Question

A parallel plate capacitor is charged by a battery to V potential difference, when air is between the plates. Without removing the capacitor from battery, air is removed by a dielectric medium of dielectric constant eε_r. What is the effect on following explain with reasons:
(i) Potential difference?
(ii) Electric field between plates?
(iii) Capacitance?
(iv) Charge?
(v) Energy?

Answer 1

When battery remains connected:

(i) When the battery remains connected then the potential between the plates is equal to the potential of battery. This, the potential remains constant.
(ii) Capacitance becomes K times by the relation (C_m) = KC₀
(iii) Charge will increase (according to q = CV) or it means it become K times.
(iv) Electric field

Thus no effect on the electric field.
(v) The energy of capacitor
U = \(\frac{1}{2}\)CV_m
put V = const
∴ U ∝ C
Hence the energy stored by the capacitor is also increased K times.