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

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.

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.

An ebomiote rod (K = 3), 6 mm thick is introduced between the plates of a parallel plate capacitor of plate area 4xx10^(-2) m^(2) and plate separation 0.01m. Find the capacitance.

Figure shows a parallel plate capacitor with its plate area A=lb and plate separation d at left end of the plates. Upper plate of capacitor is slightly tilted by a very small angle 8 as shown. Find the capacitance of this capacitor.

Briefly explain the principle of a capacitor. Derive an expression for the capacitance of a parallel plate capacitor, whose plates are separated by a dielectric medium.

If a dielectric slab of thickness t and area A is inserted in between the plates of a parallel plate capacitor of plate area A and distance between the plates d(d gt t) then find out capacitance of system. What do you predict about the dependence of capacitance on location of slab?

What will be the change in the capacitance of a parallel plate capacitor, if the surface area of its plates is doubled and separation between plates doubled?