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?

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.