is along OB, $\vec{F}_{Q3}$ is along OC. Oh, the text says "along AO" and "along BO" and "along CO". If the charges are repulsive, and Q is at O, the force on Q due to q at A is away from A, i.e., along OA. The force on Q due to q at B is away from B, i.e., along OB. The force on Q due to q at C is away from C, i.e., along OC. The angles between these are 120 degrees. The vector sum of three equal magnitude forces separated by 120 degrees is zero. The text's diagram agrees with repulsive forces (Q at O, forces away from vertices). The text's solution then states the resultant of F2 and F3 is along OA, which is correct, and then the total force is (F_m along OA) + (F_m along OA) = 2F_m. This would only be true if F2 and F3 added up to be along OA, but they add up to a vector along OA. Ah, the text in the answer says "resultant of forces F2 and F3 is (magnitude 3Qq/4pe0 l^2) along OA". And F1 is also (magnitude 3Qq/4pe0 l^2) along OA. No, F1 is along OA, F2 is along OB, F3 is along OC. The sum is F1+F2+F3=0. The text's answer seems to imply the resultant of F2 and F3 is along OA and F1 is along OA, and F1+F2+F3=0. Let's re-read carefully. "Force F1 on Q due to charge q at A = ... along AO". "Force F2 on Q due to charge q at B = ... along BO". "Force F3 on Q due to charge q at C = ... along CO". These directions (AO, BO, CO) are radially outwards from the centroid towards the vertices A, B, C. The text then says "The resultant of forces F2 and F3 is ... along OA, by the parallelogram law". The resultant of vectors along OB and OC should indeed be along OA, and its magnitude is $F_m$. So the resultant of F2 and F3 is a vector of magnitude $F_m$ pointing along OA. The force F1 is also of magnitude $F_m$ pointing along OA. The total force is F1 + (F2+F3) = (F_m along OA) + (F_m along OA) = 2F_m along OA. Wait, the final result says =0. There is a contradiction in the text's solution steps. Let's trust the diagram and the symmetry argument. The forces are $\vec{F}_{Q1}$ (along OA), $\vec{F}_{Q2}$ (along OB), $\vec{F}_{Q3}$ (along OC). They have equal magnitude $F_m$. The angle between any two is 120°. The vector sum is zero.

By symmetry, the three forces exerted by the charges at the vertices A, B, and C on the charge Q at the centroid O are of equal magnitude and directed from O towards A, B, and C, respectively (since Q and q have the same sign, the forces are repulsive).

Let $r_0 = \frac{l}{\sqrt{3}}$ be the distance from the centroid to each vertex.

The magnitude of each force is $F = \frac{1}{4\pi\epsilon_0} \frac{Qq}{r_0^2} = \frac{1}{4\pi\epsilon_0} \frac{Qq}{(l/\sqrt{3})^2} = \frac{1}{4\pi\epsilon_0} \frac{3Qq}{l^2}$.

Let the vectors representing these forces be $\vec{F}_{\text{OA}}$, $\vec{F}_{\text{OB}}$, and $\vec{F}_{\text{OC}}$, directed along OA, OB, and OC respectively. The angle between any two of these vectors is 120°.

By the principle of superposition, the total force on Q at O is the vector sum: $\vec{F}_{\text{total}} = \vec{F}_{\text{OA}} + \vec{F}_{\text{OB}} + \vec{F}_{\text{OC}}$.

Since these three force vectors of equal magnitude are oriented at 120° to each other, their vector sum is zero.

Thus, the total force on the charge Q placed at the centroid of the triangle is **zero**.

The symmetry argument is the most direct way to see this. If the total force were non-zero, say in a certain direction, rotating the triangle by 120° would map the configuration onto itself, but the force vector would also rotate by 120°. This would imply the force points in a new direction for the same configuration, which is a contradiction, unless the force vector is the zero vector.

Example 1.7. Consider the charges q, q, and –q placed at the vertices of an equilateral triangle, as shown in Fig. 1.10. What is the force on each charge?

Diagram for Example 1.7 showing charges q, q, and -q at vertices of an equilateral triangle.

Answer:

Let the vertices of the equilateral triangle be A, B, and C with charges $q_A = q$, $q_B = q$, and $q_C = -q$. Let the side length of the triangle be $l$. The distance between any pair of charges is $l$.

The magnitude of the force between any two charges is given by Coulomb's law. The magnitude of the force between charges $q$ and $q$ or between $q$ and $-q$ (magnitude $|q(-q)| = q^2$) is the same:

$F_0 = \frac{1}{4\pi\epsilon_0} \frac{q^2}{l^2}$.

Let's calculate the force on each charge using the superposition principle:

Force on charge $q$ at A:

Force due to $q$ at B ($\vec{F}_{\text{AB}}$): Repulsive, directed along BA. Magnitude is $F_0$.
Force due to $-q$ at C ($\vec{F}_{\text{AC}}$): Attractive, directed along AC. Magnitude is $F_0$.

The angle between $\vec{F}_{\text{AB}}$ (along BA) and $\vec{F}_{\text{AC}}$ (along AC) is 120° (the external angle of the equilateral triangle vertex A). Note: BA is along the line from B to A. AC is along the line from A to C. The angle BAC is 60°. Force from B on A is along BA. Force from C on A is attractive, so along AC. Angle between BA and AC? Draw vector AB and vector AC. The angle between vector BA and vector AC is 120 degrees. The resultant force $\vec{F}_A = \vec{F}_{\text{AB}} + \vec{F}_{\text{AC}}$. The magnitude of $\vec{F}_A$ is $F_A = \sqrt{F_0^2 + F_0^2 + 2 F_0 F_0 \cos(120^\circ)} = \sqrt{2F_0^2 + 2F_0^2 (-1/2)} = \sqrt{F_0^2} = F_0$.

The direction of $\vec{F}_A$ is along the angle bisector of the 120° angle between $\vec{F}_{\text{AB}}$ and $\vec{F}_{\text{AC}}$. This direction is perpendicular to BC and points upwards. The text's diagram shows $F_1$ along BC, which is incorrect. Let's re-read the text's solution: "F12 along BA and F13 along AC...total force F1 on the charge q at A is given by F1 = F 1 rˆ where 1 rˆ is a unit vector along BC". The sum of two vectors of magnitude F0 at 120 degrees is F0. The diagram shows F1 with length F0 magnitude. The direction along BC is not correct from vector addition. Let's trust the diagram and the angle logic. $\vec{F}_{\text{AB}}$ along BA, $\vec{F}_{\text{AC}}$ along AC. Angle is 120. Resultant magnitude is $F_0$. The direction is perpendicular to BC, pointing upwards in the diagram.

Magnitude of force on charge $q$ at A is $F_A = F_0 = \frac{1}{4\pi\epsilon_0} \frac{q^2}{l^2}$. Direction is perpendicular to BC, away from BC.

Force on charge $q$ at B:

Force due to $q$ at A ($\vec{F}_{\text{BA}}$): Repulsive, directed along AB. Magnitude is $F_0$.
Force due to $-q$ at C ($\vec{F}_{\text{BC}}$): Attractive, directed along BC. Magnitude is $F_0$.

The angle between $\vec{F}_{\text{BA}}$ (along AB) and $\vec{F}_{\text{BC}}$ (along BC) is 60°. The resultant force $\vec{F}_B = \vec{F}_{\text{BA}} + \vec{F}_{\text{BC}}$. The magnitude of $\vec{F}_B$ is $F_B = \sqrt{F_0^2 + F_0^2 + 2 F_0 F_0 \cos(60^\circ)} = \sqrt{2F_0^2 + 2F_0^2 (1/2)} = \sqrt{3F_0^2} = \sqrt{3}F_0$.

The direction of $\vec{F}_B$ is along the angle bisector of the 60° angle between $\vec{F}_{\text{BA}}$ and $\vec{F}_{\text{BC}}$.

Magnitude of force on charge $q$ at B is $F_B = \sqrt{3}F_0 = \frac{\sqrt{3}}{4\pi\epsilon_0} \frac{q^2}{l^2}$. Direction is along the angle bisector of $\angle\text{ABC}$.

Force on charge $-q$ at C:

Force due to $q$ at A ($\vec{F}_{\text{CA}}$): Attractive, directed along CA. Magnitude is $F_0$.
Force due to $q$ at B ($\vec{F}_{\text{CB}}$): Attractive, directed along CB. Magnitude is $F_0$.

The angle between $\vec{F}_{\text{CA}}$ (along CA) and $\vec{F}_{\text{CB}}$ (along CB) is 60°. The resultant force $\vec{F}_C = \vec{F}_{\text{CA}} + \vec{F}_{\text{CB}}$. The magnitude of $\vec{F}_C$ is $F_C = \sqrt{F_0^2 + F_0^2 + 2 F_0 F_0 \cos(60^\circ)} = \sqrt{3F_0^2} = \sqrt{3}F_0$.

The direction of $\vec{F}_C$ is along the angle bisector of the 60° angle between $\vec{F}_{\text{CA}}$ and $\vec{F}_{\text{CB}}$.

Magnitude of force on charge $-q$ at C is $F_C = \sqrt{3}F_0 = \frac{\sqrt{3}}{4\pi\epsilon_0} \frac{q^2}{l^2}$. Direction is along the angle bisector of $\angle\text{ACB}$.

In summary:

Force on A: Magnitude $\frac{1}{4\pi\epsilon_0} \frac{q^2}{l^2}$, directed perpendicular to BC, upwards.
Force on B: Magnitude $\frac{\sqrt{3}}{4\pi\epsilon_0} \frac{q^2}{l^2}$, directed along the angle bisector of $\angle\text{ABC}$ (outwards from triangle).
Force on C: Magnitude $\frac{\sqrt{3}}{4\pi\epsilon_0} \frac{q^2}{l^2}$, directed along the angle bisector of $\angle\text{ACB}$ (inwards towards triangle).

The sum of the forces $\vec{F}_A + \vec{F}_B + \vec{F}_C = 0$. This is consistent with Newton's third law and the conservation of momentum for the entire system of charges.

Electric Field

The concept of an electric field is introduced to explain how one charge exerts a force on another without direct contact. A charge $Q$ creates an **electric field** in the space around it. When another charge $q$ is placed at some point in this field, the field at that point exerts a force on $q$.

The **electric field** $\vec{E}$ at a point $\vec{r}$ due to a source charge $Q$ is defined as the force per unit positive test charge placed at that point. Mathematically:

$\mathbf{\vec{E}(\vec{r}) = \frac{\vec{F}(\vec{r})}{q}}$

where $\vec{F}(\vec{r})$ is the force experienced by a small test charge $q$ placed at point $\vec{r}$. The test charge $q$ is considered to be negligibly small so that it does not disturb the position of the source charge $Q$. The formal definition uses a limit:

$\mathbf{\vec{E} = \lim_{q \to 0} \frac{\vec{F}}{q}}$

The unit of electric field in SI units is **Newton per Coulomb (N/C)**.

