Question

Two parallel plate capacitors of capacitances $C$ and $2C$ are connected in parallel and charged to a potential difference $V$. The battery is then disconnected and the region between the plates of the capacitor $C$ is completely filled with a material of dielectric constant $K$. The potential difference across the capacitors now becomes :

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

Answer 1

Answer: (A)

Initial total charge of the system is: $Q_i=Q_1+Q_2=CV+2CV=3CV$
When dielectric is inserted in $C$ so the capacitance becomes $KC$
Final charge, $Q_f=Q_1^{\prime }+Q_2^{\prime }=KC{V}^{\prime }+2C{V}^{\prime }=(K+2)C{V}^{\prime }$ where $V^{\prime }=$ common potential after disconnected the battery.
As battery is disconnected so total charge will remain unchanged.
Thus, $Q_i=Q_f\Rightarrow 3CV=(K+2)C{V}^{\prime }$
$\therefore V^{\prime }=\frac{3V}{K+2}$