Solution to Question 51
1. Initial State:
We have a system of three plates. The middle plate (charge $Q$) acts as the common node, and the two outer plates are connected together. This forms two capacitors in parallel.
Capacitance of each side: $C_0 = \frac{\epsilon_0 A}{d}$.
Equivalent Capacitance: $C_{eq} = C_0 + C_0 = \frac{2\epsilon_0 A}{d}$.
Initial Stored Energy:
$$ U_i = \frac{Q^2}{2 C_{eq}} = \frac{Q^2 d}{4 \epsilon_0 A} $$
2. The Shift Process (Work Done):
The middle plate is shifted “quickly” by a distance $d/2$ towards one of the outer plates.
- Charge Distribution: Because the shift is quick and the resistance $R$ is high, charge cannot redistribute between the outer plates during the movement. The charges on the inner faces of the outer plates remain $-Q/2$ each during the shift.
- Electric Field: With equal and opposite charge densities on both sides of the middle plate, the electric fields in the top and bottom gaps are equal in magnitude and opposite in direction.
- Net Force: The net electrostatic force on the middle plate is zero. Therefore, the external work done to shift the plate is zero. The system’s potential energy remains $U_i$ immediately after the shift.
3. Final State (After Redistribution):
After the shift, charge flows through resistor $R$ until potentials equalize. We now have two capacitors with different separations $d_1 = d – d/2 = d/2$ and $d_2 = d + d/2 = 3d/2$.
New Capacitances:
$$ C_1 = \frac{\epsilon_0 A}{d/2} = \frac{2\epsilon_0 A}{d}, \quad C_2 = \frac{\epsilon_0 A}{3d/2} = \frac{2\epsilon_0 A}{3d} $$
New Equivalent Capacitance (Parallel):
$$ C’_{eq} = C_1 + C_2 = \frac{2\epsilon_0 A}{d} \left( 1 + \frac{1}{3} \right) = \frac{8\epsilon_0 A}{3d} $$
Final Stored Energy:
$$ U_f = \frac{Q^2}{2 C’_{eq}} = \frac{Q^2}{2} \cdot \frac{3d}{8\epsilon_0 A} = \frac{3 Q^2 d}{16 \epsilon_0 A} $$
4. Heat Dissipated:
The heat dissipated in the resistor is the loss in the system’s potential energy.
$$ H = U_i – U_f $$
$$ H = \frac{Q^2 d}{4 \epsilon_0 A} – \frac{3 Q^2 d}{16 \epsilon_0 A} $$
$$ H = \frac{4 Q^2 d}{16 \epsilon_0 A} – \frac{3 Q^2 d}{16 \epsilon_0 A} = \frac{Q^2 d}{16 \epsilon_0 A} $$