Electrostatics

AP Physics C E&M guide to electrostatics: Coulomb's law, electric fields, Gauss's law, and electric field calculations for symmetric charge distributions.

# Electrostatics — AP Physics C E&M

AP Physics C: E&M begins with electrostatics — the study of stationary charges and the electric fields they produce. You must apply Coulomb's law, calculate fields using integration, and use Gauss's law for symmetric charge distributions.

Key Concepts

Coulomb's Law (Vector Form)

F=kq1q2r2r^\vec{F} = \frac{kq_1 q_2}{r^2}\hat{r}

Electric Field

E=Fq0=kQr2r^\vec{E} = \frac{\vec{F}}{q_0} = \frac{kQ}{r^2}\hat{r}

Continuous charge distribution: E=kdqr2r^\vec{E} = k\int \frac{dq}{r^2}\hat{r}

Common linear charge density: λ=dq/dL\lambda = dq/dL; surface: σ=dq/dA\sigma = dq/dA; volume: ρ=dq/dV\rho = dq/dV.

Gauss's Law

EdA=Qencϵ0\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}

Use when there is high symmetry (spherical, cylindrical, planar).

Common results:

  • Infinite line charge: E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}
  • Infinite plane of charge: E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0}
  • Uniformly charged sphere (outside): E=Q4πϵ0r2E = \frac{Q}{4\pi\epsilon_0 r^2} (like a point charge)
  • Uniformly charged sphere (inside): E=Qr4πϵ0R3E = \frac{Qr}{4\pi\epsilon_0 R^3}

Electric Field of a Charged Ring (on axis)

Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

Electric Field of a Charged Disk (on axis)

Ex=σ2ϵ0(1xx2+R2)E_x = \frac{\sigma}{2\epsilon_0}\left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right)

Worked Example

Problem: Use Gauss's law to find the electric field inside and outside a uniformly charged solid sphere of total charge QQ and radius RR.

Solution:

Outside (r>Rr > R): Gaussian sphere of radius rr encloses all charge QQ. E(4πr2)=Q/ϵ0E=Q4πϵ0r2E(4\pi r^2) = Q/\epsilon_0 \quad \Rightarrow \quad E = \frac{Q}{4\pi\epsilon_0 r^2}

Inside (r<Rr < R): Enclosed charge Qenc=Q(r/R)3Q_{\text{enc}} = Q(r/R)^3. E(4πr2)=Q(r/R)3ϵ0E=Qr4πϵ0R3E(4\pi r^2) = \frac{Q(r/R)^3}{\epsilon_0} \quad \Rightarrow \quad E = \frac{Qr}{4\pi\epsilon_0 R^3}

Inside, ErE \propto r; outside, E1/r2E \propto 1/r^2.

Practice Questions

  1. 1. Find the electric field at a distance rr from an infinite line of charge with linear charge density λ\lambda using Gauss's law.

    Cylindrical Gaussian surface of length LL and radius rr: E(2πrL)=λL/ϵ0E(2\pi r L) = \lambda L/\epsilon_0. E=λ/(2πϵ0r)E = \lambda/(2\pi\epsilon_0 r).

    2. A conducting sphere of radius RR carries charge QQ. What is the field at the surface?

    E=Q/(4πϵ0R2)=σ/ϵ0E = Q/(4\pi\epsilon_0 R^2) = \sigma/\epsilon_0 (since all charge is on the surface).

    3. Set up the integral for the electric field on the axis of a uniformly charged ring of radius RR and total charge QQ.

    By symmetry, only the xx-component survives. dEx=kdq(x2+R2)xx2+R2dE_x = \frac{k\,dq}{(x^2+R^2)} \cdot \frac{x}{\sqrt{x^2+R^2}}. Integrating over the ring: Ex=kQx/(x2+R2)3/2E_x = kQx/(x^2+R^2)^{3/2}.

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Summary

  • Electric field from continuous distributions: E=kdq/r2r^\vec{E} = k\int dq/r^2\,\hat{r}.
  • Gauss's law: EdA=Qenc/ϵ0\oint \vec{E} \cdot d\vec{A} = Q_{\text{enc}}/\epsilon_0 — powerful for symmetric distributions.
  • Inside a conductor: E=0E = 0; charge resides on the surface.
  • Know standard results for sphere, cylinder (line), and plane.

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