Question

A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab of dielectric constant K is inserted in between the plates. How would i) The capacitance ii) The electric field between the plates and iii) The energy stored in the capacitor be affected? Give reasons with necessary equations.

Answer 1

Let the area of the plates be A, and distance between them be d

Then capacitance before the insertion of the dielectric slab will be,

C = ε_oA/d

capacitance after the insertion of the dielectric slab will be,

C’ = ε_oKA/d

1. Let the battery charge the plate with charge Q. The electric field between the plates will be given by

Before the insertion of the dielectric slab

E = σ/2ε_o = Q/2ε_oA

After the insertion of the dielectric slab

E’ = σ/2ε_oK = Q/(2ε_oKA)

1. Energy stored in the capacitor is given by

Initial energy before insertion of dielectric

E = ½CV²

V = Q/C

=> E = ½C(Q/C)²

=> E = ½Q²/C

Again C = ε_oA/d

=> E = ½Q²/ (ε_oA/d) = ½Q²d/ ε_oA

Final energy after the insertion of dielectric

E’ = ½Q²d/ (ε_oKA)

Since K > 1 thus E’ < E, we see that energy of the capacitor decreases.