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Calculate the electric potential of a charged bar

  1. May 30, 2013 #1
    Capture1.JPG
    Consider that the bar has a linear density λ and is in the x axis. Also consider that V=0 in infinite. Determine the potential in point P.


    I can calculate it, but I don't understand how the bounds of the integral are supposed to be determined:
    [itex]dq =\lambda dx'\\ r = x + x'\\dV = \frac{kdq}{r^2} = \frac{k \lambda dx'}{x+x'} \\ V = \int dV[/itex]

    Now here's the problem. I don't understand how the bounds are supposed to be set up. What I mean is, if they should be from x' = 0 to x' = L or the opposite situation.

    [itex]\int_{x'=0}^{x'=L} \frac{k \lambda dx'}{x+x'}[/itex]

    The only difference of this is, of course, a minus or plus sign in the end, but since the electric field can be calculated by the potential, it is important to get the sign right. So, how can I determine the bounds?

    Second question: Why is it important (or relevant) that the question informs me that the potential in infinity is 0?
     
  2. jcsd
  3. May 30, 2013 #2
    Let the left end of the bar be zero. Then you should integrate from L+x to x. I assume you are integrating

    (λ/(l+x)) dl Your differential element, dl, is at a distance x +l from P.

    Does potential have a r^2 or just r?
    I hope this makes sense.
     
  4. May 30, 2013 #3
    Your second question. Potential is 0 at infinity so to find the potential, you integrate from infinity toward the point where thee is a charge. It is analogous to the fact that gravity is zero at infinity and gets bigger as you approach a mass.
     
  5. May 31, 2013 #4

    haruspex

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    You are taking the charge elements to be ##\lambda dx'##. If that is to be positive then you need dx' positive, i.e. x' must be increasing.
    Potentials are always relative. If you don't know the potential at some point then there's an unknown constant in the expression for it.
     
  6. May 31, 2013 #5

    rude man

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    By now you know dV = kdq/r (as others have hinted)
    You know dq = λdr
    So integrate dV = kλ dr/r from x to x + L.

    The order of integration is obvious since you know the answer has to be positive.
     
  7. May 31, 2013 #6

    Integration.png
    Note: I really apologize for my poor paint skills, but I tried to do my best.
    To explain the drawing. It supposed that L > 1 and basically "shows" how I see the x' increasing in my mind.

    Ok so, before I used to think "Well, how can I decide in what order to integrate, left to right or right to left? it seems arbitrary"

    But by reading your comments, it seems that the integration should always go from the element closer to the potential I desire to calculate to the most distant one. Is this correct?

    Yes, I both realize and agree that the answer has to be positive. Still, I would like to understand why I integrate in a certain order to avoid future mistakes.

    Sorry, indeed. I did calculate using just "r" however, when transferring it to latex I for some reason wrote ##r^2##

    But the integral doesn't take that "Infinity" point into account. Let me see if I got it: So if the potential at infinity was 100 V and my integration gave me 200V (completely random values), the potential at point P would be ##{V_p \infty} = V_p - V_\infty = 200 - 100 = 100 V## ? Furthermore, when anyones says ##V_p## what they actually mean is ##V_{p \infty}## ?

    Thanks for everything. And sorry for the possibly obvious questions.
     
  8. May 31, 2013 #7

    haruspex

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    No, you're missing my point. I'm not saying it's an arbitrary choice made just so that the answer comes out positive. You have to think about the sign of dx'. If you want an individual element λdx' to be positive (which you do, because the charge is positive) then you must integrate from the lower value of x' to the higher value - otherwise dx' will be inherently negative.
    [/QUOTE]
     
  9. May 31, 2013 #8
    I think you guys have the wrong idea. You should not be integrating with respect to x. In this problem, you can replace x by a constant and then find the potential at that constant. You should be integrating with respect to l, the distance from the left side of the bar, i.e.; dl. Of course the final answer will be a function of x.
     
  10. May 31, 2013 #9

    rude man

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    If you integrate that way you get V = λ ln 2x/(L+2x) which is < 0 so is wrong.

    If you invert the order of integration you get λ ln (L + 2x)/2x which I think is still wrong.
     
  11. May 31, 2013 #10
    I agree with barryj. I get
    [tex]dV=k\lambda\frac{dl}{(L-l+x)}[/tex]
    Integrating from l=0 to l=L,
    [tex]V=k\lambda \ln \left({\frac{L+x}{x}}\right)[/tex]
     
  12. May 31, 2013 #11

    rude man

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    That's what I get but it isn't what he said to do in his post #2.
     
  13. May 31, 2013 #12
    Hey Rude Man. As usual, we agree on the answer. I guess I misinterpreted what barryj was saying.

    Chet
     
  14. May 31, 2013 #13

    rude man

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    Yay team!
    rudy
     
  15. May 31, 2013 #14

    haruspex

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    Nobody in this thread was integrating wrt x. The integration was wrt x', as introduced in the OP, and completely equivalent to your l.
     
  16. May 31, 2013 #15
    Thanks ! I actually think I got it now, I tried changing the origin (placing the bar in the negative sign, for instance, to change my perspective) and finally grasped this concept.

    I still don't understand very well why I can't have ##V_a## without the information that ##V_\infty = 0##, can't I simply calculate the potential at point a?
    I always thought that ##V_\infty = 0## because [itex]\lim_{r \rightarrow \infty} \frac{kq}{r} = 0[/itex]

    So, summarizing:
    1. Is there any situation where the potential in infinity won't be 0? And can you give me an example?
    2. Why can't I have potential at point A without knowing the potential at infinity?
     
  17. May 31, 2013 #16

    rude man

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    An infinitely long wire with charge density λ. Try computing the potential at a point R away from the wire.
     
  18. May 31, 2013 #17
    Potential is not an absolute quantity. It is always expressed relative to some specific location. It is conventional to take the potential at infinity equal to zero. Then, physically, the potential represents the amount of work required to bring a unit positive charge from infinity to any specific spatial location. But, for practical purposes, there is no reason why one could not take the potential at infinity to be some arbitrary value like say 4 volts. However, the difference in potential between any two spatial locations will remain unchanged.

    The equation [itex]\lim_{r \rightarrow \infty} \frac{kq}{r} = 0[/itex] does not mean that the potential is zero at infinity. It means that the derivative of the potential is zero.

    "Why can't I have potential at point A without knowing the potential at infinity?"
    The potential at point A is not absolute, and must be expressed relative to some reference spatial location (e.g., infinity). Why isn't gravitational potential energy absolute? Because we have to express it relative to some convenient vertical datum. If we change the datum, we change the potential energy, but differences in potential energy between various heights remain unchanged.
     
  19. May 31, 2013 #18

    rude man

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    You can. In fact, with an infinitely long charged wire you have to.

    It's always potential difference. If we can assign the potential = 0 at infinity we often do.
    But sometimes, as in the case of the infinite wire, we can't do that & then we speak of the difference in potential between two points. So in that case you can still assign zero potential at point A and then speak of the potential at point B.
     
  20. May 31, 2013 #19

    TSny

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    Note that whenever you write ##V = \frac{kq}{r}## for a point charge, then that presumes that you are taking V = 0 at infinity.

    So, whenever you are integrating over a distribution of charge and you use ##dV = \frac{kdq}{r}## for an element of charge dq, then you have already imposed V = 0 at infinity.
     
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