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)`

A parallel plate air condenser has a capacity of 21 mu F . What will be the new capacity if A marble slab of dielectric constant 8 is introduced between the two 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_

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?

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.

Two plates of a parallel plate capacitor are 4 mm apart. A slab of dielectric constant 3 and thickness 3mm is introduced between the plates with the faces parallel to them. The distance between the faces is so adjusted that the capacitance of the capacitor becomes 2/3 of its original value. What is the new distance between the plates ?

A metal foil of negligible thickness is introduced between the two plates of a capacitor at the centre. The capacitance of capacitor will be

A parallel plate condenser with air as dielectric has capacity C_0 = (in_0 A)/(d) . A thin mica sheet of dielectric constant K and thickness is introduce near the first plate and then moved with constant velocity v towards the other plate . The capacity of the condenser will be

A parallel plate capacitor has an area of 4cm^2 and a plate separation of 3mm:- What is its capacitance if the space between the plates is filled completely with a dielectric having dielectric constant of 6.7.