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?

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. How much work is required to remove the dielectric slab?

Explain why the capacitance of capacitor increases when dieslectric slab is inserted between plates of the capacitor.

Explain why the capacitance of capacitor increases when dieslectric slab is inserted between plates of the capacitor.

The potential enery of a charged parallel plate capacitor is U_(0). If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be

A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be

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 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,

A parallel plate capacitor with oil between the plates has a capacitance C(dielectric constant of oil is 2). If the oil is removed, then capacitance of the capacitor becomes