The capacitance of a parallel plate capacitor is 50 pF and the distance between the plates is 4mm. It is charged to 200 V and then the charging battery is removed. Now a dielectric slab `(kappa=4)` of thickness 2mm is placed. Determine (i) final charge on each plate (ii) finial potential difference between the plates (iii) final energy is the capacitor.

A parallel plate capacitor (C=50 muF. d=4mm) is charged to 200V and then charging battery is removed. Now a dielectric slab (K=4) of thickness 2mm is placed between the plates. Find new potential difference across capacitor.

After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates

A parallel plate capacitor having plate area A and separation between the plates d is charged to potential V .The energy stored in capacitor is ....???

Separation between the plates of parallel plate capacitor is 5 mm. this capacitor, having air as the dielectric medium between the plates, is charged to a potential difference 25 V using a battery. The battery is then disconnected and a dielectric slab of thickness 3mm and dielectric constant K=10 is placed between the plates as shown. potential difference between the plates after the dielectric slab has been introduced is-

A parallel plate capacitor is charged completely and then disconnected from the battery. IF the separation between the plates is reduced by 50% and the space between the plates if filled with a dielectric slab of dielectric constant 10, then the potential difference between the plates

Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery. Now,

The space between plates of a parallel plate capacitor is filled by a dielectric and it is charged and then battery is removed. Now dielectric slab is slowly drawn out of the capacitor parallel to the plates. The variatiton of the potential of capacitor with respect to the length of the dielectric plate drawn out is