Coulomb's Law and Electric Field
Coulomb’S Law ($ F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2} $)
Coulomb's Law is a fundamental law in electrostatics that quantifies the force between two stationary point electric charges. It is analogous to Newton's Law of Gravitation, describing an inverse square force.
Statement of Coulomb's Law
Coulomb's Law states that the magnitude of the electrostatic force of attraction or repulsion between two stationary point charges is:
- Directly proportional to the product of the magnitudes of the two charges.
- Inversely proportional to the square of the distance between them.
The force acts along the straight line joining the two charges.
Let $q_1$ and $q_2$ be the magnitudes of two point charges, and $r$ be the distance between them. The magnitude of the electrostatic force ($F$) is:
$ F \propto |q_1 q_2| $
$ F \propto \frac{1}{r^2} $
Combining these proportionalities:
$ F \propto \frac{|q_1 q_2|}{r^2} $
To turn this into an equation, we introduce a constant of proportionality $k$:
$ F = k \frac{|q_1 q_2|}{r^2} $
The constant $k$ depends on the medium in which the charges are placed and the system of units used. In vacuum, $k$ is given by:
$ k = \frac{1}{4\pi\epsilon_0} $
where $\epsilon_0$ is the permittivity of free space (vacuum). $\epsilon_0 \approx 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)$.
So, the magnitude of the electrostatic force between two point charges $q_1$ and $q_2$ in vacuum, separated by a distance $r$, is:
$ F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2} $
The value of the constant $k = \frac{1}{4\pi\epsilon_0}$ is approximately $8.987 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, often rounded to $9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$ for calculations.
Direction of the Force
The force is repulsive if the charges have the same sign ($q_1 q_2 > 0$) and attractive if the charges have opposite signs ($q_1 q_2 < 0$). The force on each charge is directed along the line connecting the two charges. The force exerted by $q_1$ on $q_2$ is equal in magnitude and opposite in direction to the force exerted by $q_2$ on $q_1$ (consistent with Newton's Third Law).
Coulomb's Law in a Medium
If the charges are placed in a medium other than vacuum (like air, water, glass), the electrostatic force between them is reduced. The constant $k$ becomes $k_{medium} = \frac{1}{4\pi\epsilon}$, where $\epsilon$ is the absolute permittivity of the medium. The absolute permittivity $\epsilon$ is related to the permittivity of free space $\epsilon_0$ by the relative permittivity (or dielectric constant) $\epsilon_r$ of the medium: $\epsilon = \epsilon_0 \epsilon_r$.
$ F_{medium} = \frac{1}{4\pi\epsilon_0 \epsilon_r} \frac{|q_1 q_2|}{r^2} = \frac{F_{vacuum}}{\epsilon_r} $
The force between two charges in a medium is $\epsilon_r$ times weaker than the force in vacuum (or air) for the same distance. The value of $\epsilon_r$ is always $\ge 1$. For vacuum, $\epsilon_r = 1$. For air, $\epsilon_r \approx 1.0006$. For water, $\epsilon_r \approx 80$.
Vector Form of Coulomb's Law
Let $\vec{r}_{12}$ be the position vector from charge $q_1$ to charge $q_2$. The unit vector is $\hat{r}_{12} = \vec{r}_{12}/|\vec{r}_{12}| = \vec{r}_{12}/r$. The force exerted on $q_2$ by $q_1$, denoted by $\vec{F}_{21}$, acts along the line joining the charges. Its direction depends on the signs of $q_1$ and $q_2$.
$ \vec{F}_{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \hat{r}_{12} $
If $q_1 q_2 > 0$ (like charges), $\vec{F}_{21}$ is in the direction of $\hat{r}_{12}$ (repulsive). If $q_1 q_2 < 0$ (unlike charges), $\vec{F}_{21}$ is in the direction opposite to $\hat{r}_{12}$ (attractive). This single vector equation correctly represents both attraction and repulsion based on the signs of the charges.
Example 1. Two point charges $q_1 = +2 \times 10^{-9}$ C and $q_2 = -3 \times 10^{-9}$ C are placed 0.5 metres apart in vacuum. Calculate the magnitude and nature of the electrostatic force between them.
Answer:
Charge 1, $q_1 = +2 \times 10^{-9}$ C.
Charge 2, $q_2 = -3 \times 10^{-9}$ C.
Distance between charges, $r = 0.5$ m.
Medium: Vacuum. Constant $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$.
