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Electricity and Magnetism
Gauss's Law in Electrostatics
Gauss's law relates the net electric flux through a closed surface to the total enclosed electric charge. It is a fundamental law of electrostatics derived from Coulomb's inverse-square force law.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Calculation of electric field using symmetric charge distributions (infinite line charge, infinite plane sheet, and spherical shells) and basic flux calculations.
PGT extension
Differential form of Gauss's law (divergence theorem applications), non-uniform charge distributions, solid angle calculations, and electrostatic shielding.
Prerequisites
Vector calculus (scalar product of vectors, surface integrals, and divergence)
Concept of electric field and electric lines of force
Electric flux and solid angle
Electric Flux and Surface Integral
Electric flux measures the total number of electric field lines passing normally through a given area.
Mathematically defined as the surface integral of the electric field vector over a closed or open surface.
SI unit of electric flux is N m2 C-1 or V m.
Statement of Gauss's Law
The total electric flux through any closed hypothetical surface (Gaussian surface) is equal to 1/ε0 times the net charge enclosed by that surface.
It is independent of the shape, size, and position of the charges inside the Gaussian surface.
Charges outside the Gaussian surface contribute zero net flux because the number of field lines entering equals the number leaving.
Differential Form of Gauss's Law
Relates the divergence of the electric field directly to the volume charge density at any point in space.
Expressed vectorially as the divergence of E equals the volume charge density divided by permittivity of free space.
Represents the local or point-wise formulation of electrostatics.
Choice of Gaussian Surface
A Gaussian surface must be a closed mathematical surface chosen to exploit the symmetry of the charge distribution.
The electric field magnitude should be constant at every point on the surface where flux is evaluated.
The angle between the electric field vector and the area vector must be consistently zero, 90 degrees, or 180 degrees.
Derivations
Electric Field due to an Infinite Thin Plane Sheet of Charge
Conditions: Uniform surface charge density σ and infinite spatial extent
Construct a cylindrical Gaussian surface of cross-sectional area A piercing perpendicularly through the infinite sheet.
Evaluate the electric flux through the two flat end-caps of the cylinder as 2EA, while the curved surface contributes zero flux since E is parallel to the surface.
Equate the total flux 2EA to the enclosed charge q/ε0 where $q = \sigmaA$.
Solve for the electric field magnitude E.
Result: $E = \frac{\sigma}{2epsilon_0}$
Electric Field due to a Uniformly Charged Thin Spherical Shell
Conditions: Radius R, total charge Q distributed uniformly over the surface
For a point outside the shell (r > R), choose a concentric spherical Gaussian surface of radius r.
By spherical symmetry, the electric field E is radial and uniform in magnitude at radius r, yielding a total flux of E(4πr2).
Apply Gauss's law equating the flux to Q/ε0 to find the external field.
For a point inside the shell (r < R), the enclosed charge is zero, which results in zero electric field inside.
Result: $E = \frac{1}{4piepsilon_0} \frac{Q}{r^2} \text{}$ for } r > R \text{ and } E = 0 \text{ for } r < R
A solid insulating sphere of radius R has a non-uniform volume charge density given by $\rho = \rho_{0} (r/R)$, where r is the radial distance from the center. Find the electric field inside the sphere at distance r (< R).
Construct a spherical Gaussian surface of radius r inside the sphere (r < R).
Express the enclosed charge qencl by integrating the volume charge density: qencl = ∫ ρ dV from 0 to r.
Substitute d$V = 4\pir'^{2} dr'$ to get qencl = ∫ ρ0(r'/R)(4πr'2) dr' = (ρ0 4π / R) × (r4 / 4) = ρ0 π r4 / R.
Gauss's law is universally valid for any inverse-square central force field, but its utility for simple algebraic calculations of electric field strictly requires high geometric symmetry (spherical, cylindrical, or planar).
It assumes macroscopic electrostatics where quantum electrodynamic corrections and relativistic field fluctuations are negligible.
Common misconceptions
Common error: The electric field on a Gaussian surface is produced solely by the charges enclosed inside that surface.
Correct: The electric field at any point on the Gaussian surface is the vector sum of electric fields due to ALL charges, both inside and outside the surface.
