Question

8. The capacitance of a parallel plate capacitors
60 MF. If the distance betweon the plates is
tripled and the area doubled then the new
capacitance will be​

Answer 1

Answer:

Let the distance between the plates initially be d.

Thus initial capacitance C

1

=ϵ

o

A/d=8 pF

Substance of dielectric constant K=6 is inserted between the plates.

Also, now the distance between the plates is d

′

=0.5d

Thus new capacitance C

2

=Kϵ

o

A/d

′

∴ C

2

=6ϵ

o

A/0.5d=12C

1

=96 pF