JEE PYQ: Capacitance Question 36
Question 36 - 2019 (10 Jan Shift 2)
A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is:
(1) 692 pJ
(2) 508 pJ
(3) 560 pJ
(4) 600 pJ
Show Answer
Answer: (2)
Solution
$W = -\Delta U = (-1)\left[\frac{(ce)^2}{2kc} - \frac{(ce)^2}{2c}\right] = \frac{e^2c}{2} \cdot \frac{k-1}{k} = \frac{(10)^2 \times 12}{2} \times \frac{5.5}{6.5} = 508$ pJ.
BETA
