Calculate the mutual force of attraction between two parallel plates 1 cm apart and maintained at a potential differnce of 100 V Aera of rach plate `=10^(-2) m^(2)`

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

A charged drop of mass 3.2 xx 10^(-12) g floats between two horizontal parallel plates maintained at potential difference of 980 V and separation between the plates is 2 cm. The number of excess or deficient electrons on the drop is

The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d with a potential difference V between the plates, is.

The plates of a parallel-plate capacitor in vacuum are 5.00 mm apart and 2.00 m^2 in area. A potential difference of 10,000 V is applied across the capacitor. Compute (a) the capacitance (b) the charge on each plate, and (c) the magnitude of the electric field in the space between them.

Two protons A and B are placed in between the two plates of a parallel plate capacitor charged to a potential difference V as shown in the figure. The forces on the two protons are identical.

An electron enters the region between the plates of a parallel plate capacitor at a point equidistance from either plate. The capacitor plates are 2 xx 10^(-2)m apart and 10^(-1)m long. A potential difference of 300 V is kept across the plates. Assuming that the initial velocity of the electron is parallel to the capacitor plates, calculate the largest value of the velocity of the electrons so that they do not fly out of the capacitor at the other end.

A parallel plate air filled capacitor having area 40 cm^2 and plate spacing 1.0 mm is charged to a potential difference of 500V. Find (a) the capacitance, (b) the magnitude of the charge on each plate ( c) the stored energy, (d) the electric field between the plates and (e ) the energy density between the plates.