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Homework Help: Calculating Electric Potential: That Tricky Dot Product

  1. May 12, 2007 #1
    I have become exceedingly confused by the various sign changes involved in computing the electric potential produced by a charge distribution, and I am sure I am simply forgetting a negative sign somewhere and am going crazy. However,

    V = -int from infty to r of E . ds
    ds is the infinitisemal vector along to path from infty to r
    so ds = - dr
    and E . ds = - Edr
    which contradicts everything i've ever seen, and ends up with
    V = - Q/{4 pi e_o r}
    instead of V being positive, which it should be.
    This is one of those things which i'll never figure out unless someone else shows me, because I know how to do this problem, but for the life of me I can't find where I went wrong. Can someone please help?

  2. jcsd
  3. May 12, 2007 #2
    Okay, let me clarify that.

    [tex]V(r) = - \int_{\infty}^r \vec{E} \cdot d \vec{s} [/tex]

    If S is the path from [tex]{\infty}[/tex] to r, then [tex]d \vec{s} = - dr \hat{r}[/tex] in polar coordinates.
    So [tex]\vec{E} \cdot d \vec{s} = E \hat{r} \cdot -dr \hat{r} = -E dr [/tex]

    [tex]V(r) = - \int_{\infty}^r -E dr [/tex]
    [tex]V(r) = \int_{\infty}^r E dr [/tex]
    et cetera
    This result being the negative of the correct result. Where did I err?
  4. May 12, 2007 #3


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    What's the actual question? I don't understand why you are integrating from infinity to r. That doesn't make much sense to me!
  5. May 12, 2007 #4
    hm, i thought that was supposed to when finding the potential, because infinity is where the potential is zero.
  6. May 12, 2007 #5
    the actual question was "where am i messing up"
  7. May 12, 2007 #6
    or how to compute electric potential using a line integral
  8. May 12, 2007 #7
    If it's just a sign problem, are you sure you have r and infinity the right way 'round?
  9. May 12, 2007 #8
    every single place i check says that i have the right formula
    [tex]V_B - V_A = - \int_a^b \vec{E} \cdot \vec{ds} [/tex]
    I know this is the correct formula, however, I don't know how to deduce the correct answer when E = Q/(4 pi e_0 r^2) in the positive r direction, b is a point R, and a is at infinity, the point of zero potential. I should be getting Q / (4 pi e_0 R), but im getting - Q / (4 pi e_0 R).
  10. May 12, 2007 #9
    It's like this:

    [tex] V = - \int_{\infty}^R \left( \hat R \frac{q}{4 \pi \epsilon_0 R^2} \right) \left( \hat R dR \right) [/tex]

    This becomes:
    [tex] V = \frac{q}{4 \pi \epsilon_0 R} \,\,\,\, V [/tex]

    Your expression for voltage is fine, your differential line element direction is not. The differential should point to infinity, i.e. you travel towards infinity from the reference point of the point charge.
  11. May 12, 2007 #10
    thank you!
    whats a general rule for knowing what direction the differential vector should point? i.e, when calculating potential difference between two points for a given charge distribution?
  12. May 12, 2007 #11
    Well when you go against the E-field the voltage should be negative. (1)

    I think if you use the expression you are given then you always set it up so the differential follows the electric field. Just make sure you satisfy (1) above.
  13. May 13, 2007 #12
    okay, I think I have this figured out. The problem was that I didn't know how to do line integrals, because we were never taught them. I stared at a calculus book for a long time before I kind of figured it out.

    So [tex]- \int_C \vec{E} \cdot d \vec{s} = \int_a^b \vec{E(s)} \cdot \vec{s}'(t) dt [/tex]
    The kicker is the E(s) -- when you take the path from infinity to R, the easiest s(t) to set up is s = A - t (you eventually take the limit as A approaches infinity). This s is plugged into the equation for E (q/(4 pi e_0 r^2)), and it takes care of the extranegative sign when the integral is performed.
    So my calculus book says that this is the right way to do it, making it a parametric thingamijig.
  14. May 14, 2007 #13
    Dude I feel you. I never really understood line (path) or surface integrals, parametric equations, or any of the later calc III stuff until I took an emag class. I could do it, but I never really understood it. If you are interested in emag, I would definitely recommend picking up: https://www.amazon.com/Field-Wave-E...195/ref=pd_bxgy_b_text_b/103-8303814-4577433"

    You can probably get it from your school's technical library. It has a great math primer.
    Last edited by a moderator: May 2, 2017
  15. May 14, 2007 #14
    Hey thanks for you help. I'll look into that book.
    The thing about these integrals is sometimes it looks like something you can sort of do mentally - like really simple applications of Gauss' Law - but it isn't - like the problem I just had.
  16. May 14, 2007 #15
    I don't understand the confusion over here. The integration is pretty straightforward. :biggrin:
    Let me start with the original equation for the electric potential:
    [tex]V = -\int_{\infty}^{r} \vec E \cdot d\vec r [/tex]

    Plug in the expression for the Electric Field:

    [tex]V = -\int_{\infty}^{r} \frac{Q}{4 \pi \epsilon_0 r^2} \hat r \cdot \hat r dr = -\frac{Q}{4 \pi \epsilon_0} \int_{\infty}^{r}\frac{dr}{r^2} = -\frac{Q}{4 \pi \epsilon_0} \left[\frac{-1}{r}\right]_{\infty}^{r}[/tex]

    Final evaluation gives expression for V.
    Last edited: May 14, 2007
  17. May 14, 2007 #16
    thats how i had been doing it for a very long time, and it is correct
    but then I started trying to learn the more formal way to do line integrals
    and I couldn't understand why [tex]d \vec{s} [/tex] was not equal to -dr, because it starts at infinity and goes to r. However, with frogpad's excellent help i eventually figured out that [tex]d \vec{s} [/tex] is a vector valued function (ihope that is the right term) that encompasses all the vectors from the origin to points on the path from infinity to r -- so it is pointing in the positive r direction. this doesn't change the way i'll be solving the integral, but it does clear things in my head conceptually.
  18. May 18, 2007 #17
    Hey there is no question of confusion.Tbonepower07 you are not at all messing up anywhere.Let's talk about simple terms.We know that work done per unit charge define electric potential i.e. V=dw/dq.Also V=W/q not.Therefore,
    V=(-q/q not)/(4 x pie x abselon not x R x q not).So,q not cancelles from Nr and Dr.Hence V= -q / (4 x pie x abselon not x R ).IN CONCEPTUAL terms work done on the system will be -ve and work done by the system will be +ve.
  19. May 21, 2007 #18
    Your pronunciations are hilarious. :rofl:
    It is not "not". It is "naught".
    Epsilon [itex]\epsilon[/itex],[itex]\epsilon_0[/tex]
    Pi [itex]\pi[/itex]
    Click on the symbols.
    Last edited: May 21, 2007
  20. May 21, 2007 #19
    OK ! Thanks Reshma .I will keep that in mind from now on.
  21. Aug 21, 2011 #20
    Is it possible to also flip the negative integral here so rather than it being the -(integral) between the limits r and infinity it becomes the positive integral and the limits switch?

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