The electric field $\vec{E}(\vec{r})$ due to a single point source charge $Q$ located at the origin is given by applying Coulomb's law for the force on a test charge $q$ at position $\vec{r}$: $\vec{F}(\vec{r}) = \frac{1}{4\pi\epsilon_0} \frac{Qq}{r^2} \hat{r}$. Dividing by $q$ gives:

$\mathbf{\vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \hat{r}}$

where $r$ is the distance from the source charge $Q$ to the point $\vec{r}$, and $\hat{r}$ is the unit vector pointing from $Q$ towards the point $\vec{r}$.

Key points about electric field:

The electric field $\vec{E}$ depends on the source charge $Q$ and the position $\vec{r}$, but is independent of the test charge $q$.
Electric field is a **vector quantity** and exists at every point in space around a charge configuration.
For a positive source charge ($Q > 0$), $\vec{E}$ is directed radially outwards from the charge.
For a negative source charge ($Q < 0$), $\vec{E}$ is directed radially inwards towards the charge.
The magnitude of the electric field due to a point charge depends only on the distance $r$ from the charge, giving it **spherical symmetry** (magnitude is constant on any sphere centered at the charge).

Electric Field Due To A System Of Charges

By the **superposition principle**, the electric field at a point due to a system of multiple charges is the **vector sum** of the electric fields produced by each individual charge at that point, calculated as if each charge were alone.

If a system has charges $q_1, q_2, ..., q_n$ at positions $\vec{r}_1, \vec{r}_2, ..., \vec{r}_n$, the total electric field $\vec{E}(\vec{r})$ at a point P with position vector $\vec{r}$ is:

$\mathbf{\vec{E}(\vec{r}) = \vec{E}_1(\vec{r}) + \vec{E}_2(\vec{r}) + ... + \vec{E}_n(\vec{r})}$

where $\vec{E}_i(\vec{r})$ is the electric field at point P due to charge $q_i$ alone. $\vec{E}_i(\vec{r}) = \frac{1}{4\pi\epsilon_0} \frac{q_i}{r_{iP}^2} \hat{r}_{iP}$, where $r_{iP}$ is the distance from $q_i$ to P, and $\hat{r}_{iP}$ is the unit vector from $q_i$ to P.

So, the total electric field is:

$\mathbf{\vec{E}(\vec{r}) = \sum_{i=1}^{n} \vec{E}_i(\vec{r}) = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_{iP}^2} \hat{r}_{iP}}$

The vector summation is performed using standard vector addition methods.

Physical Significance Of Electric Field

While for electrostatics, the force between charges could be calculated directly using Coulomb's law and superposition, the concept of electric field is introduced for several reasons:

It provides an elegant way to characterise the electrical environment of a system of charges, independent of any test charge. It tells us the force a unit positive charge would experience at any point.
Its true significance appears in **time-dependent electromagnetic phenomena**. When charges are in accelerated motion, their effect (force) on distant charges is not instantaneous but propagates as electromagnetic waves at the speed of light ($c$). The electric (and magnetic) field concept naturally accounts for this time delay. The field is seen as a physical entity that carries energy and momentum and evolves according to its own laws.

The concept of field, pioneered by Faraday, is now central to physics, representing a quantity defined at every point in space.

Example 1.8. An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude $2.0 \times 10^4$ N C–1 [Fig. 1.13(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. 1.13(b)]. Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’.

Diagram for Example 1.8 showing an electron falling in an upward electric field and a proton falling in a downward electric field.

Answer:

The force on a charge $q$ in a uniform electric field $\vec{E}$ is $\vec{F} = q\vec{E}$. The acceleration is $\vec{a} = \vec{F}/m = q\vec{E}/m$.

The electron has charge $q_e = -e$ and mass $m_e = 9.11 \times 10^{-31}$ kg. The proton has charge $q_p = +e$ and mass $m_p = 1.67 \times 10^{-27}$ kg.

Magnitude of electric field $E = 2.0 \times 10^4$ N/C. Distance of fall $h = 1.5$ cm $= 1.5 \times 10^{-2}$ m.

In a uniform field, the force and acceleration are constant. Since the particles start from rest ($v_0 = 0$), the distance covered in time $t$ is $h = v_0 t + \frac{1}{2} a t^2 = \frac{1}{2} a t^2$. So, the time of fall is $t = \sqrt{\frac{2h}{a}}$.

For the electron [Fig. 1.13(a)], the field $\vec{E}$ is upwards. The force on the electron is $\vec{F}_e = (-e)\vec{E}$, which is downwards (opposite to $\vec{E}$). The acceleration is $a_e = |\vec{F}_e|/m_e = eE/m_e$.

$a_e = \frac{(1.602 \times 10^{-19} \text{ C}) \times (2.0 \times 10^4 \text{ N/C})}{9.11 \times 10^{-31} \text{ kg}} \approx \frac{3.204 \times 10^{-15}}{9.11 \times 10^{-31}} \text{ m/s}^2 \approx 3.52 \times 10^{15} \text{ m/s}^2$.

Time of fall for the electron $t_e = \sqrt{\frac{2h}{a_e}} = \sqrt{\frac{2 \times 1.5 \times 10^{-2} \text{ m}}{3.52 \times 10^{15} \text{ m/s}^2}} = \sqrt{\frac{3 \times 10^{-2}}{3.52 \times 10^{15}}} \text{ s} = \sqrt{0.852 \times 10^{-17}} \text{ s} = \sqrt{8.52 \times 10^{-18}} \text{ s}$.

$t_e \approx 2.92 \times 10^{-9} \text{ s}$.

For the proton [Fig. 1.13(b)], the field $\vec{E}$ is downwards. The force on the proton is $\vec{F}_p = (+e)\vec{E}$, which is also downwards (in the direction of $\vec{E}$). The acceleration is $a_p = |\vec{F}_p|/m_p = eE/m_p$.

$a_p = \frac{(1.602 \times 10^{-19} \text{ C}) \times (2.0 \times 10^4 \text{ N/C})}{1.67 \times 10^{-27} \text{ kg}} \approx \frac{3.204 \times 10^{-15}}{1.67 \times 10^{-27}} \text{ m/s}^2 \approx 1.92 \times 10^{12} \text{ m/s}^2$.

Time of fall for the proton $t_p = \sqrt{\frac{2h}{a_p}} = \sqrt{\frac{2 \times 1.5 \times 10^{-2} \text{ m}}{1.92 \times 10^{12} \text{ m/s}^2}} = \sqrt{\frac{3 \times 10^{-2}}{1.92 \times 10^{12}}} \text{ s} = \sqrt{1.56 \times 10^{-14}} \text{ s}$.

$t_p \approx 3.95 \times 10^{-7} \text{ s}$.

Contrast with free fall under gravity: In free fall under gravity, the acceleration ($g$) is constant for all objects, regardless of their mass ($a = g = 9.8 \text{ m/s}^2$). Thus, the time of fall for a given distance is independent of mass. Here, the acceleration is $a = qE/m$. Since $q$ and $E$ are the same magnitude for both particles, the acceleration is inversely proportional to mass ($a \propto 1/m$). The proton is much more massive than the electron ($m_p \gg m_e$), so $a_p \ll a_e$. Consequently, the time of fall is inversely proportional to the square root of acceleration ($t \propto 1/\sqrt{a}$), and thus proportional to the square root of mass ($t \propto \sqrt{m}$). The heavier particle (proton) takes a much longer time to fall the same distance than the lighter particle (electron) in this electric field.

The magnitude of the acceleration due to the electric field for both particles is vastly larger than the acceleration due to gravity ($g \approx 9.8 \text{ m/s}^2$). For the proton, $a_p \approx 1.92 \times 10^{12} \text{ m/s}^2 \gg 9.8 \text{ m/s}^2$. For the electron, $a_e \approx 3.52 \times 10^{15} \text{ m/s}^2 \gg 9.8 \text{ m/s}^2$. Therefore, the effect of gravity is negligible compared to the electric force in this scenario and can be ignored in the calculation of the time of fall.

Example 1.9. Two point charges q1 and q2, of magnitude +10–8 C and –10–8 C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig. 1.14.

Diagram for Example 1.9 showing charges q1 and q2 separated by 0.1 m and points A, B, C.

Answer:

Given charges $q_1 = +10^{-8}$ C and $q_2 = -10^{-8}$ C, placed 0.1 m apart. Let's place $q_1$ at the origin (0, 0) and $q_2$ at (0.1 m, 0). Points A, B, and C are located as shown in the diagram.

A is at $x = 0.05$ m, B is at $x = -0.05$ m, C is at $(0.05, 0.0866)$ m relative to $q_1$. Note that points A, B, C are not necessarily on the x-axis in a general 2D coordinate system, but the diagram shows A and B on the line joining the charges, and C forming an equilateral triangle with $q_1$ and $q_2$. Let's assume $q_1$ is at (-0.05m, 0) and $q_2$ is at (+0.05m, 0). Then the origin O is at (0,0). Point A is at (+0.05m + 0.05m = +0.1m, 0). Point B is at (-0.05m - 0.05m = -0.1m, 0). Point C is located such that it forms an equilateral triangle with $q_1$ and $q_2$, meaning distance $q_1q_2 = q_1C = q_2C = 0.1$ m. The x-coordinate of C would be 0 (midpoint of $q_1q_2$ line), and the y-coordinate would be the height of the equilateral triangle with side 0.1m, which is $\sqrt{(0.1)^2 - (0.05)^2} = \sqrt{0.01 - 0.0025} = \sqrt{0.0075} \approx 0.0866$ m. So C is at (0, 0.0866 m).

Let's use the original diagram's implied coordinates where $q_1$ and $q_2$ are on the x-axis, separated by 0.1m. Let $q_1$ be at $x=0$ and $q_2$ at $x=0.1$m.

Point A is 0.05m from $q_1$ towards $q_2$. Point B is 0.05m from $q_1$ away from $q_2$. Point C is on a line perpendicular to the line joining $q_1$ and $q_2$, likely at the midpoint, and at a distance forming an equilateral triangle (0.1m from both $q_1$ and $q_2$).

Let's follow the text's implied geometry and calculations more closely.

Let $q_1$ be at point O, and $q_2$ be at a point 0.1 m away on the x-axis. The point A is 0.05 m from $q_1$ along the x-axis, towards $q_2$. So distance $q_1A = 0.05$ m, distance $q_2A = 0.1 - 0.05 = 0.05$ m.

Point B is 0.05 m from $q_1$ along the x-axis, away from $q_2$. So distance $q_1B = 0.05$ m. Wait, looking at the diagram, B is on the opposite side of $q_1$ from $q_2$. Distance between $q_1$ and $q_2$ is 0.1m. A is 0.05m from $q_1$ and 0.05m from $q_2$. A is exactly in the middle. B is 0.05m from $q_1$ in the opposite direction of $q_2$. So distance $q_1B = 0.05$ m, distance $q_2B = 0.1 + 0.05 = 0.15$ m.

