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

A parallel plate capacitor with plate area $'A'$ and distance of separation $'d'$ is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :

$\varepsilon(x)=\varepsilon_{0}+k x, \text { for }\left(0\,<\,x \leq \frac{d}{2}\right)$

$\varepsilon(x)=\varepsilon_{0}+k(d-x)$, for $\left(\frac{d}{2} \leq x \leq d\right)$

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air cored parallel capacitor charged to a potential $V$. The two capacitors share charges and the common potential is $V$. The dielectric constant $K$ is

What will be the capacity of a parallel-plate capacitor when the half of parallel space between the plates is filled by a material of dielectric constant ${\varepsilon _r}$ ? Assume that the capacity of the capacitor in air is $C$

A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is $A\;metr{e^2}$ and the separation is $t$ $metre$. The dielectric constants are ${k_1}$ and ${k_2}$ respectively. Its capacitance in farad will be

After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :-