The sign of F (dot) dl when finding electric potential

[EquationOfMotion]

The electric potential can be defined as

V = - ∫_C E⋅dl

where we are taking the line integral along C from some convenient reference point O, where we have set V = 0, to the point r we are trying to find the potential at. Of course, C can be any curve, but it's usually the most convenient to take it as a straight line from O to r.

Doesn't this mean dl will point in the direction from O to r? If that is the case, then say we're trying to find the potential at some distance from point charge +q and we've set our reference point infinitely far away. We have E = (kq/r^2) r(hat), where r(hat) is the spherical unit vector. Then, E⋅dl = E r(hat)⋅dl. But dl points from infinity to the point we're trying to find, and say that we've picked O so that dl points in the -r(hat) direction. Then, E⋅dl = -Edr, so evaluating the integral, we get V = -kq/r, which is evidently wrong.

I am aware that we can fix this by taking O to be somewhere else, but do you always take the line integral to be positive? If so, why? Or is there something else I'm missing?

EDIT: This seems like it'd fit better in General Physics. Woops.

 

[TSny]

Hello. Welcome to PF.
Your question is a good one. Most of us have gotten confused by this at one time or another.

Doesn't this mean dl will point in the direction from O to r?

Yes

If that is the case, then say we're trying to find the potential at some distance from point charge +q and we've set our reference point infinitely far away. We have E = (kq/r^2) r(hat), where r(hat) is the spherical unit vector.

Yes. r(hat) points radially outward from q.

Then, E⋅dl = E r(hat)⋅dl. But dl points from infinity to the point we're trying to find, and say that we've picked O so that dl points in the -r(hat) direction.

OK

Then, E⋅dl = -Edr,

This is the mistake. The dot product of two vectors A and B is A⋅B = |A| |B| cosθ. So, in your case (coming in from infinity),
E⋅dl = |E| |dl| cos(180^o) = -|E| |dl| = -E |dr|. When integrating from ∞ to r, how should you write |dr| in terms of dr? In particular, is dr a positive quantity or a negative quantity in this case? Hint: is r increasing or decreasing?

Or you can think about it more directly. When coming in from infinity, is E⋅dl a positive quantity or a negative quantity (assuming q is positive)? Now consider the expression E dr when coming in from infinity. Is this a positive quantity or a negative quantity?

Or, you can approach it by convincing yourself that dl = dr r(hat) whether you are coming in from infinity along a radial line or going out to infinity along a radial line. So, you can write E⋅dl = E⋅r(hat) dr for either incoming or outgoing.

 

[EquationOfMotion]

When integrating from ∞ to r, how should you write |dr| in terms of dr? In particular, is dr a positive quantity or a negative quantity in this case? Hint: is r increasing or decreasing?

So since r is decreasing as you come in from infinity, |dr| = -dr.

Or you can think about it more directly. When coming in from infinity, is E⋅dl a positive quantity or a negative quantity (assuming q is positive)? Now consider the expression E dr when coming in from infinity. Is this a positive quantity or a negative quantity?

E⋅dl is negative, while E dr is positive so you need a negative sign. Nice!

Or, you can approach it by convincing yourself that dl = dr r(hat) whether you are coming in from infinity along a radial line or going out to infinity along a radial line. So, you can write E⋅dl = E⋅r(hat) dr for either incoming or outgoing.

Is this because going out it's obviously r(hat) dr, and coming in it's (-r(hat))(-dr)?

Thank you very much for your help!

 

[TSny]

So since r is decreasing as you come in from infinity, |dr| = -dr.

Yes. dr represents the infinitesimal change in r. So dr is negative when coming in.

E⋅dl is negative, while Edr is positive so you need a negative sign.

E⋅dl is negative when coming in. But Edr is also negative when coming in. Remember, dr is a negative quantity since r is decreasing. So, E⋅dl has the same sign as Edr. You can check that these two expressions also have the same sign when going out. So, E⋅dl = Edr for both coming in and going out. (These statements assume that q is positive.)

going out it's obviously r(hat) dr, and coming in it's (-r(hat))(-dr)?

You are saying that dl = r(hat) dr when going out and that dl = -r(hat) (-dr) = r(hat) dr when coming in. Yes, that's true.

When coming in, dl = -r(hat) |dr| = -r(hat) (-dr) = r(hat) dr.