Question

A capacitor of capacitance $2\mu \text{F}$ is charged with a $10\text{V}$ battery. Now the battery is disconnected and a dielectric of dielectric constant $K=4$ is inserted between the plates. What is the potential difference between the plates of the capacitor ?

• A. $10\text{V}$
• B. $25\text{V}$
• C. $0.25\text{V}$
• D. $2.5\text{V}$

Answer 1

The correct option is C $0.25\text{V}$

Given:
$C=2\mu \text{F};V=10\text{V};K=4$

Charge on the plates of the capacitor is given by

$Q=CV=2\times 10=20\mu \text{C}$

When dielectric is inserted, its capacitance becomes, $C^{\prime }=KC$

$\therefore C^{\prime }=4\times 20=80\mu \text{C}$

Since the battery is disconnected, charge on the capacitor will remain $20\mu \text{C}$.

$\therefore Q=C^{\prime }V$

$\Rightarrow 20=80V$

$\Rightarrow V=\frac{1}{4}=0.25\text{V}$

Therefore, option $\left(C\right)$ is right.

Why this question ? Tip : Dielectric increases the capacitance of the capacitor