Question

A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage $\left(U\right)$ as $ϵ=\alpha U$ where $\alpha =2V^{-1}$. A similar capacitor with no dielectric is charged to $U_0=78V$. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Answer 1

Formula used: $Q=CV$

Given,

Relative permittivity of dielectric,

$ϵ=\alpha U$

Constant, $\alpha =2V^{-1}$

Initial voltage, $U_0=78V$

As we know, $Q=CV$

Suppose the final voltage be $U$.

If $C$ is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is, $Q_1=CU$

The capacitor with the dielectric has a capacitance $ϵC$. Hence the charge on the capacitor is

$Q_2=ϵCU=\alpha U^{2}C=2U^{2}C$

The initial charge on the capacitor that was charged is

$Q_0=C{U}_0$

From the conservation of charges,

$Q_0=Q_1+Q_2$

Substitute the values,

$78C=CU+2U^{2}C$

$2U^2+U-78=0$

solving for $U$, we get

$U=\frac{-1\pm \sqrt{1+624}}{4}$

$U=\frac{-1\pm \sqrt{625}}{4}$

As $U$ is positive,

$U=\frac{\sqrt{625}-1}{4}$

$=\frac{24}{4}$

$=6V$

Final Answer: $6V$