An electric field of `3 X 10^4 Vm^-1` is produced between the plates 0.05m apart, of a parallel plate capacitor when it is fully charged. Now an uncharged metal plate of thickness 0.01m is inserted between capacitor plates:- Find the potential difference after introduction of plates.

An electric field of 3X10^4 Vm^-1 is produced between the plates 0.05m apart, of a parallel plate capacitor when it is fully charged. Now an uncharged metal plate of thickness 0.01m is inserted between capacitor plates:-What would be the potential difference if a dielectric slab (k = 3) were introduced in place of metal plate

A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?

A charged parallel plate capacitor has a potential energy U. if a slab of dielectric constant k is inserted between the plates, then the new potential energy will be_

The capacity of a parallel plate air capacitor is 10 mu F and it is given a charge of 40 mu C . The electrical energy stored in the capacitor is

A parallel plate capacitor has circular plates of radius 8 cm and plate separation 1mm. What will be the charge on the plates if a potential difference of 100V is applied?

What is the area of the plates of a 2 F parallel plate capacitor,given that the separation between the plates is 0.5 cm ?

A metal foil of negligible thickness is introduced between the two plates of a capacitor at the centre. The capacitance of capacitor will be

In a parallel plate capacitor with air between the plates, each plate has an area of 6 xx 10^-3m^2 and the separation between the plate is 2mm:- Calculate the capacitance of the capacitor.

A parallel plate capacitor has an area of 4cm^2 and a plate separation of 3mm:- What is its capacitance if the space between the plates is filled completely with a dielectric having dielectric constant of 6.7.

A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has thickness (3d//4) , where d is separation of the plates.The charge in capacitance when the slab is inserted between the plates is