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.

Which one statement is correct ? A parallel plate air condenser is connected with a battery. Its charge, potential, electric field and energy are ${Q_o},\;{V_o},\;{E_o}$ and ${U_o}$ respectively. In order to fill the complete space between the plates a dielectric slab is inserted, the battery is still connected. Now the corresponding values $Q,\;V,\;E$ and $U$ are in relation with the initially stated as

Write the capacitance of parallel plate capacitor with medium of dielectric of dielectric constant $\mathrm{K}$.

A parallel plate capacitor of capacitance $5\,\mu F$ and plate separation $6\, cm$ is connected to a $1\, V$ battery and charged. A dielectric of dielectric constant $4$ and thickness $4\, cm$ is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is........$\mu C$

The distance between the plates of a parallel plate condenser is $8\,mm$ and $P.D.$ $120\;volts$. If a $6\,mm$ thick slab of dielectric constant $6$ is introduced between its plates, then

A parallel plate capacitor with air between the plates has a capacitance of $9\,pF$. The separation between its plates is $'d'$. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $K_1=3$ and thickness $\frac {d}{3}$ while the other one has dielectric constant $K_2 = 6$ and thickness $\frac {2d}{3}$ . Capacitance of the capacitor is now........$pF$