Question

the plates of a 0.6uf uncharged capacitor are connected to a 10v battery. the charge on the capacitor after a long time is

Answer 1

Answer:

Initial energy:

U

1

=

2

1

CV

2

=

2

1

×100×10

−6

×(50)

2

=12.5×10

−2

J

When the battery is kept connected, voltage across the capacitor does not change.

We know: C=

d

ϵ

0

A

=100 μF

And when distance is halved:

C

′

=

d

2ϵ

0

A

=200 μF

So, the final energy:

U

2

=

2

1

C

′

V

2

=

2

1

×200×10

−6

×(50)

2

=25×10

−2

J

∴ additional energy given = U

2

−U

1

=(25×10

−2

−12.5×10

−2

) J=12.5×10

−2

J