Gauss's Law and its Applications

Continuous Charge Distribution (Linear, Surface, Volume Density)

While Coulomb's Law is defined for point charges, real-world objects often have charge spread out over a line, a surface, or throughout a volume. When the charge distribution is continuous (or can be treated as continuous because the distances are much larger than the separation between discrete charges), we describe it using charge density. Instead of summing up the fields or forces from individual point charges, we integrate over the charge distribution.

Linear Charge Density ($\lambda$)

When charge is distributed along a line (e.g., a charged wire or rod), the charge distribution is described by the linear charge density ($\lambda$). It is defined as the amount of charge per unit length.

$ \lambda = \frac{\text{Charge}}{\text{Length}} = \frac{dQ}{dl} $

where $dQ$ is the amount of charge on an infinitesimal length element $dl$. If the charge is uniformly distributed over a length $L$ with total charge $Q$, then $\lambda = Q/L$. The units of linear charge density are Coulombs per metre (C/m).

The charge on an infinitesimal length $dl$ is $dQ = \lambda \, dl$.

Surface Charge Density ($\sigma$)

When charge is distributed over a surface (e.g., a charged plate or the surface of a charged conductor), the charge distribution is described by the surface charge density ($\sigma$). It is defined as the amount of charge per unit area.

$ \sigma = \frac{\text{Charge}}{\text{Area}} = \frac{dQ}{dA} $

where $dQ$ is the amount of charge on an infinitesimal area element $dA$. If the charge is uniformly distributed over an area $A$ with total charge $Q$, then $\sigma = Q/A$. The units of surface charge density are Coulombs per square metre (C/m$^2$).

The charge on an infinitesimal area $dA$ is $dQ = \sigma \, dA$.

Volume Charge Density ($\rho$)

When charge is distributed throughout the volume of a region (e.g., a charged sphere or a cloud of ions), the charge distribution is described by the volume charge density ($\rho$). It is defined as the amount of charge per unit volume.

$ \rho = \frac{\text{Charge}}{\text{Volume}} = \frac{dQ}{dV} $

where $dQ$ is the amount of charge in an infinitesimal volume element $dV$. If the charge is uniformly distributed throughout a volume $V$ with total charge $Q$, then $\rho = Q/V$. The units of volume charge density are Coulombs per cubic metre (C/m$^3$).

The charge in an infinitesimal volume $dV$ is $dQ = \rho \, dV$.

These charge densities allow us to treat the charge distribution as continuous, which is particularly useful when calculating the electric field or potential due to extended charged objects using integration or Gauss's Law.

Electric Flux ($ \Phi_E = \int \vec{E} \cdot d\vec{A} $)

Electric flux ($\Phi_E$) is a measure of the "flow" of the electric field through a given surface. It is a concept related to how many electric field lines pass through a surface and at what angle.

Definition of Electric Flux

For a uniform electric field $\vec{E}$ passing through a flat surface of area $A$, the electric flux $\Phi_E$ is defined as the product of the magnitude of the electric field, the area, and the cosine of the angle between the electric field vector and the area vector.

The area vector ($d\vec{A}$ or $\vec{A}$) is a vector perpendicular to the surface, with magnitude equal to the area. The direction of the area vector for a closed surface is conventionally taken as pointing outwards from the surface.

$ \Phi_E = |\vec{E}| |\vec{A}| \cos\theta = \vec{E} \cdot \vec{A} $

where $\theta$ is the angle between $\vec{E}$ and $\vec{A}$.

For a non-uniform electric field or a curved surface, the electric flux is calculated by dividing the surface into infinitesimal area elements $d\vec{A}$, finding the flux through each element ($d\Phi_E = \vec{E} \cdot d\vec{A}$), and then integrating over the entire surface:

$ \Phi_E = \int_S \vec{E} \cdot d\vec{A} $

The integral is a surface integral over the surface $S$. The units of electric flux are Newton-metre squared per Coulomb (N·m$^2$/C).

Flux through a Closed Surface

When the surface is a closed surface (like a sphere, cube, or any enclosed volume), the integral is denoted by $\oint \vec{E} \cdot d\vec{A}$, with the circle on the integral sign indicating a closed surface integral. The direction of $d\vec{A}$ for a closed surface is always taken as outwards.

