Question

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)560pJ
2)508PJ
3)692PJ
4)600pJ

Answer 1

Answer:

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

Answer 2

Step-by-step explanation:

option 2 is crct one......