Question

A parallel-plate capacitor whose capacitance is C = 13:5pF has a potential difference of V = 12:5V between its plates. The charging battery is now disconnected and a porcelain slab with dielectric constant k = 6:50 is slipped between the plates. What is the potential energy of the device before and Èafter the slab is introduced?

Answer 1

Answer:

508J

Explanation:

Intial energy of capacitor

U

i

=

2

1

c

c

2

v

2

=

2

1

×

12

120×120

=600 pJ

Since battery is disconnected so charge remain same.

Final energy of capacitor

U

f

=

2

1

Kc

c

2

v

2

=

2

1

×

12×6.5

120×120

=92

W+U

f

=U

i

W=508J