# Potential Difference and Electric Potential

## Homework Statement

A uniform electric field of magnitude 375 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. After the electron has moved 3.2 cm, what is

a. the work done by the field on the electron
b. change in potential energy associated with the electron
c. velocity of the electron

W = -qEd
KE = 1/2 mv^2

## The Attempt at a Solution

As far as math goes, I got all the answers, but I do not have the signs correct.

a. 1.92 x 10^-18 joules
b. -1.92 x 10^-18 joules
c. -2.05 x 10^6 m/s

I got those numbers, but the signs are entirely wrong. My book doesn't explain things very clearly. But what I'm confused about, is when an E field is pointing to the right, doesn't that mean the electron should move to the right? What does negative work mean? How does the electron lose potential energy?

Thanks.

Last edited:

Redbelly98
Staff Emeritus
Homework Helper

## Homework Equations

W = -qEd
KE = 1/2 mv^2

Actually, W = +qEd. (The force is qE, and W=Fd.)

Okay, understood the equation mistake, but can someone answer my questions at the bottom?

cepheid
Staff Emeritus
Gold Member
But what I'm confused about, is when an E field is pointing to the right, doesn't that mean the electron should move to the right?

No. By definition, the direction of the electric field at a point is the direction in which a positive test charge would move if it were placed at that point. A negative charge would therefore move in the opposite of the field direction.

What does negative work mean?

If force and the displacement are in the opposite direction, then it is clear from the definition of work that the work done will be negative. The force acts to impede the existing motion, decelerating the particle. By the work-energy theorem, the work done is equal to the change in kinetic energy. Therefore, if negative work is done, the kinetic energy decreases. That's not what is happening in this situation though. In this situation, positive work is done and the charge is accelerated.

How does the electron lose potential energy?

Just as a mass raised to a certain height in a gravitational field has a certain potential energy, a charge located at a certain position within an electric field results in a system having a certain amount of electric potential energy. Conceptually, this potential energy can be thought of as the work that would have had to be done against the field in order to get the charge to that location. Just as when the mass is dropped and accelerates under gravity it loses potential energy (converted to kinetic) so too does charge lose potential energy (converted to kinetic) when accelerated by a field in the direction that it "wants" to go.