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 $2x = y$.

$A$ parallel-plate capacitor of area $A,$ plate separation $d$ and capacitance $C$ is filled with four dielectric materials having dielectric constants $K_1, K_2, K_3$ and $K_4$ as shown in the figure. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor,then its dielectric constant $K$ is given by

Two capacitors,each having capacitance $40\,\mu F$,are connected in series. The space between the plates of one of the capacitors is filled with a dielectric material of dielectric constant $K$ such that the equivalent capacitance of the system becomes $24\,\mu F$. The value of $K$ is:

The space between the plates of a parallel-plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants \(K _1=2, K_2=3\), and \(K _3=5\) (as shown in the figure). If new capacitance is \(\frac{ n }{3} C\) then the value of \(n\) is ___________.

The space between the plates of a parallel plate capacitor is filled with a dielectric whose dielectric constant varies with distance $x$ as per the relation: $K(x) = K_0 + \lambda x$ (where $\lambda$ is a constant). The capacitance $C$ of the capacitor would be related to its vacuum capacitance $C_0$ by the relation:

$A$ charged conductor produces an electric field of intensity $10^3 \ V/m$ just outside its surface in vacuum. Then,it produces an electric field of intensity $E$ just outside its surface,when it is placed in a medium of dielectric constant $4$. The value of $E$ will be (in $V/m$)