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+### Statement
+
+$8.2.11.$ [Insert the problem statement]
+
+### 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:
+
+$$Q = Q_0 \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$$
+
+#### Answer
+
+a. $I = \frac{Q_0}{\varepsilon_0\rho}$
+
+b. $Q = Q_0 \exp\left(-\frac{t}{\varepsilon_0\rho}\right)$
+
+#### Answer
+
+[Insert a concise answer or boxed result]