Question

A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, the electric field, the charge stored, and voltage will increase, decrease, or remain constant.

Answer 1

Consider a parallel-plate capacitor of a plate area A and a plate separation d that is filled with a dielectric of relative permittivity (dielectric constant) k.

Its capacitance is C = `("k"ε_0"A")/"d"`   ...(1)

The charge on its plates is Q = CV if it is charged to a voltage (potential) of V.

Because the battery is disconnected after charging, the charge Q on its plates, and thus the product CV, remain constant.

When the dielectric is fully removed, the capacitance becomes Eq.(1)

C' = `(ε_0"A")/"d" = 1/"k" "C"`   ...(2)

that is, its capacitance decreases by the factor k. Since C'V' = CV, hence the new voltage is

V' = `"C"/"C'" "V"` = kV   ...(3)

so that its voltage increases by the factor k.

The stored potential energy, u = `1/2` QV, so that while Q remains constant, u increases by factor k. The electric field, E = V/d, increases by a factor of k.