Give the expression for capacitance of a capacitor when n dielectric slabs of thickness `t_1, t_2 ... t_n` and dielectric constants `k_1, k_2 ... k_n` respectively, fills the entire space between the plates.

Derive an expression for capacitance of a parallel plate capacitor without a dielectric slab between the plates.

A parallel plate capacitor consists of two identical metal plates. Two dielectric slabs having dielectric donstants k_1 and k_2 are introduced in the space between two plates as shown below. Show that capacity of capacitor in arrangement (a) is given by , C = (epsilon_0 (k_1+k_2)A)/(2d)

In a capacitor of capacitance 20muF , 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, what is the new capacitance?

A parallel plate air capacitor has a capacitance of 3 xx 10^-9F . A slab of dielectric constant 3 and thickness 3cm completely fills the space between the plates. The potential difference between the plates is maintained constant at 400 volt. What is the change in the energy of capacitor if the slab is removed?

A parallel plate air capacitor has a capacitance of 4muF . A slab of dielectric constant 4 and thickness 4 cm completely fills the space between the plates. The potential difference between the plates is maintained constant at 200 volt. What is the change in the energy of a capacitor if the slab is removed?

The capacity of a capacitor becomes 20muF when gap between the capacitor is filled completely by a dielectric slab of k = 4. What is the capacity of a capacitor with air in between the plates?

A charged parallel plate capacitor has a potential energy U. if a slab of dielectric constant k is inserted between the plates, then the new potential energy will be_

A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has thickness (3d//4) , where d is separation of the plates.The charge in capacitance when the slab is inserted between the plates is