Although outside charges contribute zero net flux over the closed surface, they do contribute to the local electric field vector at individual points on the surface.
Common error: Gauss's law cannot be used to find the electric field if there is no high symmetry.
Correct: Gauss's law always holds true; however, it only yields a simple analytical solution for E when symmetry allows the electric field magnitude to be factored out of the surface integral.
The surface integral requires E to be uniform and normal over symmetric portions of the chosen Gaussian surface.
Quick revision
Total electric flux through a closed surface depends exclusively on the net enclosed charge and is independent of the size or shape of the Gaussian surface.
Integral Form of Gauss's Law: oint mathbf{E} cdot dmathbf{A} = \frac{q_{\text{enclosed}}}{epsilon_0}
Differential Form of Gauss's Law: \nabla cdot mathbf{E} = \frac{\rho}{epsilon_0}
Must remember
The total electric flux through any closed Gaussian surface depends exclusively on the net charge enclosed inside it and is equal to $q_{\text{enclosed}} / \epsilon_0$.
Charges located outside the Gaussian surface contribute zero net flux, but they do contribute to the local electric field vector at any point on the surface.
The differential form of Gauss's law is expressed vectorially as $\nabla \cdot \mathbf{E} = \rho / \epsilon_0$, linking local divergence to volume charge density.
For an infinite line charge, the electric field varies inversely with distance as $E = \frac{\lambda}{2\pi\epsilon_0 r}$.
Just outside the surface of a charged conductor, the electric field is normal to the surface and given by $E = \frac{\sigma}{\epsilon_0}$.
Glossary
Electric Flux ($\Phi_E$)A measure of the total number of electric field lines passing normally through a given surface area, defined as $\int \mathbf{E} \cdot d\mathbf{A}$.
Gaussian SurfaceA hypothetical closed mathematical surface chosen to calculate electric flux and determine electric field using symmetry.
Volume Charge Density ($\rho$)The amount of charge per unit volume, measured in C m$^{-3}$.
Linear Charge Density ($\lambda$)The amount of charge per unit length, measured in C m$^{-1}$.
Surface Charge Density ($\sigma$)The amount of charge per unit surface area, measured in C m$^{-2}$.
Field due to Symmetric Charge Configurations
Configuration
Electric Field Formula ($E$)
Distance Dependence
Infinite Line Charge
$E = \frac{\lambda}{2\pi\epsilon_0 r}$
Proportional to $1/r$
Infinite Plane Sheet
$E = \frac{\sigma}{2\epsilon_0}$
Independent of distance ($r^0$)
Spherical Shell (Outside, $r \ge R$)
$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$
Proportional to $1/r^2$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the SI unit of electric flux?
Explanation: Electric flux is defined as $\Phi_E = \int \mathbf{E} \cdot d\mathbf{A}$, where electric field $E$ has units of $\text{N C}^{-1}$ and area $A$ has units of $\text{m}^2$, making the SI unit $\text{N m}^2 \text{ C}^{-1}$.
2. According to Gauss's law, the total electric flux through a closed surface depends upon:
Explanation: Gauss's law states that $\oint \mathbf{E} \cdot d\mathbf{A} = q_{\text{enclosed}} / \epsilon_0$. Thus, total flux depends strictly on the net charge enclosed by the Gaussian surface.
3. If a point charge $q$ is placed at the center of a cube of side length $a$, what is the total electric flux passing through all six faces of the cube?
Explanation: By Gauss's law, the total flux through any closed surface enclosing a charge $q$ is independent of its shape and equals $q / \epsilon_0$.
4. What is the differential form of Gauss's law in electrostatics?
Explanation: The differential form relates the divergence of the electric field to the volume charge density: $\nabla \cdot \mathbf{E} = \rho / \epsilon_0$.
5. Which of the following statements about charges outside a closed Gaussian surface is correct?
Explanation: Charges outside a Gaussian surface contribute zero net flux because every field line entering the surface must also leave it. However, they do contribute to the local electric field vector at points on the surface.
6. The electric field intensity at a perpendicular distance $r$ from an infinite line charge with linear charge density $\lambda$ is given by:
Explanation: Applying Gauss's law with a cylindrical Gaussian surface around an infinite line charge yields $E = \frac{\lambda}{2\pi\epsilon_0 r}$.