Point C forms an equilateral triangle with $q_1$ and $q_2$. Distance $q_1q_2 = q_1C = q_2C = 0.1$ m.

Let's calculate the electric field at each point using superposition. $\vec{E} = \vec{E}_1 + \vec{E}_2$. $\vec{E}_i = \frac{1}{4\pi\epsilon_0} \frac{q_i}{r_i^2} \hat{r}_i$. Use $\frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \text{ N m}^2/\text{C}^2$.

At point A: Located midway between $q_1$ and $q_2$. Distance from $q_1$ to A is $r_{1A} = 0.05$ m. Distance from $q_2$ to A is $r_{2A} = 0.05$ m.

Field $\vec{E}_1$ at A due to $q_1$ (+ve): Points away from $q_1$. Magnitude $E_1 = (9 \times 10^9) \frac{10^{-8}}{(0.05)^2} = 9 \times 10^9 \times \frac{10^{-8}}{0.0025} = 9 \times 10^9 \times 400 \times 10^{-8} = 3600 \times 10^1 = 3.6 \times 10^4$ N/C. Direction: Towards $q_2$.

Field $\vec{E}_2$ at A due to $q_2$ (-ve): Points towards $q_2$. Magnitude $E_2 = (9 \times 10^9) \frac{|-10^{-8}|}{(0.05)^2} = 3.6 \times 10^4$ N/C. Direction: Towards $q_2$.

Both fields are in the same direction (towards $q_2$). Total field at A: $\vec{E}_A = \vec{E}_1 + \vec{E}_2$. Magnitude $E_A = E_1 + E_2 = 3.6 \times 10^4 + 3.6 \times 10^4 = 7.2 \times 10^4$ N/C. Direction: Towards $q_2$.

At point B: Located 0.05 m from $q_1$ on the side away from $q_2$. Distance $q_1B = 0.05$ m, distance $q_2B = 0.1 + 0.05 = 0.15$ m.

Field $\vec{E}_1$ at B due to $q_1$ (+ve): Points away from $q_1$. Magnitude $E_1 = (9 \times 10^9) \frac{10^{-8}}{(0.05)^2} = 3.6 \times 10^4$ N/C. Direction: Away from $q_1$ (opposite to the direction towards $q_2$).

Field $\vec{E}_2$ at B due to $q_2$ (-ve): Points towards $q_2$. Magnitude $E_2 = (9 \times 10^9) \frac{|-10^{-8}|}{(0.15)^2} = 9 \times 10^9 \times \frac{10^{-8}}{0.0225} = 9 \times 10^9 \times 44.44 \times 10^{-8} \approx 400 \times 10^1 = 4 \times 10^3$ N/C. Direction: Towards $q_2$ (same direction as from $q_1$ to $q_2$).

Fields $\vec{E}_1$ and $\vec{E}_2$ are in opposite directions at B. Let's define the direction from $q_1$ to $q_2$ as the positive x-direction. $\vec{E}_1$ at B is in the -x direction. $\vec{E}_2$ at B is in the +x direction. Total field at B: $\vec{E}_B = \vec{E}_1 + \vec{E}_2$. Magnitude $E_B = |E_2 - E_1| = |4 \times 10^3 - 3.6 \times 10^4| = |4000 - 36000| = 32000 = 3.2 \times 10^4$ N/C. Direction: Since $E_1 > E_2$, the resultant is in the direction of $\vec{E}_1$ (away from $q_1$). The text calculation seems to have $\vec{E}_1$ at B pointing left (towards $q_1$) and $\vec{E}_2$ at B pointing right (away from $q_1$). Let's align with text calculation logic for directions. If $q_1$ is at origin (0,0) and $q_2$ at (0.1,0). Point A is at (0.05,0). Point B is at (-0.05,0). Point C is at (0.05, 0.0866). * At A(0.05,0): $\vec{r}_{1A} = (0.05,0)$, $\hat{r}_{1A} = (1,0)$. $\vec{r}_{2A} = (0.05-0.1, 0) = (-0.05,0)$, $\hat{r}_{2A} = (-1,0)$. $E_1$ at A is $\frac{1}{4\pi\epsilon_0}\frac{q_1}{r_{1A}^2}\hat{r}_{1A} = E_1 \hat{i}$. $E_2$ at A is $\frac{1}{4\pi\epsilon_0}\frac{q_2}{r_{2A}^2}\hat{r}_{2A} = E_2 (-\hat{i})$. With $q_1 = +E$, $q_2 = -E$, magnitudes $E_1=E_2=\frac{kE}{(0.05)^2}$. $\vec{E}_1$ points +x, $\vec{E}_2$ points -x. The text says $\vec{E}_1$ at A points right, $\vec{E}_2$ at A points right. Let's check my direction logic. Field from +ve charge points AWAY from the charge. Field from -ve charge points TOWARDS the charge. At A (between $q_1$ and $q_2$), field from $q_1$ (at left) points right. Field from $q_2$ (at right) points left. Wait, $q_2$ is negative, so field points TOWARDS $q_2$. At A, this is right. So both fields point right. $E_A = E_1 + E_2$ right. Yes, $7.2 \times 10^4$ right. * At B (-0.05,0): $\vec{r}_{1B} = (-0.05,0)$, $\hat{r}_{1B} = (-1,0)$. $\vec{r}_{2B} = (-0.05-0.1, 0) = (-0.15,0)$, $\hat{r}_{2B} = (-1,0)$. $E_1$ at B points away from $q_1$: left. $E_2$ at B points towards $q_2$: right. So $\vec{E}_1$ is left, $\vec{E}_2$ is right. $E_1 = \frac{k q_1}{(0.05)^2} = 3.6 \times 10^4$. $E_2 = \frac{k |q_2|}{(0.15)^2} = \frac{9 \times 10^9 \times 10^{-8}}{(0.15)^2} = \frac{90}{0.0225} = 4000 = 4 \times 10^3$. $E_B = |E_1 - E_2| = |3.6 \times 10^4 - 4 \times 10^3| = 32000$. Direction is the direction of the larger field, $E_1$, which is left. Yes, $3.2 \times 10^4$ left.

At point C: Located such that $q_1C = q_2C = 0.1$ m. Point C is above the midpoint of $q_1q_2$. $q_1$ at origin (0,0), $q_2$ at (0.1,0). C at (0.05, 0.0866).

Distance from $q_1$ to C is $r_{1C} = 0.1$ m. Distance from $q_2$ to C is $r_{2C} = 0.1$ m.

Field $\vec{E}_1$ at C due to $q_1$ (+ve): Points away from $q_1$ along OC. Magnitude $E_1 = (9 \times 10^9) \frac{10^{-8}}{(0.1)^2} = 9 \times 10^9 \times \frac{10^{-8}}{0.01} = 9 \times 10^9 \times 100 \times 10^{-8} = 900 \times 10^1 = 9 \times 10^3$ N/C.

Field $\vec{E}_2$ at C due to $q_2$ (-ve): Points towards $q_2$ along CQ. Magnitude $E_2 = (9 \times 10^9) \frac{|-10^{-8}|}{(0.1)^2} = 9 \times 10^3$ N/C.

The magnitudes are equal, $E_1=E_2=9 \times 10^3$ N/C. The directions are shown in Fig. 1.14. Let the angle $\angle\text{C}q_1q_2 = \angle\text{C}q_2q_1 = 60^\circ$ since it's an equilateral triangle. Field $\vec{E}_1$ is along $q_1$C. Field $\vec{E}_2$ is along $Cq_2$. The angle between $q_1q_2$ line and $q_1$C is 60. The angle between $q_1q_2$ line and $q_2$C is 60. The angle between $\vec{E}_1$ and the horizontal is 60 degrees upwards relative to the line connecting $q_1$ and $q_2$. The angle between $\vec{E}_2$ and the horizontal is 60 degrees upwards relative to the line connecting $q_2$ and $q_1$. Ah, $\vec{E}_1$ is along $q_1$C, $\vec{E}_2$ is along $Cq_2$. Vector from $q_1$ to C. Vector from C to $q_2$. The angle between vector $q_1$C and vector $Cq_2$ needs to be found. Let's use components. $q_1$ at (0,0), $q_2$ at (0.1,0), C at (0.05, 0.0866). Vector $q_1$ to C: $\vec{r}_{1C} = (0.05, 0.0866)$. Unit vector $\hat{r}_{1C} = (0.05, 0.0866)/0.1 = (0.5, 0.866)$. Angle is $\tan^{-1}(0.866/0.5) = 60^\circ$ with x-axis. Vector C to $q_2$: $\vec{r}_{Cq_2} = (0.1-0.05, 0-0.0866) = (0.05, -0.0866)$. Unit vector $\hat{r}_{Cq_2} = (0.05, -0.0866)/0.1 = (0.5, -0.866)$. Angle is $\tan^{-1}(-0.866/0.5) = -60^\circ$ with x-axis. $\vec{E}_1$ is along $\hat{r}_{1C}$ (away from $q_1$). $\vec{E}_2$ is along $\hat{r}_{Cq_2}$ (towards $q_2$). $\vec{E}_1 = E_1 \hat{r}_{1C} = 9 \times 10^3 (0.5, 0.866) = (4.5 \times 10^3, 7.794 \times 10^3)$ N/C. $\vec{E}_2 = E_2 \hat{r}_{Cq_2} = 9 \times 10^3 (0.5, -0.866) = (4.5 \times 10^3, -7.794 \times 10^3)$ N/C. Total field $\vec{E}_C = \vec{E}_1 + \vec{E}_2 = (4.5 \times 10^3 + 4.5 \times 10^3, 7.794 \times 10^3 - 7.794 \times 10^3) = (9 \times 10^3, 0)$ N/C. Magnitude $E_C = 9 \times 10^3$ N/C. Direction is along the positive x-axis (towards the right).

Let's re-check the text calculation: $E_C = \sqrt{E_1^2 + E_2^2 + 2 E_1 E_2 \cos\theta}$. What is $\theta$? The angle between $\vec{E}_1$ (along $q_1$C) and $\vec{E}_2$ (along $Cq_2$). This angle is $120^\circ$ as the triangle $q_1Cq_2$ is equilateral. Vector $\vec{E}_1$ points from $q_1$ to C. Vector $\vec{E}_2$ points from C to $q_2$. The angle between $q_1$C and $Cq_2$ as vectors? No, the angle between $\vec{E}_1$ and $\vec{E}_2$ vectors. $\vec{E}_1$ is away from $q_1$ along $q_1$C. $\vec{E}_2$ is towards $q_2$ along $Cq_2$. The angle between vector $q_1$C and vector $Cq_2$ is $180 - 60 = 120$. So the angle between $\vec{E}_1$ and $\vec{E}_2$ is 120°. $E_C = \sqrt{(9 \times 10^3)^2 + (9 \times 10^3)^2 + 2 (9 \times 10^3) (9 \times 10^3) \cos(120^\circ)}$ $E_C = \sqrt{2(9 \times 10^3)^2 + 2(9 \times 10^3)^2 (-1/2)} = \sqrt{2(9 \times 10^3)^2 - (9 \times 10^3)^2} = \sqrt{(9 \times 10^3)^2} = 9 \times 10^3$ N/C. The direction is along the angle bisector of the 120° angle. This bisector is perpendicular to the line connecting the tips of $\vec{E}_1$ and $\vec{E}_2$. It points right. Yes, $9 \times 10^3$ N/C towards the right.

