A parallel plate capacitor whose capacitance C is 14 pF is charged by… | JEE Main 2021 (17 Mar, Shift 1)

Q78JEE Main 2021 · 17 MAR · SHIFT 1Medium

Dielectrics

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 porcelain 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)

Numerical answer type. Work it out, then open the answer.

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Answer: 864

Battery removed, $Q$ constant. Energy lost by the field becomes the slab's mechanical energy.
$U_i=\frac{1}{2}CV^2=\frac{1}{2}(14)(12)^2=1008\text{ pJ}$ , $U_f=\frac{U_i}{k}=144\text{ pJ}$
Mechanical energy $=1008-144=864\text{ pJ}$

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