The net electric flux through a closed surface is related to the net charge enclosed within the surface, as stated by Gauss's Law.

Physical interpretation: Electric flux can be thought of as proportional to the number of electric field lines passing perpendicularly through the surface. Positive flux means field lines are directed outwards through the surface, and negative flux means field lines are directed inwards. For a closed surface, positive net flux means there is a net positive charge inside, and negative net flux means there is a net negative charge inside.

Gauss’S Law ($ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} $)

Gauss's Law is a fundamental law of electrostatics that provides a powerful relationship between the electric flux through a closed surface and the net electric charge enclosed within that surface. It is one of Maxwell's equations and is a more general and often more convenient way to calculate the electric field, especially for charge distributions with high symmetry, compared to using Coulomb's Law and integration.

Statement of Gauss's Law

Gauss's Law states that the total electric flux through any closed surface in vacuum is equal to the net electric charge enclosed within that surface divided by the permittivity of free space ($\epsilon_0$).

Mathematically:

$ \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} $

where:

$\oint_S \vec{E} \cdot d\vec{A}$ is the total electric flux through the closed surface $S$.
$Q_{enclosed}$ is the net electric charge enclosed within the closed surface $S$. Any charges outside the closed surface contribute to the electric field $\vec{E}$ at the surface but do not contribute to the net flux through the surface.
$\epsilon_0$ is the permittivity of free space.

The closed surface $S$ is an imaginary surface called a Gaussian surface. The choice of the Gaussian surface is arbitrary, but for practical applications of Gauss's Law to calculate the electric field, we choose a Gaussian surface that matches the symmetry of the charge distribution and passes through the point where we want to find the electric field.

Gauss's Law is equivalent to Coulomb's Law and is a more general statement about the relationship between electric charge and electric field. While Coulomb's Law is suitable for point charges and can be integrated for continuous distributions, Gauss's Law is particularly useful when the charge distribution possesses spherical, cylindrical, or planar symmetry.

Relationship between Gauss's Law and Coulomb's Law

Gauss's Law can be derived from Coulomb's Law and the superposition principle, and vice versa. For example, to derive Coulomb's Law from Gauss's Law, consider a single point charge $Q$ at the centre of a spherical Gaussian surface of radius $r$. The electric field due to a point charge is radial and has the same magnitude at all points on the spherical surface. Also, the area vector $d\vec{A}$ is radial and points outwards, parallel to $\vec{E}$.

$ \oint_S \vec{E} \cdot d\vec{A} = \oint_S E \, dA \cos 0^\circ = \oint_S E \, dA $

Since $E$ is constant on the sphere, we can take it out of the integral:

$ E \oint_S dA = E (4\pi r^2) $

The total charge enclosed is $Q_{enclosed} = Q$. Applying Gauss's Law:

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

$ E = \frac{Q}{4\pi\epsilon_0 r^2} $

This is the magnitude of the electric field at distance $r$ from a point charge $Q$. If we place a test charge $q_0$ at this point, the force on it is $F = q_0 E = \frac{1}{4\pi\epsilon_0} \frac{Q q_0}{r^2}$, which is Coulomb's Law. Thus, Gauss's Law implies Coulomb's Law.

Applications Of Gauss’S Law

Gauss's Law simplifies the calculation of electric fields for charge distributions with specific symmetries. By choosing an appropriate Gaussian surface, the integral $\oint \vec{E} \cdot d\vec{A}$ becomes easy to evaluate, allowing us to solve for $E$.

Field Due To An Infinitely Long Straight Uniformly Charged Wire

Consider an infinitely long straight wire with uniform linear charge density $\lambda$. Due to the cylindrical symmetry of the charge distribution, the electric field at any point is radial, pointing perpendicular to the wire, and its magnitude depends only on the perpendicular distance $r$ from the wire.

Diagram illustrating the Gaussian surface for an infinitely long charged wire.

(Image Placeholder: An infinitely long straight line representing the charged wire. A coaxial cylindrical Gaussian surface of radius r and length L is shown around the wire. Indicate electric field vectors E pointing radially outwards, perpendicular to the curved surface and parallel to the flat ends.)

