The distance between the plates of a parallel plate capacitor is "4cm" .An electric field of `2times10^(4)`V/m is established between the plates.It is then disconnected from the battery and an uncharged metal plate of thickness "1cm" is inserted into the capacitor parallel to it's plates.The potential difference between the plates of capacitor after the introduction of the metal plate is `k times10^(2)`V .The value of ' "k" s

As the distance between the plates of a parallel plate capacitor decreased

Force of attraction between the plates of a parallel plate capacitor is

The distance between the plates fo a parallel plate capacitor is d. A metal plate of thickness d//2 is placed between the plates, what will be the new capacity ?

The' distance between the plates of a parallel plate capacitor is d . A metal plate of thickness d//2 is placed between the plates. What will e its effect on the capacitance.

The distance between the plates of a parallel plate capacitor is 0.04m A field of 5000V//m is established between the plates and an unchanged metal plate of thickness 0.01m is inserted intothe condenser parallel to its plate. Find the p.d . between the plates (a) before the introducing of the metal plate and (b) after its introduction. what would be the p.d. if plate of dielectric constant K=2 is introduced in place of metal plate?

The distance between the plates of a charged plate capacitor disconnected from the batteryis 5cm and the intensity of the field in it is E=300V//cm . An uncharged metal bar 1cm thick is introduced into the capacitor parallel to its planes. The potential difference between the plates now is :

Between the plates of a parallel plate capacitor a dielectric plate in introduced just to fill the complete space between the plates. The capacitor is charged and later disconnected from the battery. The dielectric plate is then slowly drawn out of the capacitor plates. the plot of the potential difference across the plates and the length of the dielectric plate drawn out is