Question

a slab of material of dielectric constant k has the same area A as the plates capacitor and has thickness 3/4d where d is the separation of the plates the change in capacitance when the slab is inserted between the plates is​

Answer 1

Answer:

C=

d

Aϵ

0

C

′

=

d−t+

k

t

Aϵ

0

Put t=

4

3d

;

C

′

=

3+k

4k

.

d

Aϵ

0

C

′

=

3+k

4k

.C