Q for capacitor

Dear student
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.

Regards