A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential difference V = 12 V between its plates. The charging battery is now disconnected and a porcelin plate with k = 7 is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of _____________ pJ. (Assume no friction)

Solution

Physics sir · 4 steps

Battery off means charge is locked in, so the energy the capacitor loses becomes the slab's kinetic energy.

1. Charge stays fixed. With the battery disconnected, no charge can flow on or off the plates. So Q = CV = 14 × 12 = 168 pC stays the same throughout.
2. Energy before and after. Before insertion, Uᵢ = Q²/2C = 1/2CV² = 1/2 × 14 × 144 = 1008 pJ. With the slab in, the capacitance becomes kC. At fixed Q the energy drops to U_f = Q²/2kC = 1008/7 = 144 pJ.

   Δ U = Uᵢ - U_f = 1/2CV²(1 - 1/k) = 1008 × 6/7 = 864 pJ

3. Where it goes. The fringing field pulls the slab in, and the lost field energy turns into the slab's kinetic energy. The slab overshoots, gets pulled back, and with no friction it keeps oscillating with this energy.
4. So. The slab oscillates with a constant mechanical energy of 864 pJ.

The common slip: 144. Reporting the energy left in the capacitor after insertion gives 144 pJ. The slab gets the energy that was lost, which is 1008 - 144 = 864 pJ.

The idea to keep. Battery disconnected: use U = Q²/2C with Q fixed, and inserting a dielectric lowers the energy. Battery connected: V is fixed instead, so always check which one is held constant.

Answer from the NTA final answer key. Solution by PrepWiser.