The plates of a parallel plate capacitor each have an area of 0.40 m2 and are separated by a distance of 0.02 m. They are charged until the potential difference between the plates is 3000 V. The charged capacitor is then isolated. Determine the magnitude of the electric field between the capacitor plates.

 E = -1.5 10⁵ N / C

Explanation:

In a capacitor the electric field is uniform between the blades, therefore we use the expression

                   V = - E s

                   E = - V / s

let's calculate

                   E = - 3000 / 0.02

                   E = -1.5 10⁵ N / C

the sign indicates that the field and the potential are opposite, when one increases the gold decreases

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach

uio

explanation: