If a dielectric substance is introduced between the plates of a charged air-gap capacitor,the energy of the capacitor will:

$A$ parallel plate capacitor having capacity $C_0$ is charged to $V_0$. With the battery disconnected,if the separation between the plates is doubled,then the energy stored in it is $E_1$. Instead,if the separation between the plates is doubled with the battery in connection,the energy stored in it is $E_2$. Then the value of $\frac{E_2}{E_1}$ is

$A$ parallel plate capacitor is filled with $3$ dielectric materials of the same thickness,as shown in the sketch. The dielectric constants are such that $k_3 > k_2 > k_1$. Let the magnitudes of the electric field in and potential drops across each dielectric be $E_3, E_2, E_1$ and $\Delta V_3, \Delta V_2, \Delta V_1$,respectively. Which one of the following statements is true?

$A$ parallel plate capacitor has an area of $6 \, cm^2$ and a separation of $3 \, mm$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1 = 10, K_2 = 12$,and $K_3 = 14$. The dielectric constant of a material which,when fully inserted in the above capacitor,gives the same capacitance would be:

$A$ 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.......$nC$.

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: