Question

A uniform electric field of magnitude 442 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.50 cm. (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?

Answer 1

A) The electric field is constant and in the same direction as the electron's displacement, so the work done is given by:

W = Fd

W = work, F = electric force, d = displacement

The electric force on the electron is given by:

F = Eq

F = electric force, q = electron charge

Substitute F:

W = Eqd

Given values:

E = 442N/C, q = 1.6×10⁻¹⁹C, d = 3.50×10⁻²m

Plug in and solve for W:

W = 442(1.6×10⁻¹⁹)(3.50×10⁻²)

W = 2.48×10⁻¹⁸J

B) The electric field does work to move the electron. Apply the conservation of energy and you'll see that the electron's potential energy loss is equal to the work done by the field in moving the electron.

The electric potential energy change is -2.48×10⁻¹⁸J

C) Apply the work-energy theorem; the electron's kinetic energy equals the work done on it by the field.

KE = 0.5mv² = W

m = electron mass, v = velocity, W = work

Given values:

W = 2.48×10⁻¹⁸J, m = 9.11×10⁻³¹

Plug in and solve for v:

0.5(9.11×10⁻³¹)v² = 2.48×10⁻¹⁸

v = 2.33×10⁶m/s