A parallel plate capacitor has a capacitance of $4 \mu F$ with air as the dielectric. A dielectric material is inserted, increasing the capacitance to $20 \mu F$. What is the relative permittivity of the dielectric?

When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery, then charge on plates relative to earlier charge

The capacity of a parallel plate air capacitor is $2 \mu \mathrm{F}$ and voltage between the plates is changing at the rate of $3 \mathrm{~V} / \mathrm{s}.$ The displacement current in the capacitor is

The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If ${C}$ is its initial capacitance, its final capacitance is equal to:

Suppose, one wishes to construct a 1.0 farad capacitor using circular discs. If the separation between the discs be kept at $1.0 \mathrm{~mm}$, what would be the radius of the discs ?

What is the unit of capacitance?

A parallel plate capacitor with air as medium between the plates has a capacitance 10 mF. The area of the capacitor is divided into two equal halves and filled with two media having dielectric constants ${K_1} = 2$ and ${K_2} = 4$. The capacitance of the system will be

The capacitance of a parallel plate capacitor with air as medium is $6\mu F$. With the introduction of a dielectric medium, the capacitance becomes $30\mu F$. The permittivity of the medium is: ($$\epsilon_0 = 8.85 imes 10^{-12} C^2N^{-1}m^{-2}$$)

Increasing order of dielectric constant for air, rubber and copper is:

A parallel plate capacitor with air between the plates has a capacitance of $10 pF$. A dielectric slab of dielectric constant 2 is completely filled between the plates. What is the new capacitance?

In a parallel plate capacitor of capacitance $C,$ a metal sheet is inserted between the plates, parallel to them. If the thickness of the sheet is half of the separation between the plates. The capacitance will be