A composite parallel plate capacitor is made up of two different dielectric materials with different thickness $\left(t_{1}\right.$ and $\left.t_{2}\right)$ as shown in figure. The two different dielectric material are separated by a conducting foil $F$. The voltage of the conducting foil is $.....V$
$6$
$66$
$600$
$60$
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
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
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 share the charge and the common potential is $V'$. The dielectric constant $K$ is
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 having plates of area $S$ and plate separation $d$, has capacitance $C _1$ in air. When two dielectrics of different relative permittivities $\left(\varepsilon_1=2\right.$ and $\left.\varepsilon_2=4\right)$ are introduced between the two plates as shown in the figure, the capacitance becomes $C _2$. The ratio $\frac{ C _2}{ C _1}$ is