A uniform electric field of magnitude 410 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.00 cm.

Question

(a) What is the work done by the field on the electron?(b) What is the change in potential energy associated with the electron?(c) What is the velocity of the electron?
I need help with C. I have the work as 1.97e-18, b is -1.97e-18. and for C I am using the square root of the answer for v = 2/m(w). My answer is 4.85e-23. I need it to be converted to m/s. I have no idea how but the answer is supposed to be something like 1.9e+06

Answer 1

Your numbers are not useful or correct without the dimensions that go with them.

When an electron moves 0.03 m under the influence of a 410 N/C E-field, work is done on it equal to
410 N/C*1.6*10^-19C*0.03m = 1.968*10^18 J

That becomes its kinetic energy.

So your (a) and (b) are correct.

V = sqrt(2E/m), where m is the mass of the electron and E is the kinetic energy.

The mass of an electron is 9.11* 10^-31 kg

Answer 2

Awesome! Thanks it worked out, I had the mass of an electron wrong. That was my problem there.