Question

A parallel plate is charged by a battery. When the battery remains connected, a dielectric slab is inserted in the space between the plates. Explain what changes if any, occur in the values of

(i) Potential difference between the plates

(ii) Capacitance 

(iii) Charge on the plates

(iv) Energy stored in the capacitor?

(v) Electric field strength between the plates

Answer 1

(i) When battery remains connected, the potential difference remains the same.

(ii) The capacitance of capacitor increases as K > 1.

(iii) The charge Q = CV, V = same, C = increases; there, charge on plates increases.

(iv) Energy stored by capacitor U = \(\cfrac12\)CV², also increases.

(v) As electric field E = \(\cfrac Vd,\) V = constant and d = constant; therefore, electric field strength remains the same.