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Topic 3 of 15

Gauss's Law

Gauss's law is one of Maxwell's four fundamental equations of electromagnetism. It states that the net electric flux through any closed surface equals the enclosed charge divided by ε₀. While Coulomb's law and Gauss's law contain the same physics, Gauss's law combined with symmetry gives us a powerful shortcut to find electric fields for symmetric charge distributions that would be intractable with direct integration.

Key Concepts

Electric Flux

Electric flux through a surface measures how much of the electric field passes through it:

\Phi_E = \vec{E}\cdot\vec{A} = EA\cos\theta

for a uniform field and flat surface, where

\theta

is the angle between

\vec{E}

and the outward normal. Units: N·m²/C.

Gauss's Law

The net electric flux through any closed (Gaussian) surface equals the total charge enclosed divided by

\varepsilon_0

\oint\vec{E}\cdot d\vec{A} = q_{\text{enc}}/\varepsilon_0

. It is exact and always true — but only useful for calculating

E

when the charge distribution has sufficient symmetry.

Spherical Symmetry

For a uniform spherical shell or solid sphere of charge, choose a concentric spherical Gaussian surface. Outside:

E = kQ/r^2

(same as point charge). Inside a shell:

E = 0

. Inside a solid sphere:

E = kQr/R^3

Cylindrical Symmetry

For an infinite line charge (or cylinder), choose a coaxial cylindrical Gaussian surface. The result is

E = \lambda/(2\pi\varepsilon_0 r)

, where

\lambda

is the linear charge density (C/m).

Planar Symmetry

For an infinite plane of surface charge density

\sigma

(C/m²), choose a pillbox (short cylinder) straddling the plane. Field on each side:

E = \sigma/(2\varepsilon_0)

, perpendicular to the plane.

Conductors in Electrostatic Equilibrium

Inside a conductor

E = 0

(free charges rearrange until they cancel any internal field). All excess charge resides on the surface. Just outside the surface,

E = \sigma/\varepsilon_0

, perpendicular to the surface.

Key Equations

Electric flux (uniform field)

\Phi_E = EA\cos\theta

For a uniform field E through a flat surface of area A; θ is the angle between E and the outward normal.

Gauss's Law

\oint\vec{E}\cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}

Net outward flux through any closed surface equals the enclosed charge divided by ε₀ = 8.85×10⁻¹² C²/(N·m²).

Field of infinite line charge

E = \frac{\lambda}{2\pi\varepsilon_0 r}

E at perpendicular distance r from an infinite line charge with linear density λ (C/m).

Field of infinite plane

E = \frac{\sigma}{2\varepsilon_0}

Uniform field on each side of an infinite flat sheet with surface charge density σ (C/m²).

Field inside solid sphere (non-conducting)

E = \frac{kQr}{R^3}

At radius r < R inside a uniformly charged sphere of total charge Q and radius R. Grows linearly from center.

Worked Example

Electric Field of a Uniformly Charged Solid Sphere

Problem

A non-conducting solid sphere of radius $R = 0.20$ m carries total charge $Q = +8$ nC uniformly distributed. Find the field (a) at $r = 0.30$ m (outside) and (b) at $r = 0.10$ m (inside).

Solution

(a) Outside ( $r > R$ ): Draw a spherical Gaussian surface of radius 0.30 m. The full charge is enclosed, and by symmetry $E$ is uniform on the surface:

E\cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \Rightarrow E = \frac{kQ}{r^2} = \frac{(8.99\times10^9)(8\times10^{-9})}{(0.30)^2} = 799 \text{ N/C}

(b) Inside ( $r < R$ ): The Gaussian surface of radius 0.10 m encloses only the fraction $(r/R)^3$ of the charge:

q_{\text{enc}} = Q\left(\frac{r}{R}\right)^3 = (8\times10^{-9})\left(\frac{0.10}{0.20}\right)^3 = 1.0\text{ nC}

E = \frac{kq_{\text{enc}}}{r^2} = \frac{(8.99\times10^9)(1\times10^{-9})}{(0.10)^2} = 899 \text{ N/C}

Answer (a) 799 N/C outside; (b) 899 N/C inside at r = 0.10 m.

Practice

Exercises

7 problems

1 of 7

A spherical Gaussian surface encloses a charge $q = +2$ nC. What is the total electric flux (in N·m²/C) through the surface? Use $\varepsilon_0 = 8.85\times10^{-12}$ C²/(N·m²).

Your answer

N·m²/C

2 of 7

A uniformly charged sphere has total charge $Q = +5$ nC and radius $R = 0.10$ m. What is the electric field magnitude (in N/C) at $r = 0.30$ m outside it?

Your answer

N/C

3 of 7

A non-conducting solid sphere has total charge $Q = +8$ nC uniformly distributed and radius $R = 0.20$ m. Find the electric field magnitude (in N/C) at $r = 0.10$ m inside it.

Your answer

N/C

4 of 7

An infinite line charge has linear charge density $\lambda = 2$ nC/m. Find the electric field magnitude (in N/C) at a perpendicular distance $r = 0.10$ m from it. Use $\varepsilon_0 = 8.85\times10^{-12}$ C²/(N·m²).

Your answer

N/C

5 of 7

An infinite flat plane carries surface charge density $\sigma = 4$ nC/m². What is the electric field magnitude (in N/C) on either side of the plane?

Your answer

N/C

6 of 7

A charge $q = +6$ nC is placed at the center of a cube with side length $L = 0.20$ m. What is the electric flux (in N·m²/C) through one face of the cube?

Your answer

N·m²/C

7 of 7

The surface of a conducting sphere has surface charge density $\sigma = 3$ nC/m². What is the electric field magnitude (in N/C) just outside the surface?

Your answer

N/C

Key Takeaways

Gauss's law $\oint\vec{E}\cdot d\vec{A} = q_{\text{enc}}/\varepsilon_0$ is exact, but it gives $E$ easily only when symmetry makes $E$ constant (or zero) across the Gaussian surface.
Choose the Gaussian surface to match the symmetry: sphere for spherical, coaxial cylinder for line/cylinder, pillbox for plane.
Outside any spherically symmetric distribution, the field is identical to that of a point charge at the center.
Inside a conducting shell: $E = 0$ . Outside: $E = kQ_{\text{total}}/r^2$ .
The field just outside a conductor surface is $E = \sigma/\varepsilon_0$ — twice the field of an isolated infinite plane, because the conductor interior is shielded.