Question:

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)

Show Hint

With the battery disconnected the charge stays constant, so the stored energy falls from \(\frac{Q^2}{2C}\) to \(\frac{Q^2}{2kC}\); the loss appears as mechanical energy of the plate.
Updated On: Nov 4, 2026
Show Solution
collegedunia
Verified By Collegedunia

Correct Answer: 864

Solution and Explanation

Initial energy: \(U_i=\frac{1}{2}CV^2=\frac{1}{2}(14)(12)^2=1008\ \mathrm{pJ}\).
The battery is disconnected, so the charge \(Q=CV\) stays constant.
With the dielectric (k = 7) fully inside, the capacitance becomes \(kC\), so \(U_f=\frac{Q^2}{2kC}=\frac{U_i}{k}=\frac{1008}{7}=144\ \mathrm{pJ}\).
The plate is pulled in by the field and, with no friction, oscillates with constant mechanical energy equal to the energy lost by the field: \(E=U_i-U_f=1008-144=864\ \mathrm{pJ}\).
Answer: 864
Was this answer helpful?
0
0

Top JEE Main Physics Questions

View More Questions

Top JEE Main Capacitors and Capacitance Questions

View More Questions