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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2021
  • Medium
  • Numerical

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)

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Answer & Solution

Correct Answer

(D) \(864\)

Step-by-step Solution

Detailed explanation

\(U _{ i }=\frac{1}{2} \times 14 \times 12 \times 12 pJ \quad\left(\because U =\frac{1}{2} CV ^{2}\right)\)

\(=1008 pJ\)

\(U _{ f }=\frac{1008}{7} pJ =144 pJ \quad\left(\because C _{ m }= k C _{0}\right)\)

Mechanical energy \(=\Delta U\)

\(=1008-144\)

\(=864 pJ\)

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