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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; the plate would then oscillate back and forth between the plates with a constant mechanical energy of ______ pJ. (Assume no friction.)

Asked in JEE Main 17th March 1st Shift 2021 · Battery connected or removed

Answer: 864

Step-by-step solution

Energy stored before insertion: Uᵢ=1/2CV²=1/2×14×10⁻¹²×(12)²=1008 pJ.

With the battery disconnected the charge is fixed, so use U=(Q²)/(2C): fully inserting the plate raises C to KC and drops the energy to U_f=(Uᵢ)/K=(1008)/7=144 pJ.

The difference is the work the field does pulling the plate in, and with no friction it becomes kinetic energy of the plate.

Mechanical energy =Uᵢ-U_f=1008-144=864 pJ.

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