Question

A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (V) as $\varepsilon = \alpha V$ where $\alpha = 2V^{- 1}$. A similar capacitor with no dielectric is charged to $V_{o} = 78V$. 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)

6 V

Answer 2

: 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 $\varepsilon C$.

Therefore , charge on it is $q_{2} = \varepsilon CV$

As $\varepsilon = \alpha V,$ therefore, $q_{2} = \alpha CV^{2}$

The initial charge on the capacitor (without dielectric) that was charged is

$q_{0} = CV_{0}$

Form the conservation of charge,

$q_{0} = q_{1} + q_{2}$

$CV_{0} = CV + \alpha CV^{2}$or $\alpha V^{2} + V - V_{0} = 0$

$V = \frac{- 1 \pm \sqrt{1 + 4\alpha V_{0}}}{2\alpha}$

Using $\alpha = 2V^{- 1}$ and $V_{0} = 78V,$ 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, $V = \frac{\sqrt{625} - 1}{4} = \frac{24}{4} = 6V.$