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 of plate area A and separation d filled with three dielectric materials as show in figure. The dielectric constants are K_1 , K_2 and K_3 respectively. The capacitance across AB will be :

A capacitor of plate area A and separation d is filled with two dielectrics of dielectric constant K_(1) = 6 and K_(2) = 4 . New capacitance will be

The capacitance of a parallel plate capacitor with plate area A and separation d is C . The space between the plates in filled with two wedges of dielectric constants K_(1) and K_(2) respectively. Find the capacitance of resulting capacitor.

The capacitance of a parallel plate capacitor with plate area A and separation d is C . The space between the plates in filled with two wedges of dielectric constants K_(1) and K_(2) respectively. Find the capacitance of resulting capacitor.

A capacitor of plate area A and separation between plates d is half filled with dielectric of dielectric constant K. What is equivalent capacitance?

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.