A parallel plate capacitor with plate separation 'd' and plate area 'A' is filled with a di-electric slab of thickness 't'.If it is required that the energy stored by the di-electric is half of the total energy in capacitor,then t= ....( `K=3` ) a) `d/4` b) `d/2`

Fig shows a parallel plate capacitor of plate area A and plate separation d . Its entire space is filled with three different dielectric slabs of same thickness. Find the equivalent capacitance of the arrangment.

A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of dielectric constant given by K=K_(0)(1+alphax). . Calculate capacitance of system: (given alpha d ltlt 1) .

A parallel plate capacitor of plate area A and separation d is filled with two materials each of thickness d/2 and dielectric constants in_1 and in_2 respectively. The equivalent capacitance will be

A parallel plate capacitor, with plate separation d and plate area A, has a capacitance of 5muF . Calculate the new capacitance if the plate separation is reduced to 0.75d and the capacitor is completely filled with dielectric of dielectric constant 8.

Find out the capacitance of parallel plate capacitor of plate area A and plate separation d.

A parallel plate capacitor has plate area A and plate separation d. The space betwwen the plates is filled up to a thickness x (ltd) with a dielectric constant K. Calculate the capacitance of the system.

In a parallel-plate capacitor of plate area A , plate separation d and charge Q the force of attraction between the plates is F .

A parallel plate capacitor has plate of length 'l' width 'w' and separation of plates is 'd' . It is connected to a battery of emf V . A dielectric slab of the same thickness 'd' and of dielectric constant k=4 is being inserted between the plates of the capacitor . At what lenght of the slab inside plates , will the energy stored in the capacitor be two time the initial energy stored ?

The plates of a parallel plate capacitor are separated by a distance d = 1 cm . Two parallel sided dielectric slabs of thickness 0. 7cm and 0.3cm fill the space between the plates. If the dielectric constants of the two slabs are 3 and 5 respectively and a potential difference of 440V is applied across the plates. Find (a) The electric field intensities in each of the slabs. (b) The ratio of electric energies stored in the first to that in the second dielectric slab.