Review / Chapter 1 — Electrostatics

Potential of Continuous Distributions

Chapter 1 — Electrostatics

Because potential is a scalar, computing \(V\) from a charge distribution is almost always easier than computing \(\vec E\) directly — no component-by-component bookkeeping, no cancellations to track. Once you have \(V(\vec r)\), you can get the field (if you need it) via \(\vec E = -\nabla V\).

Master formula

For any charge distribution (volume, surface, or line),

\[V(\vec r) = \frac{1}{4\pi\epsilon_0}\int \frac{dq'}{|\vec r - \vec r\,'|}.\]

With the convention \(V(\infty) = 0\). For volume, \(dq' = \rho(\vec r\,')\,dV'\); for surface, \(dq' = \sigma\,dA'\); for line, \(dq' = \lambda\,d\ell'\).

Two strategies

Direct integration: Plug in \(dq\) and compute. Works whenever you can write down the integral in coordinates aligned with the distribution's symmetry.

Field-first, then line integral: If you already know \(\vec E\) from Gauss's law (sphere, line, plane), use

\[V(\vec r) = V(\vec r_0) - \int_{\vec r_0}^{\vec r} \vec E\cdot d\vec \ell.\]

Pick a convenient reference point and integrate along a convenient path.

Worked examples

Ring of charge (radius \(R\), total \(Q\)): on the axis at height \(z\), every element is at the same distance \(\sqrt{R^2 + z^2}\), so

\[V(z) = \frac{1}{4\pi\epsilon_0}\frac{Q}{\sqrt{R^2 + z^2}}.\]

Spherical shell of radius \(R\), total \(Q\): integrating \(\vec E\) inward from infinity,

\[V(r) = \begin{cases}\dfrac{kQ}{r} & r \ge R,\\ \dfrac{kQ}{R} & r < R.\end{cases}\]

\(V\) is constant inside (since \(\vec E = 0\)) and continuous at the surface.

Uniform solid ball of radius \(R\), total \(Q\): outside, same as point charge. Inside, integrate \(E = kQr/R^3\) from \(R\) to \(r\):

\[V(r) = \frac{kQ}{2R}\left(3 - \frac{r^2}{R^2}\right),\quad r < R.\]

Max at the center: \(V(0) = \tfrac{3}{2}kQ/R\).

Disks, rods, and other finite shapes

For finite geometries (rod of length \(L\), disk of radius \(R\)), \(V\) along a symmetry axis typically integrates cleanly:

\[V_{\text{disk}}(z) = \frac{\sigma}{2\epsilon_0}\left(\sqrt{R^2+z^2}-|z|\right)\]

along the axis of a uniformly charged disk. Take \(R\to\infty\) to recover the (reference-dependent) linear result for an infinite plane. Differentiating with respect to \(z\) reproduces the disk's axial \(E\) — a nice consistency check.

Practice Problems

Problem 1easy
Find \(V\) at the center of a ring of radius \(R\) with total charge \(Q\).
Hint
All of the ring is at distance \(R\) from the center.
Solution

\(V = kQ/R\). Note \(\vec E = 0\) at the same point.

Answer: \(V = kQ/R\).

Problem 2medium
A rod of length \(L\) with uniform linear density \(\lambda\) lies on the \(x\)-axis from \(0\) to \(L\). Find \(V\) at a point \((x_0, 0)\) with \(x_0 > L\).
Hint
\(dV = k\lambda\,dx'/(x_0 - x')\).
Solution

\(V = k\lambda \int_0^L \frac{dx'}{x_0-x'} = k\lambda \ln\frac{x_0}{x_0 - L}\).

Answer: \(V = k\lambda \ln\dfrac{x_0}{x_0-L}\).

Problem 3medium
Find \(V(r)\) outside a uniformly charged sphere of radius \(R\) and total charge \(Q\), by line-integrating \(\vec E\) from infinity.
Hint
\(V(r) = -\int_\infty^r E(r')\,dr'\).
Solution

\(V(r) = -\int_\infty^r kQ/r'^2\,dr' = kQ/r\).

Answer: \(V(r) = kQ/r\) for \(r > R\). Same as a point charge.

Problem 4hard
Find \(V\) on the axis of a uniformly charged disk of radius \(R\), surface density \(\sigma\), at height \(z > 0\).
Hint
Slice into rings of radius \(r\) and thickness \(dr\): \(dV = k(2\pi\sigma r\,dr)/\sqrt{r^2+z^2}\).
Solution

\(V = 2\pi k \sigma \int_0^R \frac{r\,dr}{\sqrt{r^2+z^2}} = 2\pi k\sigma[\sqrt{r^2+z^2}]_0^R = \frac{\sigma}{2\epsilon_0}(\sqrt{R^2+z^2} - z)\).

Answer: \(V(z) = \dfrac{\sigma}{2\epsilon_0}(\sqrt{R^2+z^2} - z)\). Differentiating gives the disk's axial \(E\).