Question

A parallel plate capacitor is charged by a battery and 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 the
(i) potential difference between the plates
(ii) electric field between the plates
(ii) energy stored in the capacitor

Answer 1

$V\to constant$ because battery nemans connected.

$E=\frac{E_0}{4}\to k$ decreases

$3.U=\frac{1}{2}C{V}^2$

$U$ is same but $C$ become $kC$

So, $U=\frac{1}{2}C{V}^2\cdot k(k$ increases)

i.e. Energy store $\uparrow$.