An electron is released from rest at the negative plate of a parallel-plate capacitor. if the distance across the plate is 2.0 mm and the potential difference across the plate is 6.0 v, with what velocity does the electron hit the positive plate? ( me = 9.1 × 10^−31 kg, e = 1.6 × 10^−19 c)

Velocity of electron will be

Explanation:

We have given distance across the plate d = 2 mm

Potential difference V = 6 volt

We know that potential difference at any distance is given by

V = Ed , here V is potential difference, E is electric field and d  is distance

So

Charge on electron

We know that expression of velocity is given by , here q is charge on electron, E is electric field and d is distance

So