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Homework Help: Infinite Line Charges

  1. May 18, 2006 #1
    Calculating Potential from Infinite Line Charge

    Hi there, this is my first post and I hope you don't mind helping me out a bit because I'm really stuck. I'm doing computer science in university and the electrotechnology part can be a bit tricky for us.

    I have a basic understanding of physics, Coloumb's Law, Voltage etc... but I don't know about this infinite line charge stuff. I'm not looking for a solution, I actually have the answer here (it's 6.24v in case you're interested) but I'd like to know how to calculate it. If this wasn't a line charge, i.e. if this was a point charge at the origin, I would be able to do it.

    Thanks in advance :biggrin:

    Edit: I noticed in the FAQ it said I should show my attempts. The thing is, I know how to finish it, just not how to begin it.

    So if I were to go about this, I would get the potential of each point using this formula:

    q/(4)(pi)(eps)(r). I derived that from integrating Coloumb's Law with respect to r. I'd then subtract point b from point a to get the voltage.

    Coloumb's law wouldn't appear to work here though because it's a line charge and not a point charge. This is my attempt, I hope it's not too pathetic. :blushing:
    Last edited: May 18, 2006
  2. jcsd
  3. May 18, 2006 #2
    I see this is being done for a computer science class, are you attempting to calculate the potential from an infinite line of charge by summing up an large number of point charges, as you might do in a computer approximation? Or are you attempting to do an integral and find the algebraic answer?

    I'm guessing you are talking about the algebraic answer:

    You have an algebraic answer for the potential from a point charge correct?

    [tex] V(r) = \frac{1}{4 \pi \epsilon_0}\frac{q}{r} [/tex]

    where r is the distance from the charge to your point:

    Now what you want to do is sum up an infinite number of these small charges, the force law is the same, but your distance will change as you move down the wire. If you are integrating down the z axis from 0 to infinity, try to think of a way to write a Coloumb's law expression for any arbitrary point on the wire z distance from the origin. Then you can integrate over all the possible z's.

    Last edited: May 18, 2006
  4. May 18, 2006 #3
    Lyuokdea, thanks so much for your help! This really clears up a few things in my head.

    Maybe you could just give me one or two more hints though. Essentially what I'm trying to get the potential due to the point charge at R to R+infinite and R to R-infinite. I would imagine integration would be a good way to do this, but I can't see a way to write that. Hmmm
  5. May 18, 2006 #4
    [tex] V(r) = \frac{q}{4 \pi \epsilon_0} \int^r_\infty \frac{1}{r}[/tex]

    This is what I thought of, but it doesn't help because essentially it yields:

    [tex]\ln\frac{r}{\infty} = \ln0[/tex]

    which is undefined. :<
  6. May 18, 2006 #5


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    I can guide you through it but I first want to make sure that I am doing it at the right level.

    There are three approaches possible:

    a) There is a formula for th potential due to an infinite line of charge. If you have seen it in class and you are allowed to use it, the calculation is just two lines. I don't want to make you do an integral if you are allowed to use the formula!

    b) At a more fundamental level, one can actually prove the formula mentioned above using calculus. Do you think this is what you are expected to do? Are you at ease with simple integrals?

    c) it can also be done purely numerically as Lyuokdea mentioned.

    So what is expected of you, do you think?

  7. May 18, 2006 #6
    Hi Patrick:
    I sincerely don't know what we're allowed to use. I would like to see the integral method for my own understanding, but if there is an easier way, I would probably use that in the exam (I am just trying to understand the concepts right now).

    As for proof. I think we just have to arrive at the correct answer after calculating it somehow. I don't think we have to worry about proving formulae, just be able to use them and understand what they do.

    I apologise that I'm very hazy about this. The whole class is on the same boat since we signed up for computer science and this electrotechnology is a little out of our league. We're scared of the lecturer so nobody asks questions. :blushing: Also, the person who is supposed to help us with these problems wasn't able to. :grumpy:
    Last edited: May 18, 2006
  8. May 18, 2006 #7


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    Ah, yes, the scary lecturer :devil: I also try to be as scary as possible when I teach...saves a lot of questions :biggrin:

    There *is* a simple formula that could be used to solve in two lines. But if this was the way to go, you would have seen it in class (I assume). In any case, we'll prove it since you know calculus.

