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