Question

A parallel-plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage $(V)$ as $\varepsilon=\alpha V$ where $\alpha=3\, V ^{-1}$. A similar capacitor with no dielectric is charged to $V _{0}=52\, V$. It is then connected to the uncharged capacitor with the dielectric. What is the final voltage (in $V$) on the capacitor?

Answer 1

On connecting the two given capacitors, let the final voltage be $V$.
If the capacitance of capacitor without the dielectric is $C$ then,
$q _{1}= CV$
The other capacitor with dielectric has capacitance $\varepsilon C$.
$\therefore q _{2} =\varepsilon CV $
$=\alpha CV ^{2} \ldots .[\because \varepsilon=\alpha V ] $
The initial charge on the capacitor (without dielectric) that was charged,
$q _{0}= CV _{0}$ from conservation of charge, $ q _{0}= q _{1}+ q _{2}$
$\therefore CV _{0}= CV +\alpha CV ^{2} $
$ \therefore \alpha V ^{2}+ V - V _{0}=0 $
$\therefore V =\frac{-1 \pm \sqrt{1+4 \alpha V _{0}}}{2 \alpha}$
$= \frac{-1 \pm \sqrt{1+4 \times(3) \times(52)}}{2 \times(3)} $
$=\frac{-1 \pm \sqrt{625}}{2 \times(3)}$
As $V$ is positive,
$\therefore V =\frac{\sqrt{625}-1}{6}$
$=\frac{24}{6}=4\, V$