Potential due to Electromagnetic field

[spark693]

Homework Statement

How much work does the electric field do in moving a proton from a point with a potential of +260 V to a point where it is -48.0 V? Express your answer in electron volts.

Homework Equations

V = p(A) - p(B)

The Attempt at a Solution

V= 260 [v] - (-48.0[V]) = 3.08×102 [V]

Right answer: -3.08×102 [V]

I got the right answer, but the wrong sign. I'm not sure if this has anything to do with a proton moving from a positive to negative potential (this would only be important if it was moving from a positive to negative charge?).
Why would the answer be negative?

 

[gneill]

The change in potential is the potential at the final location minus the potential at the initial location. The proton's starting at a location where the potential is +260V and ending up at a location where it's -48.0V. So ΔV = (-48V) - (+260V) = -308 V.

That's the change in potential. But it's not the work done by the field on the electron. To find the work done there's one more step. Do you know what that is? Hint: The Volt is a compound unit which can also be written as Joules/Coulomb.

 

[spark693]

To find work, I multiplied by 1.6x10^-19 C which is the charge of one proton, and that gives me work in Joules?

 

[gneill]

To find work, I multiplied by 1.6x10^-19 C which is the charge of one proton, and that gives me work in Joules?

Yes indeed. Work = q ΔV

 

[spark693]

One more question:
What does it mean that the work is negative? I looked up negative work, and the definition given was that it means the force is hindering movement, or that it is opposite to the displacement. I was confused, as I thought that a proton moving from higher potential to lower potential is movement in the direction that the electric field is pointing.

 

[gneill]

It means that if you wanted to move the proton from +260V do -48V slowly and without letting it accelerate, you would have to apply a continuous force F to keep it from being accelerated That force opposes the direction of motion. The quantity F*d is the work done in moving the charge from A to B. Since the direction of the force opposes the direction of motion, the work is negative.

 

[spark693]

So the electric field's work is all about not letting the proton accelerate while moving it?
I'm not sure if this is correct, but I'm picturing the proton naturally moving from the point of higher potential to the point of lower potential (requires no work from the electric field), and the electric field works against it (negative work), only so the proton does not accelerate.

 

[gneill]

I retract my above statements The field will want to accelerate the proton in its move from a location of +260V potential to one of -48.0V. Thus the force due to the field is in the direction of motion, and so the field must be doing positive work in moving the proton from the initial position to the final position. So it looks like the "right answer" that was given to you wasn't right at all, and your own solution is correct after all.

I think I was thinking about the work required by an external agent to move the charge. Here the filed is doing the job.