A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to potential difference of 12.0 V. The battery is now disconnected and a dielectric slab ( r = 6) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is 10-12 J.
Answer
750
Solution
U = Q22C with Q constant
battery is disconnected, Q stays the same and Cf = r C = 6C Uf = U6
U = 12CV2 = 12 12.5 10-12 (12)2 = 900 10-12 J
Uf = 150 10-12 J
| U| = (900 - 150) 10-12 = 750 10-12 J (energy decreases)
Answer: 750