Question

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

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

Answer 1

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

Initial charge on both the capacitor $q_1=2CV,\text{and}{q}_2=CV$

Now condenser of capacity $C$ is filled with dielectric K, therefore $C_2=KC$

Now the final charge on bo the the capacitor $q_3=2C{V}^{\prime },\text{and}{q}_4=KC{V}^{\prime }$

Where $V^{\prime }$ is the voltage after an introduction of dielectric medium.


As charge is conserved,

$\therefore q_1+q_2=q_3+q_4$

Therefore,

$2CV+CV=2C{V}^{\prime }+KC{V}^{\prime }$

$\Rightarrow V^{\prime }=\frac{3CV}{(K+2)C}=\frac{3V}{K+2}$