Question

A parallel plate capacitor of area A and separation d is charged to potential difference V and removed from the charging source. A dielectric slab of constant K = 2, thickness d and area A/2  is inserted, as shown in the figure. Let σ₁ be free charge density at the conductor-dielectric surface and σ₂ be the charge density at the conductor-vacuum surface.

(A) The electric field have the same value inside the dielectric as in the free space between the plates.

(B) The ratio σ₁/σ₂ 

(C) The new capacitance is 3∈₀A/2d

(D) The new potential difference is 2/3 V

Answer 1

Correct option (A,B,C,D)

Explanation: 

Potential for each plate remain same over whole area. If potential difference between them is, say V' then V' = Ed i.e. E is also same inside the plates. To keep E same, free charge density is changed i.e. charge redistributes itself. To find new capacitance, two capacitors can be taken as connected in parallel. Then

By Q = CV, as Q remains unchanged V is changed to 2/3 V.