A parallel-plate capacitor has fixed plate area and initial separation
PHYSICS · Subjective · CBSE · Physics · Electrostatic Potential and Capacitance · Electrostatic Potential and Capacitance
Problem
A parallel-plate capacitor has fixed plate area A and initial separation d. It is first fully charged through a battery and then the circuit is opened.
(i) Explain qualitatively how its capacitance depends on (a) the plate separation and (b) the dielectric constant of the medium between the plates.
(ii) After the circuit is opened, a dielectric slab of relative permittivity κ is slowly inserted so that it completely fills the region between the plates. State with reasons what happens to the charge on the plates, the potential difference across the plates and the energy stored in the capacitor.
(iii) Repeat part (ii) for the case in which the battery remains connected to the capacitor while the slab is inserted.
(iv) Compare the energy changes obtained in parts (ii) and (iii) and explain the physical origin of the difference in the two cases.