A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of
(i) capacitance.
(ii) potential difference between the plates.
(iii) electric field between the plates.
(iv) the energy stored in the capacitor.

Question

Toppr Admin · Accepted Answer

V-x2192-constant because battery nemans connected-E-E04-x2192-k decreases3-U-12CV2U is same but C become kCSo- U-12CV2-x22C5-k-k increases-i-e- Energy store -x2191-

V→constant because battery nemans connected.E=E04→k decreases3.U=12CV2U is same but C become kCSo, U=12CV2⋅k(k increases)i.e. Energy store ↑.

$V \to c o n s t a n t$ 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 $k C$
So, $U = \frac{1}{2} C V^{2} \cdot k (k$ increases)
i.e. Energy store $↑$ .