Question

Two capacitors of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is removed and the capacitor of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors will now be :

• A. $\frac{3V}{K}$
• B. $\frac{V}{K}$
• C. $\frac{3V}{K+2}$
• D. $\frac{V}{K+2}$

Answer 1

The correct option is C $\frac{3V}{K+2}$

Just after removing the battery, net charges on the positive plates of the capacitors is given by,


$Q_1+Q_2=\left(2C\right)V+CV=3CV$

After insertion of dielectric material,

Let $V_c$ be the common potential difference across both capacitors.

$Q_1^{\prime }+Q_2^{\prime }=\left(2C\right)V_c+\left(KC\right)V_c=C{V}_c(2+K)$

Applying conservation of charges,

$Q_1+Q_2=Q_1^{\prime }+Q_2^{\prime }$

$\Rightarrow 3CV=C{V}_c(2+K)$

$\therefore V_c=\frac{3V}{K+2}$

Hence, option $\left(\text{C}\right)$ is correct.