In a parallel plate capacitor with air, the distance between the plates is reduced to half and the space is filled with dielectric of constant 4. If initial capacity of capacitor is `2muF`, then final value of capacity is

In a parallel plate air capacitor the distance between plates is reduced to one fourth and the space between them is filled with a dielectric medium of constant 2. If the initial capacity of the capacitor is 4muF . then its new capacity is .

The space between the capacitor plates is filled with a dielectric of dielectric constant K. How will the capacity change ?

A parallel plate capacitor with air between the plates has a capacitance of 8pF. The separation between the plates is now reduced by half and the space between them is filled with medium of dielectric constant 5. The value of capacitance of a capacitor in the second case is

The capacity of a parallel plate capacitor with no dielectric substance but with a separation is reduced to half and it is filled with a dielectirc substance of value 2.8 . The final capacity of the capacitor is

The capacity of a parallel plate capacitor with no dielectric but with a separation 0.4 cm is 2 mu F. The separation is reduced to half and it is filled with a dielectric of value 2.8. The final capacity of the capacitor is

A parallel plate capacitor with air inbetween the paltes has a capacitance of 8 pF. The separation between the plates is now reduced by half and the space between them is filled with a medium of dielectric constant 5. Calculate the new value of capacitance.

In a parallel plate air capacitor of capacitance 4 F if the lower half of air space is filled with a material of dielectric constant 3, its capacitance changes to

If the distance between the plates of parallel plate capacitor is halved and the dielectric constant of dielectric is doubled, then its capacity will