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

  1. Oct 31, 2018 #1
    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.
    Last edited: Oct 31, 2018
  2. jcsd
  3. Oct 31, 2018 #2


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    Hello. Welcome to PF.
    Your question is a good one. Most of us have gotten confused by this at one time or another.
    Yes. r(hat) points radially outward from q.
    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(180o) = -|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.
  4. Oct 31, 2018 #3
    So since r is decreasing as you come in from infinity, |dr| = -dr.

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

    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!
  5. Oct 31, 2018 #4


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

    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.)

    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.
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