A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential difference V = 12V 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)

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Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Initial charge Q = CV = 14 × 10? ¹² × 12 = 168 × 10? ¹² C

Initial energy U_in = ½ CV² = ½ (14 × 10? ¹²) × 12² = 1008 pJ

When the battery is disconnected and a dielectric (k=7) is inserted, the new capacitance is C' = kC.

The charge Q remains constant.

Final energy U_f = Q²/2C' = Q²/ (2kC) = (CV)²/ (2kC) = CV²/ (2k)

U_f = (14 × 10? ¹² × 12²) / (2 × 7) = 144 pJ

Mechanical energy available for oscillation

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