The battery remains connected to a parallel plate capacitor and a dielectric slab is inserted between the plates. What will be effect on its (i) potential difference, (ii) capacity, (iii) electric field, and (iv) energy stored?

Question

Philoid · Accepted Answer

Capacitance is given by

(i) After having a dielectric slab

K is the dielectric constant

(ii) If K>1 then Capacitance C will increase

If C increases the charge on the capacitor plates also increase.

(iii) When the battery remains connected voltage V is always constant

(iv) Here Electric field E is decreased by the factor of 1/C

Electric energy of the capacitor is given by

(v) Voltage V is constant.

If Capacitance increases, then energy will also increase.