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:

Two metal plates each of area $A$ form a parallel plate capacitor with air in between the plates. The distance between the plates is $d$. $A$ metal plate of thickness $\frac{d}{2}$ and of same area $A$ is inserted between the plates to form two capacitors of capacitances $C_1$ and $C_2$ as shown in the figure. If the effective capacitance of the two capacitors is $C^{\prime}$ and the capacitance of the capacitor initially is $C$,then $\frac{C^{\prime}}{C}$ is

$A$ parallel plate capacitor has a capacitance of $2\ \mu F$ with a separation of $0.4\ cm$ between the plates. If the distance between the plates is halved and the space is filled with a dielectric material of constant $K = 2.8$,the final capacitance of the capacitor will be .....$\mu F$.

$A$ parallel plate capacitor has capacitance $C$,when there is vacuum within the parallel plates. $A$ sheet having thickness $t = d/3$ (where $d$ is the separation between the plates) and relative permittivity $K$ is introduced between the plates. The new capacitance of the system is:

$A$ parallel plate capacitor has plates with area $A$ and separation $d$. $A$ battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced,is

If the distance between the plates of a parallel plate capacitor is halved and the dielectric constant of the dielectric is doubled,then its capacity will