Question

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 1.5 x 10-7 C/m2, and the plates are separated by a distance of 1.4 x 10-2 m. How fast is the electron moving just before it reaches the positive plate?



THANK YOU for any help or hints or anything that you can give me!

Answer 1

I would find the voltage on the capacitor, or Electric field first. Using Gauss Law, then E= sigma/2epsilion

Force/q= E
Force=Eq
force*distance=Eq*distance
1/2 m v^2=Eq*distance
solve for velocity v.

Answer 2

The E field between the plates is proportional to the charge per unit area, sigma (1.5 x 10-7 C/m2).

E = sigma/epsilon

where epsilon is the permittivity of free space, which you should look up.

The kinetic energy that is given to the electrion while crossing the gap of length X is

KE = e E X

Set that equal to the final kinetic energy and solve for the velocity.

Answer 3

Dr WLS is correct, E between the plates is sigma/episilon, not sigma/2epsilon.