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

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

The potential energy of a charged parallel plate capacitor is $U_0$. If a slab of dielectric constant $K$ is inserted between the plates,then the new potential energy will be

Between the plates of a parallel plate capacitor,a plate of thickness $t_1$ and dielectric constant $k_1$ is placed. In the rest of the space,there is another plate of thickness $t_2$ and dielectric constant $k_2$. The potential difference across the capacitor will be

Between the plates of a parallel plate capacitor of plate area $A$ and capacity $0.025 \mu F$,a metal plate of area $A$ and thickness equal to $1/3$ of the separation between the plates of the capacitor is introduced. If the capacitor is charged to $100 \ V$,then the amount of work done to remove the metal plate from the capacitor is (in $\mu J$)

When a parallel plate capacitor is charged up to $95 \ V$,its capacitance is $C$. If a dielectric slab of thickness $2 \ mm$ is inserted between the plates and the distance between the plates is increased by $1.6 \ mm$ such that the same potential difference is maintained,the dielectric constant of the material (slab) is: