A parallel plate capacitor of capacity is connected to a battery of emf

PHYSICS · Single correct · CBSE · Physics · Electrostatic Potential and Capacitance · Electrostatic Potential and Capacitance

Problem

A parallel plate capacitor of capacity

C

is connected to a battery of emf

V

. The force between the plates is

F_0

. Now a dielectric slab of dielectric constant

K

is inserted between the plates. The new force between the plates becomes

F

. The relation between

F

and

F_0

is (A)

F = K^2 F_0

(B)

F = K F_0

(C)

F = \frac{F_0}{K^2}

(D)

F = F_0