Physics

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

(A)

F = K^2 F_0

(B)

F = K F_0

(C)

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

(D)

F = F_0

Intuitive explanation

The key insight is that the battery stays connected, so the voltage across the plates remains constant at $V$ . When you insert a dielectric, the capacitance increases by a factor of $K$ , which means more charge flows onto the plates from the battery. Since the attractive force between plates depends on the square of the charge (or equivalently, the square of the electric field), and the field strength increases by factor $K$ while voltage stays fixed, the force scales as $K^2$ .

Intuitive explanation

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