7. What is the electric field just outside the surface of a charged conductor in electrostatic equilibrium with surface charge density $\sigma$?
Explanation: Using a tiny pillbox Gaussian surface straddling the conductor surface, the electric field just outside is $E = \sigma / \epsilon_0$, whereas inside the conductor it is zero.
8. For an infinite thin plane sheet of charge with uniform surface charge density $\sigma$, how does the electric field vary with distance $r$ from the sheet?
Explanation: The electric field due to an infinite plane sheet is given by $E = \sigma / (2\epsilon_0)$, which is completely uniform and independent of the distance from the sheet.
9. What is the electric field inside a uniformly charged solid insulating sphere of radius $R$ and total charge $Q$ at a distance $r < R$ from the center?
Explanation: Using a spherical Gaussian surface of radius $r < R$, the enclosed charge is $q_{\text{encl}} = Q (r/3 / R^3)$, leading to $E = \frac{Qr}{4\pi\epsilon_0 R^3}$.
10. What is the electric field inside a hollow spherical conducting shell carrying a net charge $Q$?
Explanation: Since all excess charge on a conductor resides on its outer surface, any Gaussian surface drawn inside the conducting material or cavity (with no enclosed charge) yields an electric field of zero.
11. Why is a symmetric Gaussian surface chosen when applying Gauss's law?
Explanation: Symmetry ensures that the magnitude of $\mathbf{E}$ is constant and the angle between $\mathbf{E}$ and $d\mathbf{A}$ is uniform over parts of the surface, allowing $\oint \mathbf{E} \cdot d\mathbf{A}$ to simplify to $E \oint dA$.
12. If the electric field in a region is given by $\mathbf{E} = c x \hat{\mathbf{i}}$, what is the volume charge density $\rho$ at any point in this region?
Explanation: Using the differential form $\rho = \epsilon_0 \nabla \cdot \mathbf{E} = \epsilon_0 \frac{\partial}{\partial x}(cx) = c \epsilon_0$.
13. A charge $Q$ is placed at one corner of a cube. What is the total electric flux through the entire cube?
Explanation: A corner of a cube is shared by 8 adjacent cubes of a symmetrical lattice. Thus, only $1/8$th of the total flux from the charge passes through the given cube, giving $Q / (8\epsilon_0)$.
14. Which of the following conditions is required for the surface integral $\oint \mathbf{E} \cdot d\mathbf{A}$ to yield $E \times A$ directly?
Explanation: When $\mathbf{E}$ and $d\mathbf{A}$ are parallel ($\cos\theta = 1$) and $E$ is constant, $\oint \mathbf{E} \cdot d\mathbf{A} = E \oint dA = EA$.
15. What happens to the electric flux through a closed surface if the magnitude of the enclosed charge is doubled?
Explanation: By Gauss's law, $\Phi_E = q_{\text{enclosed}} / \epsilon_0$, so the flux is directly proportional to the enclosed charge.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A spherical Gaussian surface encloses a point charge $q$. If the radius of the Gaussian surface is doubled, how does the total electric flux through the surface change?
Explanation: Total electric flux depends only on the net enclosed charge ($q / \epsilon_0$) and is independent of the size or radius of the Gaussian surface.
2. The divergence of the electric field vector $\mathbf{E}$ in a charge-free region is:
Explanation: In a charge-free region ($\rho = 0$), the differential form of Gauss's law $\nabla \cdot \mathbf{E} = \rho / \epsilon_0$ implies that $\nabla \cdot \mathbf{E} = 0$.
3. If a closed surface encloses no net charge, which of the following must be true regarding the electric field on the surface?
Explanation: Zero net enclosed charge means the net flux is zero ($\oint \mathbf{E} \cdot d\mathbf{A} = 0$), but individual field lines can enter and leave, meaning the electric field itself is not necessarily zero.
4. Two infinite parallel plane sheets carry uniform surface charge densities $+\sigma$ and $-\sigma$ respectively. What is the electric field in the region between the sheets?
