Solution
Let the charge on the inner surface ($r_1$) at time $t$ be $q_{in}(t)$ and the charge on the outer surface ($r_2$) be $q_{out}(t)$. Initially, the shell is neutral, so:
$$ q_{in}(0) = 0, \quad q_{out}(0) = 0 $$Since the bulk of the material remains neutral (assuming the relaxation time ensures no volume charge accumulation), charge conservation requires:
$$ q_{in}(t) + q_{out}(t) = 0 \implies q_{out}(t) = -q_{in}(t) $$Applying Gauss’s Law in the Dielectric Medium:
Consider a Gaussian spherical surface of radius $r$ where $r_1 < r < r_2$. The total free charge enclosed is the sum of the central charge $q_0$ and the charge accumulated on the inner surface $q_{in}(t)$.
$$ Q_{enclosed} = q_0 + q_{in}(t) $$The electric displacement field $\vec{D}$ is given by:
$$ \oint \vec{D} \cdot d\vec{A} = Q_{enclosed} \implies D(4\pi r^2) = q_0 + q_{in}(t) $$ $$ D = \frac{q_0 + q_{in}(t)}{4\pi r^2} $$The electric field $\vec{E}$ is related to $\vec{D}$ by the permittivity $\varepsilon = \varepsilon_r \varepsilon_0$:
$$ E = \frac{D}{\varepsilon} = \frac{q_0 + q_{in}(t)}{4\pi \varepsilon r^2} $$Current Density and Charge Flow:
Using Ohm’s law in point form $\vec{J} = \vec{E}/\rho$, the leakage current density flowing radially outward is:
$$ J = \frac{E}{\rho} = \frac{q_0 + q_{in}(t)}{4\pi \varepsilon \rho r^2} $$The total current $I$ crossing any spherical shell of radius $r$ is:
$$ I = J \cdot 4\pi r^2 = \frac{q_0 + q_{in}(t)}{\rho \varepsilon} $$This current $I$ represents the rate at which positive charge flows away from the inner surface towards the outer surface. Therefore, the charge on the inner surface decreases (becomes more negative) at this rate:
$$ \frac{dq_{in}}{dt} = -I = -\frac{q_0 + q_{in}(t)}{\rho \varepsilon} $$Solving the Differential Equation:
Let $Q(t) = q_0 + q_{in}(t)$. Then $dQ/dt = dq_{in}/dt$. The equation becomes:
$$ \frac{dQ}{dt} = -\frac{Q}{\rho \varepsilon} $$Integrating with respect to time:
$$ \int_{Q(0)}^{Q(t)} \frac{dQ}{Q} = -\int_{0}^{t} \frac{dt}{\rho \varepsilon} $$ $$ \ln\left(\frac{Q(t)}{Q(0)}\right) = -\frac{t}{\rho \varepsilon} \implies Q(t) = Q(0) e^{-\frac{t}{\rho \varepsilon}} $$At $t=0$, $q_{in}(0) = 0$, so $Q(0) = q_0$. Thus:
$$ q_0 + q_{in}(t) = q_0 e^{-\frac{t}{\rho \varepsilon}} $$ $$ q_{in}(t) = q_0 \left( e^{-\frac{t}{\rho \varepsilon}} – 1 \right) $$Charge on Outer Surface:
Using the neutrality condition $q_{out}(t) = -q_{in}(t)$:
$$ q_{out}(t) = – \left[ q_0 \left( e^{-\frac{t}{\rho \varepsilon}} – 1 \right) \right] $$ $$ q_{out}(t) = q_0 \left( 1 – e^{-\frac{t}{\rho \varepsilon}} \right) $$