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 of capacitance $90\ pF$ is connected to a battery of $emf$ $20\ V$. If a dielectric material of dielectric constant $K = \frac{5}{3}$ is inserted between the plates, the magnitude of the induced charge will be.......$n $ $C$

Define dielectric constant.

A parallel plate capacitor is made of two square plates of side $a$, separated by a distance $d\,(d  < < a)$. The lower triangular portion is filled with a dielectric of dielectric constant $K$, as shown in the figure. Capacitance of this capacitor is

Consider the arrangement shown in figure. The total energy stored is $U_1$ when key is closed. Now the key $K$ is made off (opened) and two dielectric slabs of relative permittivity ${ \in _r}$ are introduced between the plates of the two capacitors. The slab tightly fit in between the plates. The total energy stored is now $U_2$. Then the ratio of $U_1/U_2$ 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 $C$