Summary of results:

Electric field at A: $7.2 \times 10^4$ N/C, directed towards $q_2$.
Electric field at B: $3.2 \times 10^4$ N/C, directed away from $q_1$.
Electric field at C: $9.0 \times 10^3$ N/C, directed towards the right (parallel to the line joining $q_1$ and $q_2$).

Electric Field Lines

Electric field lines, also called lines of force, are a visual tool for representing the electric field in space. They are imaginary curves drawn to map the direction and strength of the electric field around a charge configuration.

An electric field line is a curve such that the **tangent** to the curve at any point gives the **direction of the net electric field** vector at that point. An arrow on the line indicates the field's direction.

The **relative closeness** or **density** of the field lines indicates the **relative strength** of the electric field. Where the field is strong, lines are closer together (crowded); where the field is weak, lines are farther apart.

For a point charge, the magnitude of the field decreases with the inverse square of the distance ($1/r^2$). If we consider the number of field lines passing through a small area perpendicular to the field, this number is proportional to the field strength. For a point charge, the number of lines passing through any spherical surface centered on the charge is constant, regardless of the radius. Since the surface area of a sphere increases as $r^2$ ($4\pi r^2$), the density of lines (lines per unit area) decreases as $1/r^2$, correctly representing the field strength decrease.

Properties of electric field lines:

(i) Field lines originate from positive charges and terminate at negative charges. If there is a single charge, they are considered to start or end at infinity.
(ii) They are continuous curves without breaks in a charge-free region.
(iii) **Two field lines never cross each other**. If they did, the electric field at the intersection point would have two different directions simultaneously, which is physically impossible.
(iv) Electrostatic field lines **do not form closed loops**. This is related to the conservative nature of the electrostatic force/field.

Examples of field lines around simple charge configurations:

Single positive charge: Lines are radially outward.
Single negative charge: Lines are radially inward.
Two equal positive charges: Lines show mutual repulsion, bending away from each other.
Electric dipole (equal and opposite charges): Lines start from the positive charge and end at the negative charge, showing mutual attraction.

Electric field lines for a positive charge, a negative charge, two positive charges, and an electric dipole.

Electric Flux

Electric flux ($\Phi_E$) is a measure of the "flow" of the electric field through a given surface. It is analogous to the flux of liquid flow in fluid dynamics (volume of liquid crossing a surface per unit time), but for electric fields, it represents the number of electric field lines passing through the surface.

To define electric flux through a surface element, the area element itself is treated as a vector quantity. The direction of a planar area vector $\Delta \vec{S}$ is taken along its **normal**. For a curved surface, it is divided into small planar elements. For a **closed surface**, the convention is that the area vector $\Delta \vec{S}$ for every element points in the direction of the **outward normal**.

The electric flux $\Delta \Phi_E$ through a small planar area element $\Delta \vec{S}$ in a uniform electric field $\vec{E}$ is defined as the dot product of the electric field vector and the area vector:

$\mathbf{\Delta \Phi_E = \vec{E} \cdot \Delta \vec{S} = E \Delta S \cos\theta}$

where $E$ is the magnitude of the electric field, $\Delta S$ is the magnitude of the area element, and $\theta$ is the angle between the electric field vector $\vec{E}$ and the area vector $\Delta \vec{S}$ (the normal to the surface). The flux is maximum when $\vec{E}$ is perpendicular to the surface ($\theta = 0^\circ$, $\cos 0^\circ = 1$), zero when $\vec{E}$ is parallel to the surface ($\theta = 90^\circ$, $\cos 90^\circ = 0$), and negative when $\vec{E}$ enters the closed surface ($\theta > 90^\circ$, $\cos \theta < 0$).

The unit of electric flux is **N C$^{-1}$ m$^2$**.

The total electric flux $\Phi_E$ through a finite surface $S$ is obtained by dividing the surface into many small area elements $\Delta \vec{S}$ and summing the flux through each element:

$\mathbf{\Phi_E \approx \sum \vec{E} \cdot \Delta \vec{S}}$

For a mathematically exact calculation, this sum becomes an integral over the surface as $\Delta \vec{S} \to 0$: $\Phi_E = \oint_S \vec{E} \cdot d\vec{S}$. The circle on the integral sign denotes integration over a closed surface.

Electric Dipole

An electric dipole is a system consisting of **two equal and opposite point charges**, $+q$ and $-q$, separated by a fixed small distance, usually denoted as $2a$. The line passing through the two charges is called the dipole axis. The direction from the negative charge to the positive charge is conventionally taken as the direction of the dipole axis.

The total electric charge of an electric dipole is $(+q) + (-q) = 0$. However, since the charges are separated, the electric field produced by a dipole is not zero. At distances much larger than the separation $2a$ (i.e., $r \gg 2a$), the fields due to $+q$ and $-q$ almost cancel out, but not completely. The electric field of a dipole falls off with distance faster than that of a single charge.

The **electric dipole moment** $\vec{p}$ is a vector quantity defined for an electric dipole. Its magnitude is the product of the magnitude of either charge and the distance of separation: $\mathbf{p = q \times 2a}$. Its direction is along the dipole axis, pointing from the negative charge (–q) to the positive charge (+q).

$\mathbf{\vec{p} = q (2a) \hat{p}}$

where $\hat{p}$ is the unit vector from –q to q.

The Field Of An Electric Dipole

The electric field of an electric dipole can be calculated at any point in space by vectorially adding the fields due to the individual charges $+q$ and $-q$ using Coulomb's law and superposition.

(i) **For points on the dipole axis:** At a point P on the axis at a distance $r$ from the dipole center, on the side of $+q$ (i.e., along the direction of $\vec{p}$).

The magnitudes of fields due to $+q$ and $-q$ at P are $E_{+q} = \frac{1}{4\pi\epsilon_0} \frac{q}{(r-a)^2}$ (away from $+q$) and $E_{-q} = \frac{1}{4\pi\epsilon_0} \frac{q}{(r+a)^2}$ (towards $-q$). Both fields are along the dipole axis. The net field is the difference.

For distances much larger than the separation ($r \gg a$), the electric field at a point on the axis simplifies to:

$\mathbf{\vec{E}_{\text{axis}} \approx \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}}{r^3}}$ (for $r \gg a$)

The direction of the field on the axis is the same as the direction of the dipole moment $\vec{p}$.

(ii) **For points on the equatorial plane:** At a point Q on the plane perpendicular to the dipole axis and passing through its center, at a distance $r$ from the center.

The magnitudes of fields due to $+q$ and $-q$ at Q are equal: $E_{+q} = E_{-q} = \frac{1}{4\pi\epsilon_0} \frac{q}{\sqrt{r^2+a^2}^2} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2+a^2}$. The fields are directed away from $+q$ and towards $-q$. The components perpendicular to the dipole axis cancel, while the components parallel to the axis add up. The resultant field is parallel to the dipole axis but points in the opposite direction to $\vec{p}$.

For distances much larger than the separation ($r \gg a$), the electric field on the equatorial plane simplifies to:

$\mathbf{\vec{E}_{\text{equatorial}} \approx \frac{1}{4\pi\epsilon_0} \frac{-\vec{p}}{r^3}}$ (for $r \gg a$)

The electric field of a dipole falls off as $1/r^3$ at large distances, which is faster than the $1/r^2$ dependence of a point charge field.

A **point dipole** is a theoretical concept where the dipole size $2a$ approaches zero while the charge $q$ approaches infinity such that the dipole moment $p = q \times 2a$ remains finite. For a point dipole, the $1/r^3$ formulas are exact for any distance $r > 0$.

Physical Significance Of Dipoles

In molecules, the distribution of positive and negative charges determines if they have a permanent electric dipole moment.

Most molecules like $\textsf{CO}_2$ or $\textsf{CH}_4$ have the centers of positive and negative charge coinciding, resulting in a net dipole moment of zero. These are **non-polar molecules**. They can develop an induced dipole moment when an external electric field is applied, as the field separates the charge centers.
Some molecules, like $\textsf{H}_2\text{O}$, have the centers of positive and negative charge naturally separated, giving them a **permanent electric dipole moment** even without an external field. These are **polar molecules**.

The behavior of polar and non-polar molecules in electric fields leads to interesting properties and applications of materials.

The attraction of a charged comb to uncharged pieces of paper can be explained by dipoles. The charged comb creates a non-uniform electric field. This field induces dipole moments in the paper pieces (polarising them). Because the field is non-uniform, the dipole experiences a net force, pulling the paper towards the comb.

Example 1.10. Two charges $\pm 10$ μC are placed 5.0 mm apart. Determine the electric field at (a) a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge, as shown in Fig. 1.21(a), and (b) a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in Fig. 1.21(b).

Diagram for Example 1.10 showing a dipole with points P on axis and Q on equatorial line.

Answer:

Given charges $q = +10 \mu\text{C} = +10^{-5}\text{ C}$ and $-q = -10 \mu\text{C} = -10^{-5}\text{ C}$. The distance between charges is $2a = 5.0 \text{ mm} = 5.0 \times 10^{-3}\text{ m}$, so $a = 2.5 \text{ mm} = 2.5 \times 10^{-3}\text{ m}$.

The dipole moment magnitude is $p = q \times 2a = (10^{-5}\text{ C}) \times (5.0 \times 10^{-3}\text{ m}) = 5.0 \times 10^{-8}\text{ C m}$. The direction of $\vec{p}$ is from $-q$ to $+q$.

Distance of points P and Q from the center O is $r = 15 \text{ cm} = 0.15 \text{ m}$.

In both cases, $r = 0.15 \text{ m}$ and $a = 0.0025 \text{ m}$. The ratio $r/a = 0.15 / 0.0025 = 60$. Since $r \gg a$ (60 times larger), we can use the approximate formulas for electric field at large distances from a dipole, and also calculate the exact value for verification.

(a) Electric field at point P on the axis: P is on the axis, on the side of the positive charge, at distance $r$ from the center. The distance from $+q$ to P is $r_+ = r - a = 0.15 - 0.0025 = 0.1475$ m. The distance from $-q$ to P is $r_- = r + a = 0.15 + 0.0025 = 0.1525$ m.

Field due to $+q$ at P: $\vec{E}_+ = \frac{1}{4\pi\epsilon_0} \frac{q}{r_+^2}$ along OP (away from $+q$). Magnitude $E_+ = (9 \times 10^9) \frac{10^{-5}}{(0.1475)^2} \approx 9 \times 10^9 \times \frac{10^{-5}}{0.021756} \approx 9 \times 10^9 \times 45.96 \times 10^{-5} \approx 413.6 \times 10^4 \approx 4.14 \times 10^6$ N/C. Direction is along the dipole axis (from $-q$ to $+q$).

