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A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as = where =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.

Vidyadip · Accepted Answer

Both capacitors will be connected in parallel, hence the potential difference across both capacitors should be same. Assuming the required final voltage. be $U.$ If $C$ is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is given by $Q_1 = CU.$ As the capacitor with the dielectric has a capacitance $\epsilon\text{C}$.
Hence, the charge on the capacitor is given by
$\text{Q}_2=\epsilon\text{CU}=(\alpha\text{U})\text{CU}=\alpha\text{CU}$
The initial charge on the capacitor is given by
$\text{Q}_0=\text{CU}_0$
From the conservation of charges,
$\text{Q}_0=\text{Q}_1+\text{Q}_2\Rightarrow\ \text{CU}_0=\text{CU}+\alpha\text{CU}^2$
$\Rightarrow\ \alpha\text{U}_2+\text{U}-\text{U}_0=0$
$\therefore \text{U} = \frac{-1\pm\sqrt{1+4\alpha\text{U}_0}}{2\alpha}$
On solving for $U_0 = 78V$ and $\alpha=2\text{V}^{-1}$
$\text{U}=\frac{-1\pm\sqrt{1+4\times2\times78}}{2\times2}=\frac{-1\pm\sqrt{1+624}}{4}$
$=\frac{-1\pm\sqrt{625}}{4}=\frac{-1\pm25}{4}=-\frac{26}{4}V$ and $6 V$
Hence final voltage. $U = 6V.$

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