Using Coulomb's Law for the magnitude of the force:
$ F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2} $
$ F = (9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \times \frac{|(+2 \times 10^{-9} \text{ C}) \times (-3 \times 10^{-9} \text{ C})|}{(0.5 \text{ m})^2} $
$ F = (9 \times 10^9) \times \frac{|-6 \times 10^{-18}|}{0.25} $ N
$ F = (9 \times 10^9) \times \frac{6 \times 10^{-18}}{0.25} $ N
$ F = (9 \times 10^9) \times (24 \times 10^{-18}) $ N
$ F = (9 \times 24) \times 10^{(9-18)} $ N
$ F = 216 \times 10^{-9} $ N.
This can also be written as $2.16 \times 10^{-7}$ N or 216 nanoNewtons (nN).
Nature: Since the charges have opposite signs ($+q_1$ and $-q_2$), the force between them is attractive.
The magnitude of the electrostatic force between the charges is $2.16 \times 10^{-7}$ Newtons, and it is an attractive force.
Forces Between Multiple Charges (Superposition Principle)
In reality, we often deal with systems containing more than two charges. To find the net force on a particular charge due to all other charges in the system, we use the Principle of Superposition.
Statement of the Principle of Superposition
The principle of superposition states that the net electrostatic force on any charge in a system of multiple charges is the vector sum of the forces exerted on it by all other charges, calculated individually as if each other charge were the only other charge present.
The presence of other charges does not affect the force between any given pair of charges. The total force is simply the vector addition of all pairwise forces.
Applying the Principle
Consider a system of $N$ point charges $q_1, q_2, ..., q_N$ located at positions $\vec{r}_1, \vec{r}_2, ..., \vec{r}_N$. To find the net force on charge $q_i$ (located at $\vec{r}_i$), we calculate the force exerted on $q_i$ by each of the other charges $q_j$ ($j \ne i$) using Coulomb's Law, and then vectorially add these forces.
The force exerted on $q_i$ by $q_j$, $\vec{F}_{ij}$, is given by Coulomb's Law (vector form):
$ \vec{F}_{ij} = \frac{1}{4\pi\epsilon_0} \frac{q_j q_i}{|\vec{r}_i - \vec{r}_j|^2} \hat{r}_{ji} $
where $\hat{r}_{ji}$ is the unit vector pointing from $q_j$ to $q_i$ ($\hat{r}_{ji} = \frac{\vec{r}_i - \vec{r}_j}{|\vec{r}_i - \vec{r}_j|}$).
The net force on charge $q_i$, $\vec{F}_{i, net}$, is the vector sum of the forces due to all other charges:
$ \vec{F}_{i, net} = \sum_{j=1, j \ne i}^{N} \vec{F}_{ij} = \sum_{j=1, j \ne i}^{N} \frac{1}{4\pi\epsilon_0} \frac{q_j q_i}{|\vec{r}_i - \vec{r}_j|^2} \hat{r}_{ji} $
This summation must be a vector sum, meaning the forces must be added using vector addition (e.g., by resolving them into components along coordinate axes and summing the components). The order of summation does not matter.
Example 1. Three point charges $q_1 = +1 \, \mu\text{C}$, $q_2 = +2 \, \mu\text{C}$, and $q_3 = -3 \, \mu\text{C}$ are placed at the vertices of an equilateral triangle of side 0.1 m. Find the net force on charge $q_1$. (Take $k = 9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$ and $1 \, \mu\text{C} = 10^{-6}$ C).
Answer:
Charges: $q_1 = +1 \times 10^{-6}$ C, $q_2 = +2 \times 10^{-6}$ C, $q_3 = -3 \times 10^{-6}$ C.
Distance between any pair of charges, $r = 0.1$ m.
We need to find the net force on $q_1$. This force is the vector sum of the force on $q_1$ due to $q_2$ ($\vec{F}_{12}$) and the force on $q_1$ due to $q_3$ ($\vec{F}_{13}$).
Let's place the charges at the vertices of the triangle. Let $q_1$ be at the top vertex, $q_2$ at the bottom-left, and $q_3$ at the bottom-right. Let's use coordinates, e.g., $q_1$ at $(0, h)$, $q_2$ at $(-0.05, 0)$, $q_3$ at $(0.05, 0)$, where $h$ is the height of the equilateral triangle, $h = 0.1 \sin(60^\circ) = 0.1 \times \sqrt{3}/2 \approx 0.0866$ m.
The distance between any pair is $r=0.1$ m.
