(a) Maximum Current within the Dielectric

The system consists of an air gap of thickness $d$ and a leaky dielectric slab of thickness $d$, area $A$, resistivity $\rho$, and relative permittivity $\varepsilon_r$. The electrical model is an ideal air capacitor $C_a$ in series with a leaky capacitor (modeled as $C_d$ in parallel with a resistor $R_d$).

The maximum current flows immediately after the switch is closed ($t=0$). At this instant, the capacitors are uncharged, and the voltage distribution is determined by the inverse ratio of their capacitances (capacitive divider action).

Capacitances:

$$ C_a = \frac{\varepsilon_0 A}{d} \quad , \quad C_d = \frac{\varepsilon_0 \varepsilon_r A}{d} $$

Since the same charge $Q$ accumulates on both initially:

$$ V = \frac{Q}{C_a} + \frac{Q}{C_d} \implies Q = \frac{V}{\frac{1}{C_a} + \frac{1}{C_d}} = \frac{V C_a C_d}{C_a + C_d} $$

The initial electric field in the dielectric $E_d(0)$ is given by $V_d / d$, where $V_d = Q/C_d$.

$$ V_d = \frac{Q}{C_d} = V \frac{C_a}{C_a + C_d} = V \frac{\frac{\varepsilon_0 A}{d}}{\frac{\varepsilon_0 A}{d} + \frac{\varepsilon_0 \varepsilon_r A}{d}} = V \frac{1}{1 + \varepsilon_r} $$

Using Ohm’s law in point form $J = E / \rho$, the current $I$ is:

$$ I = J A = \frac{E_d}{\rho} A = \frac{V_d / d}{\rho} A = \frac{V_d A}{\rho d} $$

Substituting $V_d$:

(b) Initial and Final Charges

Initial Charge ($t=0$):

The initial free charge on the plates $Q_i$ is determined by the equivalent capacitance of the series combination.

$$ Q_i = V C_{eq} = V \frac{C_a C_d}{C_a + C_d} = V \frac{(\frac{\varepsilon_0 A}{d}) (\frac{\varepsilon_0 \varepsilon_r A}{d})}{\frac{\varepsilon_0 A}{d} (1 + \varepsilon_r)} $$

Final Charge ($t \to \infty$):

At steady state, the leaky dielectric acts as a resistor. Since the air capacitor blocks DC current, the steady-state current in the circuit is zero ($I=0$). Consequently, there is no voltage drop across the resistive dielectric slab. The entire potential $V$ appears across the air gap.

Charge on the capacitor plates depends on the voltage across the air gap ($V$):

$$ Q_{final} = C_a V = \frac{\varepsilon_0 A V}{d} $$

Charge on the faces of the dielectric:
At the interface between the air and the dielectric, free charge accumulates because current flowed through the resistive slab but could not cross the air gap. Using Gauss’s Law at the interface (where $E_{air} = V/d$ and $E_{slab} = 0$):

$$ E_{air} – E_{slab} = \frac{\sigma_{net}}{\varepsilon_0} \implies \frac{V}{d} – 0 = \frac{Q_{face}}{\varepsilon_0 A} $$

(c) Time Constant

The time constant $\tau$ is found by determining the Thevenin resistance and equivalent capacitance seen by the relaxing component. We look at the circuit from the perspective of the resistance $R$ of the slab.

Shorting the voltage source $V$ connects the air capacitor $C_a$ in parallel with the dielectric capacitor $C_d$.

$$ C_{total} = C_a + C_d = \frac{\varepsilon_0 A}{d} + \frac{\varepsilon_0 \varepsilon_r A}{d} = \frac{\varepsilon_0 A}{d} (1 + \varepsilon_r) $$

The resistance of the slab is $R = \frac{\rho d}{A}$. The time constant is:

$$ \tau = R C_{total} = \left( \frac{\rho d}{A} \right) \left( \frac{\varepsilon_0 A}{d} (1 + \varepsilon_r) \right) $$