Field due to $-q$ at P: $\vec{E}_- = \frac{1}{4\pi\epsilon_0} \frac{|-q|}{r_-^2}$ along PO (towards $-q$). Magnitude $E_- = (9 \times 10^9) \frac{10^{-5}}{(0.1525)^2} \approx 9 \times 10^9 \times \frac{10^{-5}}{0.023256} \approx 9 \times 10^9 \times 43.0 \times 10^{-5} \approx 387 \times 10^4 \approx 3.87 \times 10^6$ N/C. Direction is opposite to the dipole axis.

Net electric field at P: $\vec{E}_P = \vec{E}_+ + \vec{E}_-$. Since $\vec{E}_+$ and $\vec{E}_-$ are along the axis and in opposite directions, the magnitude is $E_P = |E_+ - E_-| = |4.14 \times 10^6 - 3.87 \times 10^6| = 0.27 \times 10^6 = 2.7 \times 10^5$ N/C. The direction is the direction of the stronger field, which is $\vec{E}_+$, along the dipole axis (from $-q$ to $+q$).

Using the approximate formula for axial field at large distance ($r \gg a$): $E_{\text{axis}} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}$.

$E_{\text{axis}} = (9 \times 10^9) \frac{2 \times (5.0 \times 10^{-8})}{(0.15)^3} = 9 \times 10^9 \frac{10 \times 10^{-8}}{0.003375} = 9 \times 10^9 \frac{10^{-7}}{3.375 \times 10^{-3}} = 9 \times \frac{10^2}{3.375} \times 10^6 \approx 2.66 \times 10^6$. Wait, the calculation in text is $2.6 \times 10^5$. Let's re-calculate $r^3 = (0.15)^3 = 0.003375$. $2p = 10^{-7}$. $E_{\text{axis}} = 9 \times 10^9 \times \frac{10^{-7}}{0.003375} = \frac{900}{3.375} \times 10^6 \approx 266.67 \times 10^6$. My calculation does not match the text example value. Let's recheck the text's approximate value $2.6 \times 10^5$. $r^3 = (15 \text{ cm})^3 = (0.15 \text{ m})^3 = (15 \times 10^{-2} \text{ m})^3 = 3375 \times 10^{-6} \text{ m}^3 = 3.375 \times 10^{-3} \text{ m}^3$. $2p = 10^{-7}$ C m. $E = 9 \times 10^9 \times \frac{10^{-7}}{3.375 \times 10^{-3}} = \frac{900}{3.375} \times 10^6$. Ah, $r = 15 \text{ cm} = 0.15 \text{ m}$. Text uses $(15) \times 10^{-3}$ m, suggesting the distance is 15 mm? No, it says 15 cm away. Let's assume the text's calculation in the answer section has a typo in the distance unit conversion or the power of 10. Assuming $r=0.15$m and $p=5 \times 10^{-8}$ Cm, $E_{\text{axis}} = 9 \times 10^9 \times \frac{2 \times 5 \times 10^{-8}}{(0.15)^3} = 9 \times 10^9 \times \frac{10^{-7}}{(0.15)^3} = 9 \times 10^9 \times \frac{10^{-7}}{0.003375} \approx 2.66 \times 10^5$ N/C. Yes, this matches the text's approximate value ($2.6 \times 10^5$). The approximate formula gives a result close to the exact calculation ($2.7 \times 10^5$ N/C). Direction is along the dipole moment vector $\vec{p}$ (from $-q$ to $+q$).

(b) Electric field at point Q on the equatorial plane: Q is 15 cm away from O on a line normal to the axis. Distance $r=0.15$ m from O. Distance from $+q$ to Q is $r' = \sqrt{r^2 + a^2} = \sqrt{(0.15)^2 + (0.0025)^2} = \sqrt{0.0225 + 0.00000625} = \sqrt{0.02250625} \approx 0.15002$ m. Distance from $-q$ to Q is also $r'$.

Field due to $+q$ at Q: $\vec{E}_+$ points away from $+q$. Magnitude $E_+ = (9 \times 10^9) \frac{10^{-5}}{(0.15002)^2} \approx 9 \times 10^9 \times \frac{10^{-5}}{0.022506} \approx 3.999 \times 10^6$ N/C.

Field due to $-q$ at Q: $\vec{E}_-$ points towards $-q$. Magnitude $E_- = (9 \times 10^9) \frac{10^{-5}}{(0.15002)^2} \approx 3.999 \times 10^6$ N/C.

These two vectors have equal magnitude. Their resultant is parallel to the dipole axis but points in the opposite direction to $\vec{p}$. The magnitude of the resultant is $E_Q = (E_+ + E_-) \cos\theta$, where $\theta$ is the angle between $\vec{E}_+$ (or $\vec{E}_-$) and the direction opposite to $\vec{p}$. This angle is such that $\cos\theta = a/\sqrt{r^2+a^2} = a/r'$.

Magnitude $E_Q = 2 E_+ \times \frac{a}{r'} = 2 \times (3.999 \times 10^6) \times \frac{0.0025}{0.15002} \approx 7.998 \times 10^6 \times 0.01666 \approx 0.133 \times 10^6 = 1.33 \times 10^5$ N/C. Direction is opposite to $\vec{p}$.

Using the approximate formula for equatorial field at large distance ($r \gg a$): $E_{\text{equatorial}} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$.

$E_{\text{equatorial}} = (9 \times 10^9) \frac{5.0 \times 10^{-8}}{(0.15)^3} = 9 \times 10^9 \frac{5 \times 10^{-8}}{0.003375} = \frac{450}{3.375} \times 10^6 \approx 133.33 \times 10^6$. Again, power of 10 seems off. $(9 \times 10^9) \frac{5 \times 10^{-8}}{(0.15)^3} = 9 \times 10^9 \times \frac{5 \times 10^{-8}}{3.375 \times 10^{-3}} = \frac{45}{3.375} \times 10^4 \approx 13.33 \times 10^4 = 1.33 \times 10^5$ N/C. Yes, this matches the exact calculation and the text's approximate value ($1.33 \times 10^5$). Direction is opposite to $\vec{p}$.

Summary of electric fields:

Electric field at P (on axis): Magnitude $2.7 \times 10^5$ N/C, direction along the dipole axis (from $-q$ to $+q$). Approximate: $2.6 \times 10^5$ N/C.
Electric field at Q (equatorial): Magnitude $1.33 \times 10^5$ N/C, direction opposite to the dipole axis (from $+q$ to $-q$). Approximate: $1.33 \times 10^5$ N/C.

Dipole In A Uniform External Field

Consider an electric dipole with dipole moment $\vec{p}$ placed in a uniform external electric field $\vec{E}$. A uniform field means the field strength and direction are the same at all points in space.

The charge $+q$ experiences a force $\vec{F}_+ = +q\vec{E}$ in the direction of $\vec{E}$. The charge $-q$ experiences a force $\vec{F}_- = -q\vec{E}$ in the direction opposite to $\vec{E}$.

The net force on the dipole is $\vec{F}_{\text{net}} = \vec{F}_+ + \vec{F}_- = q\vec{E} - q\vec{E} = \vec{0}$. Thus, a dipole in a uniform electric field experiences **no net force**.

However, since the two equal and opposite forces act at different points (the locations of $+q$ and $-q$), they form a couple and exert a **torque** on the dipole. The torque $\vec{\tau}$ is given by the vector product of the dipole moment $\vec{p}$ and the electric field $\vec{E}$:

$\mathbf{\vec{\tau} = \vec{p} \times \vec{E}}$

The magnitude of the torque is $\tau = p E \sin\theta$, where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$. The torque tends to align the dipole moment $\vec{p}$ with the electric field $\vec{E}$. The torque is zero when $\vec{p}$ is parallel or antiparallel to $\vec{E}$ ($\theta = 0^\circ$ or $180^\circ$).

Diagram of a dipole in a uniform electric field experiencing a torque.

If the external electric field is **non-uniform**, the forces on $+q$ and $-q$ will have different magnitudes or directions, resulting in a **net force** on the dipole in addition to a torque. When the dipole moment $\vec{p}$ is parallel to a non-uniform field $\vec{E}$, the net force is in the direction of increasing field strength. When $\vec{p}$ is antiparallel to $\vec{E}$, the net force is in the direction of decreasing field strength.

This explains why a charged comb attracts uncharged paper. The charged comb creates a non-uniform field. This field polarises the neutral paper pieces (induces dipole moments in them). Since the field is stronger closer to the comb, and the induced dipole is aligned with the field (or opposite to it, but attracted more on the closer side), there is a net attractive force towards the comb.

Example 1.10 (continued). An electric dipole with dipole moment $4 \times 10^{–9}$ C m is aligned at 30° with the direction of a uniform electric field of magnitude $5 \times 10^4$ NC–1. Calculate the magnitude of the torque acting on the dipole.

Answer:

Given: Magnitude of dipole moment $p = 4 \times 10^{-9}$ C m. Magnitude of uniform electric field $E = 5 \times 10^4$ N C$^{-1}$. The angle between $\vec{p}$ and $\vec{E}$ is $\theta = 30^\circ$.

The magnitude of the torque acting on the dipole in a uniform electric field is given by $\tau = p E \sin\theta$.

$\tau = (4 \times 10^{-9} \text{ C m}) \times (5 \times 10^4 \text{ N C}^{-1}) \times \sin(30^\circ)$

$\tau = (20 \times 10^{-5} \text{ N m}) \times (1/2)$

$\tau = 10 \times 10^{-5} \text{ N m} = 1.0 \times 10^{-4} \text{ N m}$.

The magnitude of the torque acting on the dipole is $1.0 \times 10^{-4}$ N m. This torque tends to rotate the dipole to align its dipole moment vector with the electric field vector.

Continuous Charge Distribution

For charge configurations involving a large number of closely spaced charges (like charges on the surface of a conductor), it is often more practical to treat the charge distribution as continuous rather than discrete. This involves defining charge densities over macroscopic elements that are still microscopically large (containing many elementary charges).

Surface charge density ($\sigma$): For charge distributed over a surface. Defined as the charge $\Delta Q$ on a small surface element $\Delta S$ divided by the area of that element: $\mathbf{\sigma = \frac{\Delta Q}{\Delta S}}$. Units: C/m$^2$. Macroscopic $\sigma$ is a smoothed-out average ignoring microscopic discontinuities.
Linear charge density ($\lambda$): For charge distributed along a line or wire. Defined as the charge $\Delta Q$ on a small line element $\Delta l$ divided by the length of that element: $\mathbf{\lambda = \frac{\Delta Q}{\Delta l}}$. Units: C/m.
Volume charge density ($\rho$): For charge distributed throughout a volume. Defined as the charge $\Delta Q$ in a small volume element $\Delta V$ divided by the volume of that element: $\mathbf{\rho = \frac{\Delta Q}{\Delta V}}$. Units: C/m$^3$. Sometimes simply called charge density.

