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Deriving expression for potential of a point charge

  1. Apr 5, 2017 #1
    1. The problem statement, all variables and given/known data
    I am trying to derive an expression for the potential of a positive point charge by bringing in another positive test charge in from infinity to a point at a distance R from the point charge.

    2. Relevant equations
    $$V_f - V_i = - \int \vec E \cdot d \vec r \, dr$$

    3. The attempt at a solution
    Being that the test charge is coming in from infinity ##\vec E \cdot d \vec r = Edrcos180 = -Edr##
    So then $$V_f - V_i = \int E \, dr$$
    Then taking ##V_i = 0## at infinity
    $$ V= \frac q {4\pi\epsilon_o}
    \int_\infty^R \ {r^{-2}} \, dr =
    \left. - \frac q {4\pi\epsilon_o r} \right|_\infty^R$$

    This gives me ##V = - \frac q {4\pi\epsilon_o R}##

    Which is almost right except I have that negative sign that says the potential is decreasing...
    I thought since the potential is taken to be 0 at infinity and that the test charge is getting closer to the charge I would get a positive result saying the potential is increasing.

    Could someone help explain why I got this negative result?
     
  2. jcsd
  3. Apr 5, 2017 #2

    TSny

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    Hello, Welcome to PF!
    This should be written ##\vec E \cdot d \vec r = E|dr|cos180##. What happens if you now remove the absolute value sign on dr? Keep in mind that when integrating from ∞ to R, dr will be a negative quantity.

    Alternately, before doing the dot product, reverse the limits on your integral so that you integrate from R to ∞. Then dr will be a positive quantity.
     
  4. Apr 5, 2017 #3
    Thank you for replying.
    However, I am still not getting it.

    Wouldn't dr be a negative quantity since it is pointing radially inward? So then its magnitude would still be positive? Why would I need to remove the absolute value sign?

    Thanks.
     
  5. Apr 6, 2017 #4

    TSny

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    I think part of the confusion is using ##d\vec{r}## for the displacement along a path and also using ##dr## for the change in the radial coordinate ##r## (which increases in the radially outward direction). Suppose we let ##d\vec{s}## be the displacement along an arbitrary path and imagine doing ##\int_a^b \vec{E}\cdot d\vec{s}## along some arbitrary path connecting points ##a## and ##b##. If we agree to let ##E## stand for the magnitude of ##\vec{E}## and ##ds## stand for the magnitude of ##d\vec{s}##, then the path integral may be written as ##\int_a^b E ds \cos\theta## .

    Now suppose we apply this to a radial path that starts at infinity and goes to a radial distance ##R## from the origin. Let ##r## be the radial distance from the origin as used in the expression ##E = \frac{kq}{r^2}##. Then if we choose ##r## as the integration variable along the radial path from infinity to ##R##, ##dr## will be negative for each infinitesimal step along the path. So, the magnitude of the step, ##ds##, would be ##ds = -dr##. So, you would have ##\int_\infty^R E ds \cos\theta = -\int_\infty^R E dr \cos\theta##. In this integral, ##dr## does not represent the magnitude of the displacement (i.e., a positive quantity); rather, ##dr## represents the actual change in the radial distance ##r## when making a step along the path. So, ##dr## is a negative quantity while ##ds## is a positive quantity.
     
    Last edited: Apr 6, 2017
  6. Apr 6, 2017 #5
    ahh ok, yeah, that clears up a lot of my confusion. Thank you.
    So with ##d\vec s,## being the displacement vector along the path could it be represented by ##dr\hat r## ?
    In this case since it is in the decreasing radial direction ##d\vec s = -dr\hat r##
    I may just be confusing these two again, but I am trying to find how to mathematically arrive at
    ds = -dr.
     
  7. Apr 6, 2017 #6

    TSny

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    Yes, this is correct for a radial path
    No. Whether you are going outward along a radial path or inward along a radial path, you would always have ##d\vec s = dr\hat r## (assuming that ##\hat r## is always defined in the radially outward direction). When going outward, ##dr## is positive. When going inward, ##dr## is negative.

    ##ds## represents the magnitude of the displacement. So, ##ds## is always positive. ##dr## represents the change in the radial coordinate ##r##. So, ##dr## is positive when moving radially outward, it is negative when moving radially inward. For moving outward or inward along a radial path, you could always write ##ds = |dr|##. When moving inward, ##|dr| = -dr##. So, when moving inward, ##ds = -dr##.
     
  8. Apr 6, 2017 #7

    kuruman

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