A parallel plate condenser with oil between the plates (dielectric constant of oil $k =2$ ) has a capacitance $C.$ If the oil is removed, then capacitance of the capacitor becomes

Effective capacitance of parallel combination of two capacitor $C_1$ and $C_2$ is $10 \mathrm{\mu F}$. When we inserted a dielectric slab between the plates the new capacitance will be ( dielectric constant $K_1=2, K_2=4$)

A parallel plate condenser is filled with two dielectrics as shown in the figure. The area of each plate is $A{ \mathrm{m}}^2$ and the separation is d metre. The dielectric constants are $K_1$and $K_2$ respectively. Its capacitance in farad will be-

Capacity of air capacitor (parallel plate) is $10 \mathrm{\mu C}.$ Now a dielectric of dielectric constant $4$ is filled in the half space between the plates, then new capacity will be-

Putting a dielectric substance between two plates of condenser: capacity, potential and potential energy, respectively