The electric field due to a continuous charge distribution can be calculated by dividing the distribution into infinitesimally small elements, treating each element as a point charge ($dq = \rho dV$ or $\sigma dS$ or $\lambda dl$), finding the electric field $d\vec{E}$ due to this element using Coulomb's law, and then integrating $d\vec{E}$ over the entire distribution. Using vector notation, if a volume element $dV$ at position vector $\vec{r}'$ has charge $dq = \rho(\vec{r}') dV$, the field $d\vec{E}$ at point P with position vector $\vec{R}$ is $d\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{dq}{r'^2} \hat{r}'$, where $r'$ is the distance from the element to P, and $\hat{r}'$ is the unit vector in that direction. The total field $\vec{E}$ at P is the integral of $d\vec{E}$ over the volume.

$\mathbf{\vec{E}(\vec{R}) = \int \frac{1}{4\pi\epsilon_0} \frac{\rho(\vec{r}') dV}{|\vec{R} - \vec{r}'|^2} \frac{\vec{R} - \vec{r}'}{|\vec{R} - \vec{r}'|}}$

Similar integrals apply for surface and line charge distributions.

Example 1.11. The electric field components in Fig. 1.27 are $E_x = ax^{1/2}$, $E_y = E_z = 0$, in which $a = 800$ N/C m$^{1/2}$. Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that $a = 0.1$ m.

Diagram for Example 1.11 showing a cube in a non-uniform electric field Ex = ax^1/2.

Answer:

The electric field is given by $\vec{E} = ax^{1/2} \hat{i}$. The cube has side length $a = 0.1$ m. One face is at $x=a$ (left face), and the opposite face is at $x=2a$ (right face). The other four faces are parallel to the x-axis (top, bottom, front, back).

(a) Flux through the cube: The electric flux $\Phi_E$ through a closed surface is the sum of the flux through each face. $\Phi_E = \oint \vec{E} \cdot d\vec{S}$.

For the faces perpendicular to the x-axis (left and right), the normal vectors are parallel or antiparallel to $\hat{i}$. For the other four faces, the normal vectors are perpendicular to $\hat{i}$ (along $\pm \hat{j}$ or $\pm \hat{k}$), so $\vec{E} \cdot d\vec{S} = (ax^{1/2} \hat{i}) \cdot (dS \hat{j/k}) = 0$. The flux through these four faces is zero.

Left face: Located at $x = a$. Electric field $\vec{E}_L = aa^{1/2} \hat{i}$. The outward normal for the left face is $\hat{n}_L = -\hat{i}$. The area vector is $\vec{S}_L = a^2 (-\hat{i})$. Flux $\Phi_L = \vec{E}_L \cdot \vec{S}_L = (aa^{1/2} \hat{i}) \cdot (a^2 (-\hat{i})) = -aa^{1/2}a^2 = -a^{5/2}$.

Right face: Located at $x = 2a$. Electric field $\vec{E}_R = a(2a)^{1/2} \hat{i} = a \sqrt{2} a^{1/2} \hat{i}$. The outward normal for the right face is $\hat{n}_R = +\hat{i}$. The area vector is $\vec{S}_R = a^2 (+\hat{i})$. Flux $\Phi_R = \vec{E}_R \cdot \vec{S}_R = (a\sqrt{2}a^{1/2} \hat{i}) \cdot (a^2 (+\hat{i})) = a\sqrt{2}a^{5/2} = \sqrt{2}a^{5/2}$.

The net flux through the cube is $\Phi_{net} = \Phi_L + \Phi_R = -\alpha a^{5/2} + \sqrt{2} \alpha a^{5/2} = \alpha a^{5/2} (\sqrt{2} - 1)$.

Given $a = 800 \text{ N/C m}^{1/2}$ and $a = 0.1 \text{ m}$.

$\Phi_{net} = 800 \times (0.1)^{5/2} (\sqrt{2} - 1) = 800 \times (10^{-1})^{5/2} (\sqrt{2} - 1) = 800 \times 10^{-5/2} (\sqrt{2} - 1) = 800 \times 10^{-2.5} \times (1.414 - 1) = 800 \times 10^{-2.5} \times 0.414$.

$10^{-2.5} = 10^{-2} \times 10^{-0.5} = 10^{-2} / \sqrt{10} \approx 10^{-2} / 3.16 \approx 0.316 \times 10^{-2} = 3.16 \times 10^{-3}$.

$\Phi_{net} \approx 800 \times 3.16 \times 10^{-3} \times 0.414 \approx 2528 \times 10^{-3} \times 0.414 \approx 2.528 \times 0.414 \approx 1.047$ N m$^2$/C.

The text's value is $1.05 \text{ N m}^2\text{ C}^{-1}$, which matches this calculation.

(b) Charge within the cube: According to Gauss's law, the total electric flux through a closed surface is equal to the net charge enclosed ($q_{enc}$) divided by the permittivity of free space ($\epsilon_0$).

$\Phi_{net} = \frac{q_{enc}}{\epsilon_0}$

$q_{enc} = \Phi_{net} \epsilon_0$. Using $\epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$.

$q_{enc} = (1.047 \text{ N m}^2/\text{C}) \times (8.854 \times 10^{-12} \text{ C}^2\text{ N}^{-1}\text{ m}^{-2})$

$q_{enc} \approx 9.26 \times 10^{-12}$ C.

The text's value is $9.27 \times 10^{-12}$ C, which is close.

The net flux through the cube is $1.05 \text{ N m}^2\text{ C}^{-1}$, and the charge within the cube is $9.27 \times 10^{-12}$ C.

Example 1.12. An electric field is uniform, and in the positive x direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x. It is given that $\vec{E} = 200 \hat{i}$ N/C for $x > 0$ and $\vec{E} = –200 \hat{i}$ N/C for $x < 0$. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at $x = +10$ cm and the other is at $x = –10$ cm (Fig. 1.28). (a) What is the net outward flux through each flat face? (b) What is the flux through the side of the cylinder? (c) What is the net outward flux through the cylinder? (d) What is the net charge inside the cylinder?

Diagram for Example 1.12 showing a cylinder centered at the origin with axis along x-axis in a non-uniform E field.

Answer:

The cylinder has length $L = 20$ cm $= 0.2$ m, radius $r = 5$ cm $= 0.05$ m. Its axis is along the x-axis, centered at the origin. One face is at $x = +10$ cm $= +0.1$ m (right face), the other is at $x = -10$ cm $= -0.1$ m (left face).

The electric field is $\vec{E} = 200 \hat{i}$ for $x > 0$ and $\vec{E} = -200 \hat{i}$ for $x < 0$. The magnitude of the field is uniform, $E = 200$ N/C, but the direction changes at $x=0$.

(a) Net outward flux through each flat face: The flat faces are at $x = -0.1$ m (left face) and $x = +0.1$ m (right face).

Left face (at $x = -0.1$ m): Since $x < 0$, $\vec{E}_L = -200 \hat{i}$ N/C. The outward normal to the left face is $\hat{n}_L = -\hat{i}$. The area of the face is $A = \pi r^2 = \pi (0.05 \text{ m})^2 = \pi \times 0.0025 \text{ m}^2$. The area vector is $\vec{S}_L = A \hat{n}_L = A (-\hat{i})$. Flux $\Phi_L = \vec{E}_L \cdot \vec{S}_L = (-200 \hat{i}) \cdot (A (-\hat{i})) = (-200)(-A) (\hat{i} \cdot \hat{i}) = 200 A = 200 \times \pi (0.05)^2 = 200 \times \pi \times 0.0025 = 0.5 \pi$ N m$^2$/C.

Right face (at $x = +0.1$ m): Since $x > 0$, $\vec{E}_R = +200 \hat{i}$ N/C. The outward normal to the right face is $\hat{n}_R = +\hat{i}$. The area vector is $\vec{S}_R = A \hat{n}_R = A (+\hat{i})$. Flux $\Phi_R = \vec{E}_R \cdot \vec{S}_R = (200 \hat{i}) \cdot (A (+\hat{i})) = (200)(A) (\hat{i} \cdot \hat{i}) = 200 A = 200 \times \pi (0.05)^2 = 0.5 \pi$ N m$^2$/C.

Using $\pi \approx 3.14$, $0.5 \pi \approx 0.5 \times 3.14 = 1.57$ N m$^2$/C. So, flux through each flat face is +1.57 N m$^2$/C (outward).

(b) Flux through the side of the cylinder: For any point on the curved surface (side) of the cylinder, the outward normal vector $\hat{n}_{\text{side}}$ is perpendicular to the x-axis (i.e., it lies in the yz-plane). The electric field $\vec{E}$ is always parallel to the x-axis (either $+200 \hat{i}$ or $-200 \hat{i}$). Therefore, the angle between $\vec{E}$ and $\hat{n}_{\text{side}}$ is always 90°. The dot product $\vec{E} \cdot \hat{n}_{\text{side}} = 0$.

The flux through the side of the cylinder is zero.

(c) Net outward flux through the cylinder: The total flux through the closed cylindrical surface is the sum of the flux through the left face, the right face, and the curved side.

$\Phi_{net} = \Phi_L + \Phi_R + \Phi_{\text{side}}$

$\Phi_{net} = (0.5 \pi) + (0.5 \pi) + 0 = \pi$ N m$^2$/C.

Using $\pi \approx 3.14$, $\Phi_{net} \approx 3.14$ N m$^2$/C.

(d) Net charge inside the cylinder: According to Gauss's law, the net outward flux through a closed surface is equal to the net charge enclosed divided by $\epsilon_0$.

$\Phi_{net} = \frac{q_{enc}}{\epsilon_0}$

$q_{enc} = \Phi_{net} \epsilon_0$. Using $\epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$.

$q_{enc} = (\pi \text{ N m}^2/\text{C}) \times (8.854 \times 10^{-12} \text{ C}^2\text{ N}^{-1}\text{ m}^{-2})$

$q_{enc} \approx 3.14 \times 8.854 \times 10^{-12}$ C.

$q_{enc} \approx 27.81 \times 10^{-12}$ C $= 2.781 \times 10^{-11}$ C.

The text gives $2.78 \times 10^{-11}$ C, which matches.

The net outward flux through each flat face is $1.57 \text{ N m}^2\text{ C}^{-1}$. The flux through the side is $0 \text{ N m}^2\text{ C}^{-1}$. The total net outward flux is $3.14 \text{ N m}^2\text{ C}^{-1}$. The net charge inside the cylinder is $2.78 \times 10^{-11}$ C.

Gauss’S Law

Gauss's law is a fundamental law in electrostatics that relates the total electric flux through any closed surface to the total electric charge enclosed within that surface. It provides a powerful alternative method to Coulomb's law for calculating electric fields, especially for charge distributions with high symmetry.