Force on $q_1$ due to $q_2$ ($\vec{F}_{12}$):
$q_1$ and $q_2$ are both positive, so the force is repulsive. It acts along the line joining $q_2$ to $q_1$. The magnitude is:
$ F_{12} = k \frac{|q_1 q_2|}{r^2} = (9 \times 10^9) \frac{(1 \times 10^{-6})(2 \times 10^{-6})}{(0.1)^2} = 9 \times 10^9 \frac{2 \times 10^{-12}}{0.01} $
$ F_{12} = 9 \times 10^9 \times 200 \times 10^{-12} = 1800 \times 10^{-3} = 1.8 $ N.
The direction of $\vec{F}_{12}$ is along the line from $q_2$ to $q_1$. This line makes an angle of $60^\circ$ with the horizontal axis (if we place $q_2 q_3$ on x-axis) and $90 - 60 = 30^\circ$ with the vertical axis. Let's resolve into components. The line from $q_2$ to $q_1$ makes an angle of $60^\circ$ above the negative x-axis, or $120^\circ$ with the positive x-axis. Or angle of $30^\circ$ left from vertical line upwards.
Force on $q_1$ due to $q_3$ ($\vec{F}_{13}$):
$q_1$ is positive, $q_3$ is negative, so the force is attractive. It acts along the line joining $q_1$ to $q_3$. The magnitude is:
$ F_{13} = k \frac{|q_1 q_3|}{r^2} = (9 \times 10^9) \frac{(1 \times 10^{-6})|-3 \times 10^{-6}|}{(0.1)^2} = 9 \times 10^9 \frac{3 \times 10^{-12}}{0.01} $
$ F_{13} = 9 \times 10^9 \times 300 \times 10^{-12} = 2700 \times 10^{-3} = 2.7 $ N.
The direction of $\vec{F}_{13}$ is along the line from $q_1$ to $q_3$. This line makes an angle of $60^\circ$ with the horizontal axis and $30^\circ$ with the vertical axis, pointing downwards and to the right from $q_1$. Or angle $60^\circ$ below the positive x-axis, or $-60^\circ$ with the positive x-axis. Or $30^\circ$ right from vertical line downwards.
Let's use components, taking the y-axis vertically upwards through $q_1$, and the x-axis horizontally. $q_1$ is at origin $(0,0)$, $q_2$ at $(-r \cos 30, -r \sin 30)$, $q_3$ at $(r \cos 30, -r \sin 30)$. No, let $q_2, q_3$ be on x-axis. $q_2$ at $(-0.05, 0)$, $q_3$ at $(0.05, 0)$, $q_1$ at $(0, 0.0866)$. We want force ON $q_1$. Position of $q_1$ is $\vec{r}_1 = (0, 0.0866)$. Position of $q_2$ is $\vec{r}_2 = (-0.05, 0)$. Position of $q_3$ is $\vec{r}_3 = (0.05, 0)$.
Force on $q_1$ due to $q_2$: $\vec{F}_{12}$. Direction from $q_2$ to $q_1$. Angle is $60^\circ$ above the positive x-axis (from perspective of $q_2$). Angle $120^\circ$ from +x axis if origin at $q_1$. Easier way: use symmetry. The triangle vertices are $q_1=(0,h)$, $q_2=(-a/2,0)$, $q_3=(a/2,0)$. $a=0.1$. $h = a\sqrt{3}/2$. Force on $q_1$ due to $q_2$ is repulsive, pointing from $q_2$ to $q_1$. This direction makes angle $60^\circ$ with -x axis. Force on $q_1$ due to $q_3$ is attractive, pointing from $q_1$ to $q_3$. This direction makes angle $60^\circ$ with +x axis.
Let's resolve forces relative to $q_1$ location. $\vec{F}_{12}$ is repulsive from $q_2$. Angle with horizontal is $60^\circ$. X-component: $F_{12x} = -F_{12} \cos(60^\circ) = -1.8 \times 0.5 = -0.9$ N. Y-component: $F_{12y} = F_{12} \sin(60^\circ) = 1.8 \times \sqrt{3}/2 \approx 1.8 \times 0.866 = 1.5588$ N.
$\vec{F}_{13}$ is attractive to $q_3$. Angle with horizontal is $60^\circ$. X-component: $F_{13x} = F_{13} \cos(60^\circ) = 2.7 \times 0.5 = 1.35$ N. Y-component: $F_{13y} = -F_{13} \sin(60^\circ) = -2.7 \times \sqrt{3}/2 \approx -2.7 \times 0.866 = -2.3382$ N.
Net force components on $q_1$:
$ F_{1, net, x} = F_{12x} + F_{13x} = -0.9 \text{ N} + 1.35 \text{ N} = 0.45 $ N.
