JEE PYQ: Capacitance Question 4
Question 4 - 2021 (17 Mar Shift 1)
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, then the plate would oscillate back and forth between the plates with a constant mechanical energy of ____pJ
(Assume no friction)
Show Answer
Answer: 864
Solution
$U_i = \frac{1}{2} \times 14 \times 12^2 \times 12$ pJ $= 1008$ pJ. $U_f = \frac{1008}{7}$ pJ $= 144$ pJ ($\because C_m = kC_0$). Mechanical energy $= \Delta U = 1008 - 144 = 864$ pJ.
BETA