To find the electric field at a distance $r$ from the wire, choose a cylindrical Gaussian surface of radius $r$ and arbitrary length $L$, coaxial with the wire. This closed surface has three parts: the curved cylindrical surface and the two flat end caps.

The electric field $\vec{E}$ is radial, so it is perpendicular to the area vectors of the end caps ($\vec{E} \cdot d\vec{A}_{cap} = 0$). The electric field is parallel to the area vectors on the curved surface ($\vec{E} \cdot d\vec{A}_{curved} = E \, dA_{curved} \cos 0^\circ = E \, dA_{curved}$). The magnitude of $\vec{E}$ is constant on the curved surface at a fixed radius $r$.

The total electric flux through the Gaussian surface is the sum of the flux through the curved surface and the end caps:

$ \oint \vec{E} \cdot d\vec{A} = \int_{curved} \vec{E} \cdot d\vec{A}_{curved} + \int_{caps} \vec{E} \cdot d\vec{A}_{cap} $

$ \oint \vec{E} \cdot d\vec{A} = \int_{curved} E \, dA_{curved} + 0 = E \int_{curved} dA_{curved} = E (2\pi r L) $

The charge enclosed within the cylindrical Gaussian surface of length $L$ is the charge on the length $L$ of the wire, $Q_{enclosed} = \lambda L$.

Applying Gauss's Law: $ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} $

$ E (2\pi r L) = \frac{\lambda L}{\epsilon_0} $

The length $L$ cancels out:

$ E (2\pi r) = \frac{\lambda}{\epsilon_0} $

$ E = \frac{\lambda}{2\pi\epsilon_0 r} $

The electric field magnitude at a distance $r$ from an infinitely long uniformly charged wire is $E = \frac{\lambda}{2\pi\epsilon_0 r}$. The direction is radially outwards if $\lambda$ is positive, and radially inwards if $\lambda$ is negative. The field falls off as $1/r$, which is slower than the $1/r^2$ field of a point charge or finite line charge.

Field Due To A Uniformly Charged Infinite Plane Sheet

Consider an infinitely large plane sheet with uniform surface charge density $\sigma$. Due to the planar symmetry, the electric field at any point is perpendicular to the sheet, pointing away from it if $\sigma$ is positive, and its magnitude depends only on the perpendicular distance from the sheet (and is uniform if the sheet is infinite).

Diagram illustrating the Gaussian surface for an infinite charged plane sheet.

(Image Placeholder: A plane representing the charged sheet. A cylindrical Gaussian surface with its axis perpendicular to the sheet is shown passing through the sheet. It has two flat end caps parallel to the sheet and a curved surface. Indicate electric field vectors E pointing perpendicularly outwards from the sheet, parallel to the axis of the cylinder.)

To find the electric field, choose a cylindrical Gaussian surface with its axis perpendicular to the sheet and its two flat end caps of area $A$ located symmetrically on either side of the sheet at equal distances. The curved surface of the cylinder is parallel to the field (if we consider the field from infinite sheet is perpendicular to sheet). The electric field $\vec{E}$ is perpendicular to the sheet, so it is parallel to the area vectors of the end caps ($\vec{E} \cdot d\vec{A}_{cap} = E \, dA_{cap} \cos 0^\circ = E \, dA_{cap}$ on the side field points out, $\cos 180^\circ = -1$ on the side field points in, but we choose surface vector outwards, so always E dot dA is positive on both caps if field points outwards). On the curved surface, the field is perpendicular to the area vectors ($\vec{E} \cdot d\vec{A}_{curved} = E \, dA_{curved} \cos 90^\circ = 0$).

The total electric flux through the Gaussian surface is the sum of the flux through the two end caps and the curved surface:

$ \oint \vec{E} \cdot d\vec{A} = \int_{cap1} \vec{E} \cdot d\vec{A}_1 + \int_{cap2} \vec{E} \cdot d\vec{A}_2 + \int_{curved} \vec{E} \cdot d\vec{A}_{curved} $

$ \oint \vec{E} \cdot d\vec{A} = E A + E A + 0 = 2EA $ (assuming the magnitude of E is the same at both end caps, which is true for an infinite sheet by symmetry).