Gauss's law states that the **total electric flux** ($\Phi_E$) through any **closed surface** (often called a **Gaussian surface**) is directly proportional to the **net electric charge enclosed** ($q_{enc}$) by the surface, and inversely proportional to the permittivity of free space ($\epsilon_0$).

Mathematically:

$\mathbf{\Phi_E = \oint_S \vec{E} \cdot d\vec{S} = \frac{q_{enc}}{\epsilon_0}}$

Here, the integral is over the entire closed surface $S$, $\vec{E}$ is the electric field at each point on the surface, $d\vec{S}$ is the outward normal area vector element, and $q_{enc}$ is the algebraic sum of all charges inside the surface $S$.

A simple illustration: Consider a positive point charge $q$ at the center of a sphere of radius $r$. The electric field on the sphere is radial and has magnitude $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$. The outward normal $d\vec{S}$ is parallel to $\vec{E}$. The total flux is $\oint \vec{E} \cdot d\vec{S} = \oint E dS = E \oint dS = E (4\pi r^2) = (\frac{1}{4\pi\epsilon_0} \frac{q}{r^2}) (4\pi r^2) = \frac{q}{\epsilon_0}$. This matches Gauss's law.

Important points about Gauss's law:

It is valid for **any** closed surface, regardless of its shape or size.
$q_{enc}$ is the total charge *algebraically summed* and *enclosed* by the surface. Charges outside the surface do not contribute to the net flux, but they *do* contribute to the electric field $\vec{E}$ at points on the surface.
The Gaussian surface is an imaginary mathematical surface. It is chosen strategically to simplify the calculation of the flux integral, especially when the charge distribution has symmetry. It should not pass through any discrete point charge (where the field is infinite), but can pass through continuous charge distributions.
Gauss's law is equivalent to Coulomb's law and is based on the inverse square dependence of the electric field on distance.
If the net flux through a closed surface is zero, it implies the total charge enclosed by the surface is zero (though there might be charges outside the surface). If the charge enclosed is zero, the net flux must be zero, even if the field itself is non-zero at points on the surface (e.g., uniform field through a closed cylinder, shown in Figure 1.26).

Diagram illustrating zero net flux through a cylinder in a uniform electric field.

Applications Of Gauss’S Law

Gauss's law is particularly useful for calculating the electric field $\vec{E}$ for charge distributions that possess a high degree of symmetry (spherical, cylindrical, or planar). The strategy is to choose a Gaussian surface that matches the symmetry of the problem, so that the electric field is either constant in magnitude and perpendicular to the surface (flux is $E \times \text{Area}$), or parallel to the surface (flux is zero).

Field Due To An Infinitely Long Straight Uniformly Charged Wire

Consider an infinitely long straight wire with a uniform linear charge density $\lambda$ (charge per unit length). Due to the cylindrical symmetry, the electric field $\vec{E}$ at any point P can only depend on the perpendicular distance $r$ from the wire to P, and its direction must be radially outward (for $\lambda > 0$) or inward (for $\lambda < 0$) from the wire, in the plane perpendicular to the wire through P. The field magnitude is constant on any cylinder centered on the wire.

To apply Gauss's law, choose a cylindrical Gaussian surface of radius $r$ and length $l$, coaxial with the wire. The total flux through this closed cylinder is the sum of the flux through the two flat ends and the curved cylindrical surface.

Flux through the ends: The electric field is radial, so it is perpendicular to the normal vectors of the flat ends. The dot product $\vec{E} \cdot d\vec{S} = 0$. Flux through each end is zero.
Flux through the curved surface: At every point on the curved surface, the electric field $\vec{E}$ is radial and perpendicular to the wire, so it is parallel to the outward normal vector $d\vec{S}$. The magnitude of $\vec{E}$ is constant ($E$) everywhere on this surface (since $r$ is constant). The total area of the curved surface is $2\pi r l$. Flux $\Phi_{\text{curved}} = \oint \vec{E} \cdot d\vec{S} = \oint E dS = E \oint dS = E (2\pi r l)$.

Total flux through the Gaussian surface is $\Phi_{net} = 0 + 0 + E(2\pi r l) = E(2\pi r l)$.

The charge enclosed by the cylindrical Gaussian surface is the charge on the length $l$ of the wire, which is $q_{enc} = \lambda l$.

Applying Gauss's law: $\Phi_{net} = \frac{q_{enc}}{\epsilon_0} \implies E(2\pi r l) = \frac{\lambda l}{\epsilon_0}$.

Solving for $E$: $\mathbf{E = \frac{\lambda}{2\pi\epsilon_0 r}}$.

Vectorially, the electric field at a distance $r$ from the wire is $\mathbf{\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \hat{n}}$, where $\hat{n}$ is the radial unit vector outward from the wire in the plane perpendicular to the wire.

Field Due To A Uniformly Charged Infinite Plane Sheet

Consider an infinite thin plane sheet with uniform surface charge density $\sigma$ (charge per unit area). Due to the planar symmetry, the electric field $\vec{E}$ must be perpendicular to the sheet, uniform in magnitude at any given distance from the sheet, and point away from the sheet on both sides (for $\sigma > 0$).

To apply Gauss's law, choose a cylindrical (or rectangular parallelepiped) Gaussian surface with its axis perpendicular to the plane sheet and with its two flat ends of area $A$ parallel to the sheet, one on each side at equal distances from the sheet.

Flux through the side: The electric field is perpendicular to the sheet and parallel to the axis of the cylinder. The normal vectors to the curved side surface are perpendicular to $\vec{E}$. The dot product $\vec{E} \cdot d\vec{S} = 0$. Flux through the side is zero.
Flux through the ends: The electric field $\vec{E}$ is perpendicular to the sheet and parallel to the outward normal vectors of the flat ends on both sides. The magnitude of $\vec{E}$ is the same ($E$) at both ends (due to symmetry and infinite extent). The area of each end is $A$. Flux through the left end is $\Phi_{\text{left}} = EA$ (since $\vec{E}$ is constant and parallel to $d\vec{S}$). Flux through the right end is $\Phi_{\text{right}} = EA$.

Total flux through the Gaussian surface is $\Phi_{net} = \Phi_{\text{left}} + \Phi_{\text{right}} + \Phi_{\text{side}} = EA + EA + 0 = 2EA$.

The charge enclosed by the Gaussian surface is the charge on the area $A$ of the sheet intercepted by the cylinder, which is $q_{enc} = \sigma A$.

Applying Gauss's law: $\Phi_{net} = \frac{q_{enc}}{\epsilon_0} \implies 2EA = \frac{\sigma A}{\epsilon_0}$.

Solving for $E$: $\mathbf{E = \frac{\sigma}{2\epsilon_0}}$.

Vectorially, the electric field is $\mathbf{\vec{E} = \frac{\sigma}{2\epsilon_0} \hat{n}}$, where $\hat{n}$ is a unit vector normal to the plane and pointing away from it on either side. Note that the magnitude of the electric field is constant and does not depend on the distance from the sheet.

Field Due To A Uniformly Charged Thin Spherical Shell

Consider a thin spherical shell of radius $R$ with a uniform surface charge density $\sigma$. Due to spherical symmetry, the electric field $\vec{E}$ at any point P can only depend on the radial distance $r$ from the center of the shell O to P, and its direction must be radial (outward if $\sigma > 0$, inward if $\sigma < 0$).

To apply Gauss's law, choose a spherical Gaussian surface of radius $r$, centered at O and passing through P.

(i) **Field outside the shell (r > R):** The spherical Gaussian surface encloses the entire charge of the shell. The total charge on the shell is $q = \sigma \times (\text{Surface Area of shell}) = \sigma (4\pi R^2)$. So $q_{enc} = q$.

On the Gaussian surface, $\vec{E}$ is radial and perpendicular to the surface (parallel to $d\vec{S}$), and its magnitude $E$ is constant at all points (since $r$ is constant). Total flux $\Phi_{net} = \oint \vec{E} \cdot d\vec{S} = \oint E dS = E \oint dS = E (4\pi r^2)$.

Applying Gauss's law: $\Phi_{net} = \frac{q_{enc}}{\epsilon_0} \implies E(4\pi r^2) = \frac{q}{\epsilon_0}$.

Solving for $E$: $\mathbf{E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}}$.

Vectorially, $\mathbf{\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}}$, where $\hat{r}$ is the unit radial vector from the center. This is the same as the field of a point charge $q$ located at the center of the shell. For points outside, the shell behaves like a point charge.

(ii) **Field inside the shell (r < R):** The spherical Gaussian surface is inside the shell and encloses no charge. So $q_{enc} = 0$.

The total flux through the Gaussian surface is still $\Phi_{net} = E(4\pi r^2)$ (since $\vec{E}$ is radial and its magnitude depends only on $r$, even inside). However, since $r \ne 0$, $4\pi r^2 \ne 0$.

Applying Gauss's law: $\Phi_{net} = \frac{q_{enc}}{\epsilon_0} \implies E(4\pi r^2) = \frac{0}{\epsilon_0} = 0$.

Since $4\pi r^2 \ne 0$, we must have $\mathbf{E = 0}$ for $r < R$.

The electric field is zero at all points inside a uniformly charged thin spherical shell. This is a crucial result of Gauss's law and consistent with Coulomb's inverse square law.

Example 1.13. An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus?

Diagram for Example 1.13 showing a point nucleus and a uniformly distributed negative charge cloud.

Answer:

The atom consists of a point positive charge $+Ze$ at the center (nucleus) and a uniformly distributed negative charge within a sphere of radius $R$. The atom is neutral, so the total negative charge must be $-Ze$. Let the volume charge density of the negative charge cloud be $\rho$. The total volume of the negative charge cloud is $\frac{4}{3}\pi R^3$.

Total negative charge = $\rho \times \frac{4}{3}\pi R^3 = -Ze$.

So, the charge density $\rho = \frac{-Ze}{\frac{4}{3}\pi R^3} = -\frac{3Ze}{4\pi R^3}$. This density is uniform for $0 \le r \le R$, and zero for $r > R$.

We want to find the electric field $\vec{E}(r)$ at a distance $r$ from the nucleus. Due to spherical symmetry, $\vec{E}$ will be radial and its magnitude will depend only on $r$. Choose a spherical Gaussian surface of radius $r$ centered at the nucleus.

Case 1: Field inside the charge cloud (r < R).

The Gaussian surface is a sphere of radius $r < R$. Total flux through this surface is $\Phi = \oint \vec{E} \cdot d\vec{S} = E(r) (4\pi r^2)$, where $E(r)$ is the magnitude of the electric field at distance $r$.

The charge enclosed by the Gaussian surface consists of the positive nucleus charge at the center ($+Ze$) and the negative charge within the sphere of radius $r$. The volume of the negative charge cloud within radius $r$ is $\frac{4}{3}\pi r^3$.

