Question

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area of each plate is $\sigma$ = 2.0 x 10^-7 C/m², and the plates are seperated by a distance of 2.1 x 10^-2 m. How fast is the electron moving just before it reaches the positive plate?

Answer 1

potential difference between the plates V = E d = sigma x d / eo, where 'sigma' is the charge density and 'eo' is the permittivity of air

then, if 'v' is the velocoty of electron at the positive plate, by conservation of energy

1 / m v^2 = V e or, v = square root of ( 2 V e / m )

= square root of ( 2 sigma d x e / eo m )

= square root of [ 2 x 2.0x10^-7 x 2.1x10^-2 x1.76x10^11 / 9 x10^-12]

ie., v = 1.28 x10^7 m /.................this is the speed at which the electron moving just before it reaches the positive plate