The electric field between the plates of a parallel plate capacitor when connected to a certain battery is $E_0$. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant $K$ without disturbing the battery connections,the field between the plates shall be

If half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $4$,the capacitance is $C_1$. If one third of the space between the plates of the capacitor is filled with the medium of dielectric constant $4$,the capacitance is $C_2$. If in both cases,the dielectric is placed parallel to the plates of the capacitor,then $C_1: C_2=$

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \,F$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \,V$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

An air-filled parallel plate capacitor has capacity $C$. If the distance between the plates is doubled and it is immersed in a liquid,the capacity becomes twice. The dielectric constant of the liquid is:

The force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$. Half the gap between the two charges is filled by a dielectric slab of dielectric constant $K=4$. Then the new force of repulsion between those two charges becomes:

$A$ parallel plate air capacitor has a capacitance of $10 \ \mu F$. As shown in the figure,if this capacitor is divided into two equal parts and these parts are filled with dielectric materials of dielectric constants $K_1 = 2$ and $K_2 = 4$,then the capacitance of this arrangement is ............. $\mu F$.