The amount of negative charge within radius $r$ is $q_{neg\_enc} = \rho \times (\text{Volume}) = \left(-\frac{3Ze}{4\pi R^3}\right) \times \left(\frac{4}{3}\pi r^3\right) = -Ze \frac{r^3}{R^3}$.

The total charge enclosed is $q_{enc} = (+Ze) + q_{neg\_enc} = Ze - Ze \frac{r^3}{R^3} = Ze \left(1 - \frac{r^3}{R^3}\right)$.

Applying Gauss's law: $\Phi = \frac{q_{enc}}{\epsilon_0} \implies E(r) (4\pi r^2) = \frac{Ze}{\epsilon_0} \left(1 - \frac{r^3}{R^3}\right)$.

Solving for $E(r)$: $E(r) = \frac{1}{4\pi\epsilon_0} \frac{Ze}{r^2} \left(1 - \frac{r^3}{R^3}\right) = \frac{Ze}{4\pi\epsilon_0} \left(\frac{1}{r^2} - \frac{r}{R^3}\right)$ for $r < R$.

The direction of $\vec{E}$ is radially outward because for $r < R$, the positive charge of the nucleus dominates the enclosed negative charge, making $q_{enc}$ positive ($1 - r^3/R^3$ is positive for $r < R$).

Case 2: Field outside the charge cloud (r > R).

The Gaussian surface is a sphere of radius $r > R$. The total charge enclosed is the sum of the nucleus charge (+Ze) and the total negative charge within the cloud of radius R (-Ze). $q_{enc} = +Ze + (-Ze) = 0$.

Applying Gauss's law: $\Phi = \frac{q_{enc}}{\epsilon_0} \implies E(r) (4\pi r^2) = \frac{0}{\epsilon_0} = 0$.

Since $r > R$, $r \ne 0$, so $4\pi r^2 \ne 0$. Thus, $E(r) = 0$ for $r > R$.

At $r=R$, the formula for $r R$.

In summary, the electric field at a distance $r$ from the nucleus is:

For $r < R$: $\mathbf{\vec{E}(r) = \frac{Ze}{4\pi\epsilon_0} \left(\frac{1}{r^2} - \frac{r}{R^3}\right) \hat{r}}$ (radially outward)
For $r \ge R$: $\mathbf{\vec{E}(r) = 0}$

Summary

This chapter introduces the fundamental concepts of electric charges and fields, laying the groundwork for electrostatics.

Electric charge is a fundamental property of matter, existing in two types: positive and negative. Like charges repel, unlike charges attract. Electrification by rubbing involves the transfer of electrons.
Materials are classified as conductors (allow charge movement), insulators (restrict charge movement), or semiconductors (intermediate). Earthing is the process of sharing charges with the Earth for safety.
Charging can occur by contact or by induction (charge separation without contact).
Basic properties of electric charge:
- Additivity: Total charge is the algebraic sum of individual charges.
- Conservation: Total charge in an isolated system is constant.
- Quantisation: Any free charge is an integral multiple of the elementary charge $e$ ($q = ne$). At the macroscopic level, charge appears continuous.
Coulomb's Law quantitatively describes the force between two point charges $q_1$ and $q_2$ separated by $r$: $F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2}$. The force is along the line joining the charges. Vector form $\vec{F}_{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{21}^2} \hat{r}_{21}$ is consistent with Newton's third law. Electric forces are vastly stronger than gravitational forces.
Superposition Principle: The force on a charge due to multiple other charges is the vector sum of individual forces from each charge.
Electric Field ($\vec{E}$) at a point is the force per unit positive test charge ($\vec{E} = \vec{F}/q$). It's a vector field created by source charges, independent of the test charge. For a point charge $q$, $E = \frac{1}{4\pi\epsilon_0} \frac{|q|}{r^2}$. For a system of charges, $\vec{E}$ is the vector sum of fields from individual charges. Fields are physical entities, especially in time-dependent scenarios.
Electric Field Lines are visual representations of $\vec{E}$. Tangent shows direction, density shows strength. They start on positive charges, end on negative charges (or infinity), are continuous in charge-free regions, never cross, and do not form closed loops.
Electric Flux ($\Phi_E$) through a surface measures the amount of electric field passing through it: $\Phi_E = \int_S \vec{E} \cdot d\vec{S}$. For a closed surface, the area vector points outward.
An **Electric Dipole** is a pair of equal and opposite charges ($\pm q$) separated by $2a$, with dipole moment $\vec{p} = q(2a)\hat{p}$. Dipole field falls off as $1/r^3$ at large distances, faster than a point charge field.
In a uniform electric field $\vec{E}$, a dipole experiences **zero net force** but a torque $\vec{\tau} = \vec{p} \times \vec{E}$ that aligns it with the field. In a non-uniform field, it experiences both net force and torque.
For distributed charges, we use continuous charge densities: linear ($\lambda$), surface ($\sigma$), volume ($\rho$). Field calculation involves integration.
Gauss's Law: The total electric flux through any closed surface is $\frac{q_{enc}}{\epsilon_0}$, where $q_{enc}$ is the total charge enclosed by the surface ($\oint \vec{E} \cdot d\vec{S} = q_{enc}/\epsilon_0$). It's powerful for calculating fields of symmetric charge distributions.
- Infinitely long wire: $E = \frac{\lambda}{2\pi\epsilon_0 r}$.
- Infinite plane sheet: $E = \frac{\sigma}{2\epsilon_0}$.
- Uniformly charged thin spherical shell: $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$ for $r \ge R$ (like point charge at center), $E = 0$ for $r < R$.

Points to Ponder

Further thoughts on key concepts:

Protons in the nucleus are held together by the strong nuclear force, overcoming electrostatic repulsion. Electrons are not allowed inside the nucleus by quantum mechanics.
Coulomb force and gravitational force are inverse square laws, but gravity is always attractive, while electric force can be attractive or repulsive. This allows electric forces to cancel out, making gravity dominant on large scales despite being much weaker.
The SI unit of charge (Coulomb) is defined via the ampere (magnetic effects). This makes the constant $k$ in Coulomb's law fixed ($\approx 9 \times 10^9$), and the unit of charge is large from an electrostatic perspective.
Charge additivity is due to its scalar nature.
Charge invariance under relative motion and conservation are separate, fundamental properties.
Charge quantisation is a basic law without a classical analogue for mass.
Superposition principle states that forces between pairs of charges are independent of other charges, and there are no additional multi-body forces.
Electric field is undefined at the location of discrete charges. For continuous distributions, it can be defined. It is discontinuous across surface charge distributions.
Charge configurations with zero total charge can still produce electric fields, but they typically fall off faster than $1/r^2$ at large distances (e.g., dipole field $\propto 1/r^3$).

Exercises

Question 1.1. What is the force between two small charged spheres having charges of $2 \times 10^{–7}$C and $3 \times 10^{–7}$C placed 30 cm apart in air?

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Question 1.2. The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

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Question 1.3. Check that the ratio $ke^2/G m_e m_p$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

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Question 1.4. (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.

(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

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Question 1.5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

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Question 1.6. Four point charges $q_A = 2 \mu C$, $q_B = –5 \mu C$, $q_C = 2 \mu C$, and $q_D = –5 \mu C$ are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

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Question 1.7. (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

(b) Explain why two field lines never cross each other at any point?

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Question 1.8. Two point charges $q_A = 3 \mu C$ and $q_B = –3 \mu C$ are located 20 cm apart in vacuum.

(a) What is the electric field at the midpoint O of the line AB joining the two charges?

(b) If a negative test charge of magnitude $1.5 \times 10^{–9}$ C is placed at this point, what is the force experienced by the test charge?

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Question 1.9. A system has two charges $q_A = 2.5 \times 10^{–7}$ C and $q_B = –2.5 \times 10^{–7}$ C located at points A: (0, 0, –15 cm) and B: (0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?

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Question 1.10. An electric dipole with dipole moment $4 \times 10^{–9}$ C m is aligned at 30° with the direction of a uniform electric field of magnitude $5 \times 10^4$ NC$^{–1}$. Calculate the magnitude of the torque acting on the dipole.

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Question 1.11. A polythene piece rubbed with wool is found to have a negative charge of $3 \times 10^{–7}$ C.

(a) Estimate the number of electrons transferred (from which to which?)

(b) Is there a transfer of mass from wool to polythene?

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Question 1.12. (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{–7}$ C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

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Question 1.13. Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

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Question 1.14. Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Three charged particles moving through a uniform electric field between two plates. Particles 1 and 2 are deflected towards the positive plate, while particle 3 is deflected towards the negative plate. Particle 3 shows the largest deflection.

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Question 1.15. Consider a uniform electric field $\textbf{E} = 3 \times 10^3 \hat{\textbf{i}}$ N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

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Question 1.16. What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

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Question 1.17. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 10^3 \text{ Nm}^2/\text{C}$. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

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Question 1.18. A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

A point charge located above the center of a square.

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Question 1.19. A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

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Question 1.20. A point charge causes an electric flux of $–1.0 \times 10^3 \text{ Nm}^2/\text{C}$ to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

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Question 1.21. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is $1.5 \times 10^3$ N/C and points radially inward, what is the net charge on the sphere?

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Question 1.22. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m$^2$. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

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Question 1.23. An infinite line charge produces a field of $9 \times 10^4$ N/C at a distance of 2 cm. Calculate the linear charge density.

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Question 1.24. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0 \times 10^{–22}$ C/m$^2$. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

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ADDITIONAL EXERCISES

Question 1.25. An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55 \times 10^4$ NC$^{–1}$ (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm$^{–3}$. Estimate the radius of the drop. (g = 9.81 m s$^{–2}$; e = $1.60 \times 10^{–19}$ C).

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Question 1.26. Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

Five diagrams (a-e) showing different patterns of electric field lines. Some show lines crossing, forming closed loops, or originating from empty space, which are incorrect representations of electrostatic field lines.

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Question 1.27. In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of $10^5$ NC$^{–1}$ per metre. What are the force and torque experienced by a system having a total dipole moment equal to $10^{–7}$ Cm in the negative z-direction ?

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Question 1.28. (a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

(a) A conductor A with a cavity. (b) A charged conductor B is placed inside the cavity of conductor A.

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Question 1.29. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $(\sigma/2\epsilon_0) \hat{\textbf{n}}$, where $\hat{\textbf{n}}$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.

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Question 1.30. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

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Question 1.31. It is now established that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + (2/3) e, and the ‘down’ quark (denoted by d) of charge (–1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

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Question 1.32. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.

(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

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Question 1.33. A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed $v_x$ (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $qEL^2/(2m v_x^2)$.

Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

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Question 1.34. Suppose that the particle in Exercise in 1.33 is an electron projected with velocity $v_x = 2.0 \times 10^6 \text{ m s}^{–1}$. If E between the plates separated by 0.5 cm is $9.1 \times 10^2$ N/C, where will the electron strike the upper plate? ($|e|=1.6 \times 10^{–19} \text{ C}, m_e = 9.1 \times 10^{–31} \text{ kg}$.)

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