The plates of a parallel plate capacitor have surface area 'A' and are initially separated by a distance 'd'.They are connected to a battery of voltage `v_(0)`. Now the plates of the capacitor are pulled apart to a separation `2d` .Then,increase in the energy of the battery is

Find out the capacitance of parallel plate capacitor of plate area A and plate separation d.

A parallel plate capacitor is charged. If the plates are pulled apart

A parallel plate capacitor has plate A and separation d between the plates. The capacitor is connected to a battery of emf V . (a) find the charge on the capacitor. (b) the plate separation is decreases to d//2 . Find the extra charge given by the battery to the positive plate. (c) work done in reducing separation between plates.

The plates of a parallel-plate capacitor are separated by a solid dielectric. This capacitor and a resistor are connected in series across the terminals ofa battery. Now the plates of the capacitor are pulled slightly further apart. After some time circuit again attains steady state.

A parallel plate capacitor of plate are A and plate separation d is charged by a battery of voltage V. The battery is then disconnected. The work needed to pull the plates to a separation 2d is

Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery. Now,

A parallel -plate capacitor having plate area 25.0 cm^ and a separation 2.00 mm between the plates .the capacitor is connected to a bettery of 12.0 V.(a)find the charge on the capacitor .(b) the plate separation is decrerased to 1.00 mm. Find the extra charge given by the bettery to the positive plate.

A pararllel plate capacitor has plates of area A and separation d and is charged to potential diference V . The charging battery is then disconnected and the plates are pulle apart until their separation is 2d . What is the work required to separate the plates?