For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectricconstant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is 3d/4 , where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation:

 

A dielectric slab of dielectric constants k is slowly inserted inside the parallel plate capacitor having plate area A and separation between plates d as shown in figure. If dimensions of dielectric slab is same as plates of capacitor and capacitor is connected through the battery, then work done by agent will be (C = ε₀ A/d)

Four metallic plates, each with a surface area A, are placed at a distance d from each other as shown in the figure. If points A and B is connected with a battery of emf V, then the charge on the plate a will be   

A cube of side a has charge q at two diagonally opposite vertices as shown in the figure. The net electric potential at the centre of the cube will be

 

If q₁ + q₂ = 3q, then the value of the q₁/q₂, for which the force of repulsion between q₁ and q₂ is maximum is

A capacitor of 20 μF is charged to 500 volt and a capacitor 10μF is charged to 200 volt and connected. Then after connecting both of them, the common potential for each will be :–