Solution
Step 1: Analyze the Initial Charge Distribution
We are given a parallel plate capacitor where initially one plate has charge $Q$ and the other is uncharged (charge $0$). When isolated plates are brought close, the charge redistributes on the inner and outer surfaces. The charge on the outer surfaces of both plates must be equal:
$$q_{outer} = \frac{Q_1 + Q_2}{2} = \frac{Q + 0}{2} = \frac{Q}{2}$$By conservation of charge, the inner surfaces will hold:
- Plate 1 Inner: $Q – Q/2 = +Q/2$
- Plate 2 Inner: $0 – Q/2 = -Q/2$
The energy stored in the electric field is localized between the inner surfaces.
Step 2: Calculate Initial Stored Energy
The energy is stored in the electric field between the plates due to the inner charges ($+Q/2$ and $-Q/2$). The outer charges create fields pointing away from the system but do not contribute to the energy stored inside the capacitor gap.
The effective charge for the capacitor energy formula is $q_{inner} = Q/2$.
$$U_{initial} = \frac{q_{inner}^2}{2C} = \frac{(Q/2)^2}{2C} = \frac{Q^2}{8C}$$Step 3: Analyze the Effect of the Resistive Slab
The slab has finite electrical resistance. It acts as a leakage path (a resistor) connected between the two inner surfaces. Because the plates are isolated (not connected to a battery), the opposite charges $+Q/2$ and $-Q/2$ on the inner faces will flow through the resistive material and neutralize each other completely.
In the final steady state, there is no separation of charge on the inner surfaces ($q_{inner, final} = 0$). Consequently, the electric field inside the slab becomes zero.
Step 4: Calculate Energy Loss
The final energy stored in the internal field is zero.
$$U_{final} = 0$$The total energy lost in the slab (dissipated as heat) is the difference between the initial and final stored energies.
$$\Delta U = U_{initial} – U_{final} = \frac{Q^2}{8C} – 0 = \frac{Q^2}{8C}$$Correct Option: (b) $\frac{Q^2}{8C}$