A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as $\epsilon=2 U$. A similar capacitor with no dielectric is charged to $U_0=78 \mathrm{~V}$. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Let $C_0$ be the capacitance of the capacitor without the dielectric. The charge on this capacitor is $Q_0 = C_0 U_0$, where $U_0 = 78$ V.

When the two capacitors are connected in parallel, the total charge $Q_0$ is shared between them. Let $U$ be the final common voltage. The charge on the first capacitor is $Q_1 = C_0 U$.

The second capacitor has a dielectric with relative permittivity $\epsilon = 2U$. The capacitance of this capacitor is $C = \epsilon C_0 = (2U) C_0$. The charge on this capacitor is $Q_2 = C U = (2U) C_0 U = 2 C_0 U^2$.

By the principle of conservation of charge, $Q_0 = Q_1 + Q_2$.

$C_0 U_0 = C_0 U + 2 C_0 U^2$.

Dividing by $C_0$, we get $U_0 = U + 2U^2$.

Substituting $U_0 = 78$, we have $2U^2 + U - 78 = 0$.

Solving the quadratic equation for $U$ using the quadratic formula $U = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$U = \dfrac{-1 \pm \sqrt{1^2 - 4(2)(-78)}}{2(2)} = \dfrac{-1 \pm \sqrt{1 + 624}}{4} = \dfrac{-1 \pm \sqrt{625}}{4} = \dfrac{-1 \pm 25}{4}$.

Since voltage must be positive, $U = \dfrac{24}{4} = 6$ V.

Answer: 6V