Explanation: Between the two sheets, the fields due to both sheets point in the same direction from positive to negative, adding up to $E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$.
5. A solid conducting sphere of radius $R$ is given a total charge $Q$. What is the electrostatic potential and electric field inside the conductor at a distance $r < R$?
Explanation: Inside a conductor in electrostatic equilibrium, the electric field is strictly zero, and the potential remains constant throughout the volume, equal to its value at the surface.
6. What is the electric flux through a square surface of side $L$ in a uniform electric field $\mathbf{E}_0$ if the electric field vector makes an angle of $60^{\circ}$ with the normal to the surface?
Explanation: Electric flux is given by $\Phi_E = E A \cos\theta = E_0 L^2 \cos(60^{\circ}) = E_0 L^2 (1/2)$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which of the following laws or theorems is mathematically equivalent to Gauss's law in integral form?
Explanation: The divergence theorem connects the volume integral of $\nabla \cdot \mathbf{E}$ to the surface integral $\oint \mathbf{E} \cdot d\mathbf{A}$, transforming Gauss's law between its integral and differential forms.
2. A point charge $q$ is located at a distance $d$ directly above the center of a square of side $2d$. What is the electric flux through the square?
Explanation: A square of side $2d$ with a charge at distance $d$ above its center can be viewed as one face of a hypothetical cube of side $2d$ enclosing the charge. Since a cube has 6 identical faces, the flux through the square is $1/6$th of the total flux, i.e., $q / (6\epsilon_0)$.
3. If the electric potential in a region is given by $V = x^2 yz$, what is the $x$-component of the electric field at point $(1, 2, 3)$?
Explanation: $E_x = -\frac{\partial V}{\partial x} = -2xyz$. At point $(1, 2, 3)$, $E_x = -2(1)(2)(3) = -12 \text{ V m}^{-1}$.
4. A thick spherical conducting shell has an inner radius $a$ and outer radius $b$. A point charge $+q$ is placed at its center. What is the surface charge density on the inner surface of the shell?
Explanation: By Gauss's law, the electric field inside the conducting material must be zero, so a Gaussian surface inside the conductor must enclose zero net charge. This requires an induced charge of $-q$ on the inner surface of radius $a$, giving $\sigma_{\text{inner}} = -q / (4\pi a^2)$.
5. The electric field in a certain region of space is given by $\mathbf{E} = A r \hat{\mathbf{r}}$ in spherical polar coordinates. The volume charge density in this region is:
Explanation: Using the divergence in spherical coordinates for a purely radial field: $\nabla \cdot \mathbf{E} = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \cdot Ar) = \frac{1}{r^2} \frac{\partial}{\partial r}(A r^3) = 3A$. Thus, $\rho = \epsilon_0 (3A) = 3A\epsilon_0$.
6. What is the net electric flux through a closed cylindrical surface placed horizontally in a uniform electric field directed parallel to the axis of the cylinder?
Explanation: In a uniform electric field, the flux entering the left flat circular cap is equal and opposite to the flux leaving the right flat circular cap, while the curved surface contributes zero flux. Net flux is zero.
7. Which of the following distributions does NOT allow the straightforward calculation of electric field using Gauss's law?
Explanation: A finite rod lacks translational symmetry near its ends, preventing the electric field magnitude from being factored out of the surface integral in a simple manner.
8. An uncharged conducting shell is placed near a point charge $+Q$. Which statement is true regarding the total induced charge on the outer surface of the shell?
Explanation: Since the shell is uncharged overall and isolated, charge conservation requires that the sum of the inner and outer induced charges equals zero. Since the inner surface acquires $-Q$ to shield the interior, the outer surface must acquire $+Q$.
9. What is the unit of volume charge density $\rho$ appearing in the differential form of Gauss's law?
Explanation: Volume charge density is charge per unit volume, with standard SI units of Coulombs per cubic meter ($\text{C m}^{-3}$).
10. If a Gaussian surface encloses multiple point charges $q_1, q_2, -q_3$, what is the total electric flux through the surface?
Explanation: Gauss's law states that total flux equals the algebraic sum of all enclosed charges divided by $\epsilon_0$, which is $(q_1 + q_2 - q_3) / \epsilon_0$.