Question

Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference $V_0$. The battery is then disconnected and the region between the plates of the capacitor C is completely filled with a material of dielectric constant 2. The potential difference across the capacitors now becomes:

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$Q_1=C{V}_0\Rightarrow V_0=\frac{Q_1}{C}$

$\because$ Capacitors are in parallel

$\therefore \frac{Q_1}{C}=\frac{Q_2}{2C}$

$\Rightarrow Q_2=2Q_1\dots \dots \left(i\right)$

$V_0=\frac{Q_2}{2C}$

After disconnecting the battery and inserting the dielectric in C.

$V_1^{{}^{\prime }}=\frac{Q_1^{{}^{\prime }}}{CK}=\frac{Q_1^{{}^{\prime }}}{2C}$

$V_2^{{}^{\prime }}=\frac{Q_2^{{}^{\prime }}}{2C}$

Charge will flow from 2 to 1 till $\frac{Q_2^{{}^{\prime }}}{2C}=\frac{Q_1^{{}^{\prime }}}{2C}$; i.e., till the two potentials are equal

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

Earlier potentials $V_0=\frac{Q_1}{C}$


Now it is $\frac{Q_1^{{}^{\prime }}}{2C}$

$Q_1+Q_2=3Q_1=Q_1^{{}^{\prime }}+Q_2^{{}^{\prime }}$

$\frac{Q_2^{{}^{\prime }}}{2C}=\frac{Q_1^{{}^{\prime }}}{2C}$ or $Q_2^{{}^{\prime }}=Q_1^{{}^{\prime }}=\frac{3Q_1}{2}$

$\therefore$ New potential $=\frac{3Q_1}{4C}or\frac{3V_0}{4}$