Question

A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage $V$) as $\epsilon =\alpha V$ where $\alpha =2V^{-1}$. A similar capacitor with no dielectric is charged to $V_0=78V$. It is then connected to the uncharged capacitor with the dielectric. Final voltage on the capacitor is

• A. $2V$
• B. $3V$
• C. $5V$
• D. $6V$

Answer 1

The correct option is D $6V$

On connecting the two given capacitors, let the final voltage be $V$.

If capacity of capacitor without the dielectric is $C$, then the charge on this capacitor is $q_1=CV$

The other capacitor with dielectric has capacity $\epsilon C$.

Therefore, charge on it is $q_2=\epsilon CV$

As $\epsilon =\alpha V$, therefore $q_2=\epsilon =\alpha V$

The initial charge on the capacitor (Without dielectric)

that was charged is $q_0=C{V}_0$

From the conservation of charge, $q_0=q_1+q_2$

$C{V}_0=CV+\alpha C{V}^{2}or{V}^2+V-V_0=0$

$V=\frac{-1\pm \sqrt{1+4\alpha V_0}}{2\alpha }$

using $\alpha =2V^{-1}$ or $V_0=78V$, we get

$V=\frac{-1\pm \sqrt{1+(4\times 2\times 78)}}{2\times 2}=\frac{-1\pm \sqrt{625}}{4}$

As $V$ is positive, therefore, $V=\frac{\sqrt{625}-1}{4}=\frac{24}{4}=6V$,