A parallel plate capacitor has a capacitance of "5pF" ,a plate area of `20.0mm^(2)` and voltage across it is 10V.A dielectric slab with dielectric constant "5" is introduced in between the plates.Find the magnitude of the induced surface charge on the slab.

A parallel plate capacitor carries a harge Q. If a dielectric slab with dielectric constant K=2 is dipped between the plates, then

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K= (5)/(3) is inserted between the plates, the magnitude of the induced charge will be :

In an isolted, charged, parllel-plate air capacitor, the charge per unit area on each plate has a magnitude of sigma . A dielectric slab having the dielectric constant K is now introduced between the plates. The induced charge per unit area on the surface of the dielectric will have magnitude.

A parallel state air capacitor has capacitance of 100 muF . The plates are at a distance d apart. If a slab of thickness t (t le d) and dielectric constant 5 is introduced between the parallel plates, then the capacitance can be

In a capacitor of capacitance 20 muF ,the distance between the plates is 2 mm . If a dielectirc slab of width 1 mm and dielectric constant 12 is inserted between the plates, then the new capacitance is

A parallel plate capacitor having capacitance C farad is connected with a battery of emf V volts. Keeping the capacitor cannected with the battery, a dielectric slab of dielectric constant K is inserted between the plates. The dimensions of the slab are such that it fills the space between the capacitor plates. Then,

A parallel plate capacitor of capacitance 100muF is connected a power supply of 200 V. A dielectric slab of dielectric constant 5 is now inserted into the gap between the plates. (a) Find the extra charge flown through the power supply and the work done by the supply. (b) Find the change in the electrostatic energy of the electric field in the capacitor.

The separation between the plates of a parallel plate capacitor is d and the area of each plate is A. iff a dielectric slab of thickness x and dielectric constant K is introduced between the plates, then the capacitance will be