JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Capacitance

A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness

\frac{3}{4} d

, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :
(Given

C_{o} =

capacitance of capacitor with air as medium between plates.)

Answer: (A)

Solution

The dielectric slab of thickness

3 d /4

leaves an air gap of

d /4

. The potential difference is

Δ V = E (d /4) + (E / K) (3 d /4) = E d (1/4 + 3/ (4 K)) = E d (K + 3) / (4 K)

. Since

E = σ / ϵ_{0} = Q / (A ϵ_{0})

, we get

Δ V = (Q d / (A ϵ_{0})) (K + 3) / (4 K)

. Therefore

C = Q /Δ V = (A ϵ_{0} / d) \times 4 K / (K + 3) = 4 K C_{0} / (K + 3)

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option