Question

The capacitance of a parallel plate capacitor is C₀ when the plates has air between them. This region is now filled with a dielectric slab of dielectric constant K and capacitor is connected with battery of EMF E and zero internal resistance. Now slab is taken out, then during the removal of slab :

(A) charge EC₀ (K – 1) flows through the cell 

(B) energy E²C₀ (K – 1) is absorbed by the cell 

(C) the energy stored in the capacitor is reduced by E₂C₀ (K – 1) 

(D) the external agent has to do E²C₀ (K – 1) amount of work to take out the slab

Answer 1

Correct option (A,B,)

Explanation: 

Initial charge on capacitor = KC_oE 

Charge after removing slab = C_oE 

Amount of charge flows through the cell = KC_oE – C_oE = C_oE (K – 1) 

Energy absorbed by cell = C₀E (K – 1). E₀ = C₀E² (K – 1) 

Initial energy stored in capacitor = 1/2 kC₀ E² 

Final energy stored in capacitor = 1/2 C₀ E² 

Energy reduces in capacitor by = 1/2 C₀ E₂ (k – 1)

Work done by external agent