When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then

Question

When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then (A) charge on capacitor increases (B) potential difference across the capacitor increases (C) energy stored increases (D) capacity remains the same

Tardigrade Admin · Accepted Answer

As the battery remains connected. potential difference

(V)

of the capacitor remains constant. Suppose dielectric of dielectric constant

K

is introduced between the plates. Now, capcitance of the capacitor is

C^{'} = K C_{0}

where,

C_{0}

is original capacitance.
Initial energy stored in the capacitor

U_{i} = \frac{1}{2} C_{0} V^{2}

Final energy stored in the capacitor

U_{f} = \frac{1}{2} C V^{2} = \frac{1}{2} (K C_{0}) V^{2}

Clearly,

\frac{U _{f}}{U _{i}} = K > 1

\Rightarrow U_{f} > U_{i}

As,

Q = C V

∴ V

remains constant and

C

increases.

∴ Q

increases.

Q. When a dielectric slab is introduced between the plates of a capacitor connected to a battery, then

charge on capacitor increases

potential difference across the capacitor increases

energy stored increases

capacity remains the same