A parallel plate capacitor of capacitance `C_(0)` is charged with a charge `Q_(0)` to a potential difference `V_(0)` and the battery is then disconnected Now a dielectric slab of dielectric constant `k` is inserted between the plates ofcapacitor. The dimensions of the slab are such that it completely fills the space between the plates, then :

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 having capacitance 12pF is charged bya battery to a potential difference of 10V between its plates. The charging battery is now disonnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is

A parallel - plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected . A slab of dielectric constant K is then inserted between the plate of the capacitor so as to fill the space between the plate .Find the work done on the system in the process of inserting the slab.

A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then

A parallel capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E . If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become.

Figure shows a parallel plate capacitor with plate area A and plate separation d . A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dieletric constant K is placed in between the plates of the capacitor as shown. The electric field in the gaps between the plates and the electric slab will be

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

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-

Figue shows a parallel plate capacitor with plate area A and plate separation d . A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dieletric constant K is placed in between the plates of the capacitor as shown. The electric field in the dielectric slab is