Question

(a) How is the electric field due to a charged parallel plate capacitor affected when a dielectric slab is inserted between the plates fully occupying the intervening region? 

(b) A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness 1/2d, where d is the separation between the plates. Find the expression for the capacitance when the slab is inserted between the plates.

Answer 1

(a) Initial electric field between the plates,

After introduction of dielectric; the permittivity of medium becomes Kε₀;

so final electric field between the plates, E = q/AKε₀ = E₀/K

i.e., electric field reduces to 1/K times.

(b) Consider a parallel plate capacitor, area of each plate being A, the separation between the plates being d. Let a dielectric slab of dielectric constant K and thickness t < d be placed between the plates. The thickness of air between the plates is (d -t ). If charges on plates are + Q and - Q, then surface charge density

σ = Q/A

The electric field between the plates in air, E = σ/ε₀ = Q/ε₀A

The electric field between the plates in slab,  E2 =  σ/Kε₀ = Q/Kε₀A

∴ The potential difference between the plates 

V_AB = work done in carrying unit positive charge from one plate to another 

= ∑Ex (as field between the plates is not constant).