A capacitor with a capacitane of `50muF` is connected to 200V, if the distance between its plates is halved, what will be the potential between the plates and what will be the change in energy stored in it?

If the distance between the plates of a parallel plate 200 mu F capacitor charged to 400 V is halved. Calculate the change in energy stored in it.

If a parallel plate capacitor of capacitance C is kept connected to a supply voltage V and then to just fill the space, dielectric slab is inserted between the plates. What will be the change in the capacitance the charge, the potential difference, the electric field and the energy stored?

A parallel plate capacitor is charged to a potential difference V by a d.c. source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following will change. energy stored in the capacitor.

A parallel plate capacitor with air between the plates has a capacitance of 8 pF.What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 5?

A parallel plate capacitor is charged to a potential difference V by a d.c. source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following will change. electric field between the plates

A capacitor having capacity of 2 muF is charged to 200 V and then the plates of the capacitor are connected to a resistance wire. The heat produced in joule will be

On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved and a dielectric medium of k = 10 is introduced between the plates, without disconnectign the d.c. source. Explain, using the suitable expressions, how the capacitance

On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved and a dielectric medium of k = 10 is introduced between the plates, without disconnectign the d.c. source. Explain, using the suitable expressions, how the energy density of the capacity of change.

A parallel plate capacitor has a capacitance of 2muF . A slab of dielectric constant 5 is inserted between the plates and the capaitor is charged to 100 V and then isolated. What is the new potential difference, if the dielectric slab is removed?