A parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$ what will be capacitance if distance between plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$ ?…..$\mu F$

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100\,V$ $D.C.$ source. Capacitor $C _1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $C _2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be $………V.$ (Assuming Dielectric constant $=10$ )

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

Two identical charged spheres are suspended by string of equal lengths. The string make an angle of $37^{\circ}$ with each other. When suspended in a liquid of density $0.7 \mathrm{~g} / \mathrm{cm}^3$, the angle remains same. If density of material of the sphere is $1.4 \mathrm{~g} / \mathrm{cm}^3$, the dielectric constant of the liquid is_____$\left(\tan 37^{\circ}=\frac{3}{4}\right)$.

A parallel plate capacitor of area $A$, plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constants ${k_1},{k_2}$ and ${k_3}$ as shown. 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