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QJEEM21Phb1c2: JEE Main 2021 Physics - Class 12 JEEM21Phb1c2 Solution: A parallel plate capacitor whose capacitance C is 14 pF is…

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)

Step-by-Step Verified Solution

Ui=12cv2

=12×14×(12)2 pJ

= 1008 pJ

Uf=Q22kC

=(14×12)22×7×14

= 144 pJ

oscillating energy = Ui − Uf

= 1008 − 144

= 864 pJ

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