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)

A parallel plate capacitor is charged and then charging battery is disconnected. If the plates are now pulled apart with insulated handles

Energy stored in circuit 1 is E. If capacitors in circuit 1 and circuit 2 are connected in parallel as shown, the energy stored becomes $\frac{xE}{6}$ , find x.

Potential due to electric dipole on axial position at distance r from dipole is proportional to (assume r >> length of dipole)

The force between two charged particle placed in air at separation x is F₀. Both the charged particle immerged in a medium of dielectric constant K without changing separation between two charge, then net force on one of the particle is now