Question

A parallel-plate capacitor of capacitance 20 µF is fully charged by a battery of 12 V. The battery is then disconnected. A dielectric slab of K = 4 is slipped between the two plates of the capacitor:

(a) Find the change in potential energy of the capacitor.


(b) Does the potential energy increase or decrease? Explain

Answer 1

Step-by-step explanation:

capacitance = 20 x 10⁻⁶ F .

potential V = 12 V

charge = CV

= 20 x 10⁻⁶ x 12

Q = 240 x 10⁻⁶ C

energy = Q² / 2C

= (240 x 10⁻⁶ )² / 2 x 20 x 10⁻⁶

= 1440 x 10⁻⁶ J

b )

In this case charge will remain the same but capacity will be increased 4 times

new capacity C = 4 x 20 x 10⁻⁶

= 80 x 10⁻⁶

energy = Q² / 2C

= (240 x 10⁻⁶ )² / 2 x 4 x 20 x 10⁻⁶

= 360 x 10⁻⁶ J .

potential energy will decrease from 1440 x 10⁻⁶ J to 360 x 10⁻⁶ J