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

Answer 1

Answer: (A)

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
$U_{f}=\frac{1}{2} C V^{2}=\frac{1}{2}\left(K C_{0}\right) V^{2}$
Clearly, $\frac{U_{f}}{U_{i}}=K > 1$
$\Rightarrow U_{f} > U_{i}$
As, $Q=C V$
$\therefore V$ remains constant and $C$ increases.
$\therefore Q$ increases.