Why does electric potential energy increase if you move against the field?

[BuggyWungos]

Homework Statement

"If a proton is moved a certain distance directly opposite an external electric field, what can we say about the change in the electric potential energy of the system?"

Relevant Equations

$$U \propto \dfrac{1}{r}$$

My understanding of this question is that, if you have a proton standing against a positive electric field, and move it in the opposite direction of the field, you're putting in work and therefore should have greater electric potential energy.

But that idea breaks down when you consider a negative electric field. If you move a proton against a negative electric field (farther away), it should decrease in electric potential energy. The answer for this question is, "The electric potential energy of the system will increase." This should only apply to the first situation, right?

 

[haruspex]

There are not two sorts of field, positive and negative. Every electric field has direction which, by convention, is taken to be from positive to negative. So moving in opposition to the field means going towards the more positive end.

 

[Drakkith]

Electric fields are vector fields, meaning that at each point the force has a magnitude and a direction (a vector). There is no such thing as a 'negative electric field'. That would require a scalar field, which is a field that has a single value at each point (like the Higgs field).

As such, when you move a charge with or against the field vectors you will increase or decrease the electric potential energy. Moving a positive charge with the field vectors decreases the potential energy, as field will naturally give up energy to the charge as it moves, meaning you (you, not the field) would perform negative work on the charge to move it. Moving the charge against the field vectors would require you to perform work on the charge to overcome the electric force pushing the charge away, so you would be increasing the potential energy.

The opposite is true for a negative charge. Moving a negative charge against the field vectors decreases the potential energy and moving it with the field vectors increases the potential energy.

 

[BuggyWungos]

Electric fields are vector fields, meaning that at each point the force has a magnitude and a direction (a vector). There is no such thing as a 'negative electric field'. That would require a scalar field, which is a field that has a single value at each point (like the Higgs field).

As such, when you move a charge with or against the field vectors you will increase or decrease the electric potential energy. Moving a positive charge with the field vectors decreases the potential energy, as field will naturally give up energy to the charge as it moves, meaning you (you, not the field) would perform negative work on the charge to move it. Moving the charge against the field vectors would require you to perform work on the charge to overcome the electric force pushing the charge away, so you would be increasing the potential energy.

The opposite is true for a negative charge. Moving a negative charge against the field vectors decreases the potential energy and moving it with the field vectors increases the potential energy.

I think I understand, it's got to do with how the electric force relates to the field. If the electric force points opposite of the electric field on a point charge, and the point charge is moved alongside the electric force, we wouldn't really need to do any work (negative work) as the electric force takes care of that for us, meaning we'd end up with less potential energy than how much we started off with. And vice versa where the electric force and electric field point in the same direction.

 

[BuggyWungos]

There are not two sorts of field, positive and negative. Every electric field has direction which, by convention, is taken to be from positive to negative. So moving in opposition to the field means going towards the more positive end.

In that case, would it be better to imagine a point charge in an electric field to be inbetween 2 oppositely charged plates? I'm still struggling to wrap my head around electric fields a little bit.

A lot of diagrams show electric field lines in negative charges to be radially inwards pointing, but is this only the case due to the convention of having electric field lines point in the direction of where they'll push/pull positive charges?

If the above is the case, would those electric field lines in negatives charges 'flip' in direction if they were measured from the perspective of how they would push/pull an electron?

 

[Drakkith]

I think I understand, it's got to do with how the electric force relates to the field. If the electric force points opposite of the electric field on a point charge, and the point charge is moved alongside the electric force, we wouldn't really need to do any work (negative work) as the electric force takes care of that for us, meaning we'd end up with less potential energy than how much we started off with. And vice versa where the electric force and electric field point in the same direction.

I'd just think of it as this:

If a charge is moved against the electric force, then you are increasing the potential energy. If it is moved with the force then you are decreasing its potential energy. The direction of the force on the particle will depend on the charge and the field.

 

[Drakkith]

A lot of diagrams show electric field lines in negative charges to be radially inwards pointing, but is this only the case due to the convention of having electric field lines point in the direction of where they'll push/pull positive charges?

The direction in which we draw the arrows, from positive to negative, is a convention, yes.

If the above is the case, would those electric field lines in negatives charges 'flip' in direction if they were measured from the perspective of how they would push/pull an electron?

Yes. We could draw them as pointing away from negative charges, in which case negative charges would feel a force that points in the direction of the arrows.