Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is

$A$ parallel plate capacitor has plate area $40 \ cm^2$ and plate separation $2 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and dielectric constant $5$. The capacitance of the system is ($\varepsilon_0 =$ permittivity of vacuum)

$A$ capacitor stores $60\,\mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,an additional charge of $120\,\mu C$ flows through the battery. The dielectric constant of the dielectric inserted is

Two dielectric slabs of dielectric constants $K_1$ and $K_2$ and of the same thickness are inserted in a parallel plate capacitor. Given $K_1 = 2K_2$. If the potential differences across the slabs are $V_1$ and $V_2$ respectively,then:

Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$,then the amount of charge which will flow through the battery is (neglect internal resistance of the battery).

The variation of the electric field between the plates of a capacitor is shown in the figure. Find the relation between $K_1$,$K_2$,and $K_3$,if $2x = y$.