The charge enclosed within the cylindrical Gaussian surface is the charge on the area $A$ of the sheet intercepted by the cylinder, $Q_{enclosed} = \sigma A$.

Applying Gauss's Law: $ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} $

$ 2EA = \frac{\sigma A}{\epsilon_0} $

The area $A$ cancels out:

$ 2E = \frac{\sigma}{\epsilon_0} $

$ E = \frac{\sigma}{2\epsilon_0} $

The electric field magnitude due to a uniformly charged infinite plane sheet is $E = \frac{\sigma}{2\epsilon_0}$. The direction is perpendicular to the sheet, pointing away from it if $\sigma$ is positive, and towards it if $\sigma$ is negative. The field is uniform; its magnitude does not depend on the distance from the sheet. This is an important and unusual result, valid for an infinite sheet.

Field Due To A Uniformly Charged Thin Spherical Shell

Consider a thin spherical shell of radius $R$ with uniform surface charge density $\sigma$ and total charge $Q = \sigma (4\pi R^2)$. Due to the spherical symmetry, the electric field at any point is radial, pointing away from the centre if $Q$ is positive, and its magnitude depends only on the distance $r$ from the centre of the shell.

Diagram illustrating Gaussian surfaces for a charged spherical shell (inside and outside).

(Image Placeholder: A cross-section of a spherical shell of radius R. A point P is shown inside the shell at radius r < R. A spherical Gaussian surface of radius r is shown inside. Another point P' is shown outside the shell at radius r' > R. A spherical Gaussian surface of radius r' is shown outside.)

To find the electric field, choose a spherical Gaussian surface of radius $r$ centered at the centre of the spherical shell.

Case 1: Point P is outside the shell ($r > R$)

Choose a spherical Gaussian surface of radius $r > R$. The electric field is radial and its magnitude is constant on this surface. $\vec{E} \cdot d\vec{A} = E \, dA$.

$ \oint \vec{E} \cdot d\vec{A} = E \oint dA = E (4\pi r^2) $

The total charge enclosed within the Gaussian surface is the total charge on the spherical shell, $Q_{enclosed} = Q$.

Applying Gauss's Law: $ E (4\pi r^2) = \frac{Q}{\epsilon_0} $

$ E = \frac{Q}{4\pi\epsilon_0 r^2} $

Since $Q = \sigma (4\pi R^2)$, $E = \frac{\sigma (4\pi R^2)}{4\pi\epsilon_0 r^2} = \frac{\sigma R^2}{\epsilon_0 r^2}$.

For points outside the shell, the electric field is the same as that of a point charge equal to the total charge of the shell placed at its centre. The field falls off as $1/r^2$, similar to a point charge.

Case 2: Point P is inside the shell ($r < R$)

Choose a spherical Gaussian surface of radius $r < R$. The electric field is radial and its magnitude is constant on this surface. $\oint \vec{E} \cdot d\vec{A} = E (4\pi r^2)$.

The total charge enclosed within the Gaussian surface is zero, $Q_{enclosed} = 0$, because all the charge is on the shell at $R > r$.

Applying Gauss's Law: $ E (4\pi r^2) = \frac{0}{\epsilon_0} $

$ E (4\pi r^2) = 0 $

Since $4\pi r^2 \ne 0$ (unless $r=0$), we must have $E = 0$.

The electric field is zero everywhere inside a uniformly charged thin spherical shell. This is a remarkable result and has implications for phenomena like electrostatic shielding.

Graph showing electric field vs distance from the center of a charged spherical shell.

(Image Placeholder: A graph with distance r from the center on the x-axis. Y-axis is Electric Field E. Show E=0 for 0 <= r < R. At r=R, E jumps to a value proportional to sigma/epsilon0. For r > R, E decreases as 1/r^2.)

These examples illustrate the power of Gauss's Law in simplifying electric field calculations for symmetric charge distributions. For non-symmetric distributions, calculating the integral in Gauss's Law is as difficult as using Coulomb's Law and integration, so it is primarily a tool for understanding the relationship between charge and field in those cases.