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.

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 = εoA/d

capacitance after the insertion of the dielectric slab will be,

C’ = εoKA/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/2εo = Q/2εoA

After the insertion of the dielectric slab

E’ = σ/2εoK = Q/(2εoKA)

  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 = εoA/d

=> E = ½Q2/ (εoA/d) = ½Q2d/ εoA

Final energy after the insertion of dielectric

  E’ = ½Q2d/ (εoKA)

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

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