Solution to Question 41
Physics Principle: Electrostatic Induction and Energy Method.
Analysis:
We have a system where a central plate C (carrying a fixed volume charge \(Q\)) is placed between two grounded parallel plates A and B. The system acts effectively as two capacitors in parallel (since A and B are at the same potential, \(V=0\)) connected to the central node C.
Let the plate C be shifted by a distance \(x\) towards plate A.
Distance CA = \(d – x\)
Distance CB = \(d + x\)
Step 1: Calculate the Potential Energy
This configuration is equivalent to two stored energy components (left and right gaps). Whether C is a conductor or a uniform dielectric slab with fixed charge, the force can be derived from the energy of the fields.
The total energy \(U\) stored in the electric fields is given by the standard formula for a charge \(Q\) between two grounded plates: $$U(x) = \frac{Q^2}{2 C_{eq}}$$ However, for a fixed charge distribution (dielectric), we can also integrate the field energy density. The result for the force is identical to the conductor case. The equivalent capacitance \(C_{eq}\) of the two gaps in parallel is: $$C_{eq} = \frac{\epsilon_0 A}{d-x} + \frac{\epsilon_0 A}{d+x} = \epsilon_0 A \left( \frac{d+x + d-x}{d^2 – x^2} \right) = \frac{2d \epsilon_0 A}{d^2 – x^2}$$
Substituting this into the energy equation: $$U(x) = \frac{Q^2}{2 \left( \frac{2d \epsilon_0 A}{d^2 – x^2} \right)} = \frac{Q^2 (d^2 – x^2)}{4 d \epsilon_0 A}$$
Step 2: Calculate the Force
The electrostatic force \(F\) is the negative gradient of the potential energy with respect to the displacement \(x\): $$F = – \frac{dU}{dx}$$ $$F = – \frac{Q^2}{4 d \epsilon_0 A} \frac{d}{dx}(d^2 – x^2)$$ $$F = – \frac{Q^2}{4 d \epsilon_0 A} (-2x) = \frac{Q^2 x}{2 d \epsilon_0 A}$$
Since the derivative \(\frac{dU}{dx}\) is negative (energy decreases as x increases? No, energy is \( \propto d^2 – x^2 \), so energy decreases as x increases. The force acts in the direction that lowers potential energy. Shifting towards a plate reduces the system energy, so the force is attractive towards the nearer plate.