Question

A parallel plate capacitor of capacitance c area A and
plate separation d is charged to a charge Q so that
the potential difference between an plates is V.
1. What will be the maghitude and direction of the electric field
inside the capacitor?
11) If a dielectric medium of dielectric constant k is introduced
between the plates of the capacitor keeping the source
connected, what will be the effect on the capacitance and
the potential difference between the plates of capacitor​

Answer 1

after inserting slab

newcapacitanceC

′

=KC=

d

Kϵ

o

A

newpotentialdifferenceV

′

=

K

V

newelectricfieldE

′

=

d

V

′

=

kd

V

newchargeQ

′

C

′

=

d

ϵ

o

AV

work=final−initial

=

2

1

C

′

V

′

−

2

1

CV

2

2

1

(KC)=(1−

K

1

)

[ω]=

2d

ϵ

o

AV

2

(1− K/1 )