A parallel plate capacitor with a dielectric slab with dielectric constant `k= 3` filling the space between the plates is charged to potential `V` and isolated. Then the dielectric slab is drawn out and another dielectric slab of equal thickness but dielectric constant `k'= 2` is introduced between the plates. The ratio of the energy stored in the capacitor later to that initially is:

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

If a dielectric slab of thickness 5 mm and dielectric constant K=6 is introduced between the plates of a parallel plate air capacitor, with plate separation of 8 mm, then its capacitance is

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

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 has a capacity c. If a medium of dielectric constant K is introduced between plates, the capacity of capacitor becomes

The plates of a capacitor are connected to a source of emf and the energy stored is U_(0) . When dielectric material dielectric constant K is introduced between the plates filling the whole space, the energy stored becomes U Now.