$ F_{1, net, y} = F_{12y} + F_{13y} \approx 1.5588 \text{ N} - 2.3382 \text{ N} = -0.7794 $ N.
Net force vector $\vec{F}_{1, net} = (0.45 \hat{i} - 0.7794 \hat{j})$ N.
Magnitude of net force: $ F_{1, net} = \sqrt{F_{1, net, x}^2 + F_{1, net, y}^2} = \sqrt{(0.45)^2 + (-0.7794)^2} = \sqrt{0.2025 + 0.60746} = \sqrt{0.80996} \approx 0.8999 $ N.
The magnitude of the net force on charge $q_1$ is approximately 0.9 N. The direction is in the fourth quadrant (positive x, negative y).
The net force on charge $q_1$ is approximately 0.9 N, directed at an angle $\theta = \arctan(F_{1, net, y} / F_{1, net, x}) = \arctan(-0.7794/0.45) \approx \arctan(-1.732) \approx -60^\circ$ with the positive x-axis.
Electric Field ($ \vec{E} = \vec{F}/q $)
Just as gravity can be described using a gravitational field, electric charges create an electric field in the space around them. An electric field is a region where an electric charge would experience an electrostatic force.
Definition of Electric Field Strength
The electric field strength (or electric field intensity) at a point in space is defined as the electrostatic force experienced by a unit positive test charge placed at that point.
If a test charge $q_0$ is placed at a point and experiences an electrostatic force $\vec{F}$, the electric field strength $\vec{E}$ at that point is:
$ \vec{E} = \frac{\vec{F}}{q_0} $
The test charge $q_0$ must be infinitesimally small and positive so that it does not disturb the charge distribution creating the field. The units of electric field strength are Newtons per Coulomb (N/C).
The direction of the electric field vector $\vec{E}$ at any point is the direction of the force that would act on a positive test charge placed at that point. The electric field points away from positive charges and towards negative charges.
Electric Field due to a Point Charge
Consider a point charge $Q$ located at the origin. To find the electric field at a point P at a distance $r$ from $Q$, we place a small positive test charge $q_0$ at P. The force on $q_0$ due to $Q$ is given by Coulomb's Law:
$ \vec{F} = \frac{1}{4\pi\epsilon_0} \frac{Q q_0}{r^2} \hat{r} $
where $\hat{r}$ is the unit vector pointing from $Q$ to P (i.e., radially outwards).
The electric field at point P is $\vec{E} = \vec{F}/q_0$:
$ \vec{E} = \frac{1}{q_0} \left(\frac{1}{4\pi\epsilon_0} \frac{Q q_0}{r^2} \hat{r}\right) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \hat{r} $
The magnitude of the electric field at distance $r$ from a point charge $Q$ is:
$ E = \frac{1}{4\pi\epsilon_0} \frac{|Q|}{r^2} $
If $Q$ is positive, $\vec{E}$ is in the direction of $\hat{r}$ (radially outwards). If $Q$ is negative, $\vec{E}$ is in the direction opposite to $\hat{r}$ (radially inwards).
Electric Field Due To A System Of Charges
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. This is the principle of superposition for electric fields.
If we have a system of $N$ point charges $q_1, q_2, ..., q_N$ at positions $\vec{r}_1, \vec{r}_2, ..., \vec{r}_N$, the net electric field $\vec{E}_{net}$ at a point P (with position vector $\vec{r}_P$) is the vector sum of the fields $\vec{E}_i$ produced by each charge $q_i$ at point P:
$ \vec{E}_{net}(\vec{r}_P) = \sum_{i=1}^{N} \vec{E}_i(\vec{r}_P) $
$ \vec{E}_{net}(\vec{r}_P) = \sum_{i=1}^{N} \frac{1}{4\pi\epsilon_0} \frac{q_i}{|\vec{r}_P - \vec{r}_i|^2} \hat{r}_{iP} $
where $\hat{r}_{iP}$ is the unit vector pointing from charge $q_i$ to point P ($\hat{r}_{iP} = \frac{\vec{r}_P - \vec{r}_i}{|\vec{r}_P - \vec{r}_i|}$).
For a continuous charge distribution (like a charged rod, plate, or volume), the summation is replaced by integration over the charge distribution.
