Question

2. Consider a capacitor created when two identical conducting plates are placed parallel and close to each in other in vacuum The surface area for each plate is e.0400 m2. The two plates are separated by a distance of e.00200 m. When an electric potential difference of 5e.e v is established between the two plates, what is the magnitude of the electrostatic force between the charged plates? Is the force attractive or repulsive?

Answer 1

here,

the surface area for each plate , A = 0.04 m^2

sepration , d = 0.002 m

potential difference , V = 50 V

the magnitude of the electrostatic force between the plates , F = Q * E

F = ( C * V) * ( V/d)

F = area * e0 /d * V * ( V/d)

F = 0.04 * 8.85 * 10^-12 * 50^2 /( 0.002^2) N

F = 2.21 * 10^-4 N

the magnitude of the electrostatic force between the plates is 2.21 * 10^-4 N