Irodov 3.53: Parabolic Potential Means Uniform Charge

Problem Irodov 3.53

Inside a charged ball the electrostatic potential depends only on the distance $r$ from the centre, as $\varphi = ar^2 + b$ with $a$ and $b$ constants. Find the volume charge density $\rho(r)$ inside the ball.

The charged ball with its centre O and a point inside at distance r from it, where the potential is ar^2 + b. — The charged ball with its centre $O$ and a point inside at distance $r$ from it, where the potential is $ar^2 + b$ .

Given

$\varphi = ar^2 + b$ inside the ball
$a$ , $b$ constants

Find

the charge density

\rho(r)

inside

Solution

Step 1The field inside the ball

The potential depends on $r$ alone, so the field is radial, and its radial component is minus the slope of $\varphi$ :

E_r = -\frac{d\varphi}{dr} = -\frac{d}{dr}\left(ar^2 + b\right) = -2ar \tag{1}

For $a > 0$ the field points toward the centre and grows in proportion to $r$ ; the constant $b$ has no effect on it.

Step 2The charge from the field

Gauss’s law in differential form, $\nabla\cdot\vect E = \rho/\varepsilon_0$ , for a radial field reads

\rho = \frac{\varepsilon_0}{r^2}\,\frac{d}{dr}\left(r^2 E_r\right) \tag{2}

Putting (1) into (2):

\begin{aligned} \rho &= \frac{\varepsilon_0}{r^2}\,\frac{d}{dr}\left(-2ar^3\right)\\ &= \frac{\varepsilon_0}{r^2}\left(-6ar^2\right) \end{aligned}

\rho = -6\varepsilon_0 a

(3)

The density is the same at every point: the ball is uniformly charged, negatively for $a > 0$ .

Check

The integral form of Gauss’s law on a sphere of radius $r$ gives the same density. The flux is $4\pi r^2 E_r = -8\pi a r^3$ , so the enclosed charge is $q(r) = -8\pi\varepsilon_0 a r^3$ , and dividing by the volume $4\pi r^3/3$ gives $-6\varepsilon_0 a$ , the same at every $r$ , as (3) says. Units: $a$ is volts per square metre, so $\varepsilon_0 a$ is $(\text{C}/\text{V m})(\text{V}/\text{m}^2) = \text{C}/\text{m}^3$ .

Remark

The converse is the familiar result: a uniformly charged ball has a field growing linearly from its centre and a potential that is parabolic in $r$ . The constant $b$ never enters $\rho$ ; it records where the zero of potential is. For an isolated ball of radius $R$ with $\varphi = 0$ at infinity, the potential outside is $q(R)/(4\pi\varepsilon_0 r)$ , which at the surface is $-2aR^2$ , and matching $aR^2 + b$ to it there gives $b = -3aR^2$ .