JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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)
- A \(872\)
- B \(972\)
- C \(784\)
- D \(864\)
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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