Physical Significance Of Electric Field
The electric field is not just a mathematical convenience; it is considered a real physical entity that exists in space. Its significance lies in:
- Mediation of Force: The electric field mediates the electrostatic force between charges. A charge creates an electric field around it. This field then exerts a force on any other charge present in the field. The interaction is not instantaneous action at a distance, but rather through the field. Changes in the position of a charge propagate outwards as changes in the electric field at the speed of light.
- Energy Storage: The electric field stores energy. The energy stored per unit volume in an electric field is proportional to the square of the electric field strength.
- Description of Electromagnetic Waves: Electric and magnetic fields are not static; oscillating electric fields produce magnetic fields, and changing magnetic fields produce electric fields. These propagating disturbances in the electric and magnetic fields constitute electromagnetic waves (light, radio waves, etc.).
Thinking in terms of electric fields is essential for understanding complex charge interactions, electromagnetic waves, and energy storage in electromagnetic fields.
Example 2. Calculate the magnitude of the electric field at a distance of 0.1 m from a point charge of $+5 \times 10^{-9}$ C in vacuum.
Answer:
Charge, $Q = +5 \times 10^{-9}$ C.
Distance from the charge, $r = 0.1$ m.
Medium: Vacuum. Constant $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$.
Using the formula for the magnitude of the electric field due to a point charge:
$ E = \frac{1}{4\pi\epsilon_0} \frac{|Q|}{r^2} $
$ E = (9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \times \frac{|+5 \times 10^{-9} \text{ C}|}{(0.1 \text{ m})^2} $
$ E = (9 \times 10^9) \times \frac{5 \times 10^{-9}}{0.01} $ N/C
$ E = (9 \times 10^9) \times (500 \times 10^{-9}) $ N/C
$ E = (9 \times 500) \times 10^{(9-9)} $ N/C
$ E = 4500 \times 10^0 = 4500 $ N/C.
The magnitude of the electric field at a distance of 0.1 m from the charge is 4500 N/C. Since the charge is positive, the direction of the field is radially outwards from the charge.
Electric Field Lines
Electric field lines (also called lines of force) are a way to visualize the electric field in a region of space. They are imaginary lines or curves drawn such that they represent the direction and relative strength of the electric field.
Properties of Electric Field Lines
- Direction: The tangent to an electric field line at any point gives the direction of the electric field vector $\vec{E}$ at that point. The direction is the same as the force a positive test charge would experience.
- Origin and Termination: Electric field lines start from positive charges (or infinity) and end on negative charges (or infinity). They do not form closed loops in electrostatics (unlike magnetic field lines).
- No Intersection: Two electric field lines can never intersect each other. If they did, it would mean the electric field at the point of intersection has two different directions simultaneously, which is impossible.
- Relative Strength: The density of electric field lines in a region represents the strength of the electric field. Where the lines are closer together, the field is stronger; where they are farther apart, the field is weaker.
- Perpendicular to Equipotential Surfaces: Electric field lines are always perpendicular to equipotential surfaces (surfaces of constant electric potential).
Electric field lines are a graphical representation and do not physically exist. The actual field exists at all points in space, not just on the lines drawn.
Field Line Patterns for Different Charge Configurations
- Isolated Positive Point Charge: Field lines are radial, pointing outwards from the charge.
(Image Placeholder: A central point with a '+' sign. Arrows radiate outwards from the point in all directions.)
- Isolated Negative Point Charge: Field lines are radial, pointing inwards towards the charge.
(Image Placeholder: A central point with a '-' sign. Arrows point inwards towards the point from all directions.)
- Electric Dipole (Pair of Equal and Opposite Charges): Field lines originate from the positive charge and terminate on the negative charge. They form curves.
(Image Placeholder: A '+' charge and a '-' charge separated by a small distance. Show field lines curving from '+' to '-'. Lines near the charges are denser.)
- Two Equal Positive Charges: Field lines originate from both charges and repel each other, bending outwards. There is a point between the charges where the field is zero.
(Image Placeholder: Two '+' charges separated by a distance. Show field lines radiating outwards from both, bending away from the line joining them. Show a region between them where field lines are less dense or seem to cancel.)
- Uniform Electric Field: In a region where the electric field is uniform (same magnitude and direction everywhere), the field lines are parallel, equally spaced, and point in the direction of the field. (e.g., between two large, parallel, oppositely charged plates).
(Image Placeholder: Two parallel horizontal lines (representing plates). The top line has '+' signs, bottom has '-' signs. Show equally spaced vertical arrows pointing downwards from the top line to the bottom line, representing uniform electric field lines.)
Electric field lines are a valuable visualization tool for understanding the concept of the electric field and predicting the direction and relative strength of the force on a charge placed in the field.