A capacitor of $10 \mu \mathrm{F}$ capacitance whose plates are separated by $10 \mathrm{~mm}$ through air and each plate has area $4 \mathrm{~cm}^2$ is now filled equally with two dielectric media of $\mathrm{K}_1=2, \mathrm{~K}_2=3$ respectively as shown in figure. If new force between the plates is $8 \mathrm{~N}$. The supply voltage is . . . .. . .V.

In a parallel plate condenser, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. There is a glass slab of $3\,mm$ thick and of radius $12\,cm$ with dielectric constant $6$ between its plates. The capacity of the condenser will be

In a capacitor of capacitance $20\,\mu \,F$, the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates, then the new capacitance is......$\mu \,F$

A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage $(U )$ as $\varepsilon  = \alpha U$ where $\alpha  = 2{V^{ - 1}}$. A similar capacitor with no dielectric is charged to ${U_0} = 78\,V$. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

The area of each plate of a parallel plate capacitor is $100\,c{m^2}$and the distance between the plates is $1\,mm$. It is filled with mica of dielectric $6$. The radius of the equivalent capacity of the sphere will be.......$m$

The capacity of a parallel plate condenser is $5\,\mu F$. When a glass plate is placed between the plates of the conductor, its potential becomes $1/8^{th}$ of the original value. The value of dielectric constant will be