Solution

Statement

$8.2.11.$ An excess charge $Q_0$ is placed at the center of a conducting ball with resistivity $\rho$.

**a.** How does the current flowing from the center of the ball to its surface depend on $Q_0$?
**b.** How will the charge at the center of the ball change over time?

Solution

**a.** Inside our conducting ball of radius $R$, let us draw a sphere of radius $r$ ($r < R$). The area of this sphere is $S = 4\pi r^2$. Let us write Gauss's law:

$$E(r) \cdot 4\pi r^2 = \frac{Q_0}{\varepsilon_0} \implies E(r) = \frac{Q_0}{4\pi\varepsilon_0 r^2}$$

Ohm's law in differential form:

$$j(r) = \frac{1}{\rho} E(r)$$
$$j(r) = \frac{1}{\rho} \cdot \frac{Q_0}{4\pi\varepsilon_0 r^2}$$

where $j(r)$ is the current density.

The total current through the sphere of radius $r$:

$$I = j(r) \cdot 4\pi r^2 = \frac{1}{\rho} \cdot \frac{Q_0}{4\pi\varepsilon_0 r^2} \cdot 4\pi r^2 = \frac{Q_0}{\varepsilon_0\rho}$$

**b.** Inside a conductor, the charge cannot remain concentrated: it will begin to flow to the surface, causing a current directed from the center to the surface. Using the result of subproblem **a**, we can write:

$$I(t) = \frac{Q(t)}{\varepsilon_0\rho}$$

But the current is the rate of decrease of charge:

$$I(t) = -\frac{dQ}{dt}$$

Let us substitute:

$$-\frac{dQ}{dt} = \frac{Q}{\varepsilon_0\rho} \implies \frac{dQ}{Q} = -\frac{dt}{\varepsilon_0\rho}$$

Integrating and then exponentiating:

$$Q(t) = C \cdot \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$$

From the initial condition $Q(0) = Q_0$, we obtain that $C = Q_0$. Then:

$$Q = Q_0 \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$$