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 charged to $V_0$= 78 V. It is then connected to the uncharged capacitor with the dielectric. Final voltage on the capacitor is

• A. 2 V
• B. 3 V
• C. 5 V
• D. 6 V

Answer 1

Answer: (D)

When capacitors are connected in parallel the voltage across both the capacitors will be same. Let the final voltage be $V$

For capacitor with no dielectric let the capacitance be $C$ hence charge on the capacitor will be $Q_1=CV$

Initial charge on the capacitor is $Q_0=C{V}_0$

For similar capacitor with dielectric capacitance $C_1=\epsilon C=\left(\alpha V\right)C$ hence charge on the capacitor will be $Q_2=C_{1}V$

From the conservation of charges

$Initialcharge=Finalcharge$

$\therefore Q_0=Q_1+Q_2$

$\therefore C{V}_0=CV+C_{1}V=CV+\left(\alpha VC\right)V$

$\therefore V_0=V+\alpha V^2$

$\therefore 78=V+2V^2$

solving we get

$V=\frac{-1\pm \sqrt{1^2-4\times 2\times -48}}{2}=6Volt$

Final voltage on the capacitor $V=6V$