    Let's say that the line of charge is on the x axis, from -infinity to plus infinity. Let's say that we want to find the potential at a point P=(0,y), that is at a distance y above the origin.

    Now consider a small piece of the line of charge of length dx located somewhere to the left of the origin. It contains a small charge dq.

    So, to be clear, this small dq is located at a position (-x, 0). Ok?

    What is the distance r between this small dq and the point P, in terms of x and y? (that's just Pythagora's theorem) That will be what we will use for "r" in the integral.
  9. May 18, 2006 #8
    Patrick, thanks, this makes a lot of sense!

    Would it be something like this then?

    [tex] \frac{q}{4 \pi \epsilon_0} \int^1_\infty \frac{1}{\sqrt{r^2 + n^2}}DN[/tex]

    Thanks very much for the help! I think I'm getting on the right track now :)

    Edit: I can't seem to get it right in Latex but that should be -infinite and +infinite for the definite integral.
    Last edited: May 18, 2006
  10. May 18, 2006 #9


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    Well, I am more used to see x and y used instead or r and n!

    But you almost have it exceot for one thing:

    where does your DN come from? (what I would call dx)
    You cannot put it by hand like this with no justification.

    To be exact, you should have

    [tex] \int_{-\infty}^{\infty} { dq \over {\sqrt{x^2 + y^2}}} [/tex]

    where y is a constant kept fixed (so we could call it "r", the perpendicular distance between the line and the point P).

    The only step missing is to relate dq to dx. You know how to do that?
  11. May 18, 2006 #10
    Patrick: Now I am stuck again. I put in DN (DX) because I thought we have to integrate the formula with respect to X (since we're summing up an infinite amount of points on the X axis). I don't see where you got your DQ from though. And why would we want that as opposed to DX?

    Sorry for being clueless on the matter, I'm familiar with integration, but I think this is the first time I've applied it to a real situation.

    I don't think I know how to relate dq to dx, unless it's that substitution with U routine.

    Thanks again! I'm really learning from this.
    Last edited: May 18, 2006
  12. May 18, 2006 #11


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    Sorry I forgot the 1/(4 pi epsilon_0).

    I slowed you down because this is the most tricky (and important) point of the derivation. And I know by experience that this is always difficult for students applying calculus to physics for the first time.

    The key point is that the potential of a point charge is
    q/(4 pi epsilon_0 r)

    Now, your small dq located at (-x,o) is a point charge, so the potential it produces at the point P (0,y) is simply

    [tex] { 1 \over 4 \pi \epsilon_0} { dq \over {\sqrt{x^2 + y^2}} } [/tex]

    This is *it*!! Just think of dq as being a point charge q...there is nothing else to add!!

    Now, the problem is that before you can integrate, you must express all the variables in terms of a single variable. What varies here is "x" (as you sum over all the small pieces of the line of charge) so you want to relate the infinitesimal dq to an infinitesimal dx.

    The key point is that dq is, by definition, the small charge contained in a small piece of length dx. Therefore, [itex] dq = \lambda dx [/itex] where lambda is the linear charge density. Now you are ready to integrate because you have
    [tex] {1 \over 4 \pi \epsilon_0} \int_{- \infty}^{\infty} \, \lambda dx \, {1 \over {\sqrt{x^2 +y^2}} } [/tex]

    Here, y and lambda are constants. So you are all set, you just have to integrate this. we have done most of the work, which is to get from the physical situation to an integral to calculate. Now it is like pure maths, you just have to integrate this (should make you think about arctan, btw).

    I hope this makes sense.

  13. May 18, 2006 #12
    Hi Patrick,
    Thanks for such a lengthy reply. Once again it's very strange because some of these ideas were never introduced to us in class and now I feel a little angry because of that. :mad:

    I'd like to ask you about two things, namely the dq and the lambda.

    You see, I don't understand why we have this dq when the infinite line charge is equal, there's [tex] {10^{-9}\over 2} C/m [/tex] which is a constant. Also, the above equation has no Q in it so how do I insert the size of the charge?

    And this lambda, what do I do with it? Is there an absolute value for it? How do I get rid of it to get my potential, or does it disappear when we subtract F(b) from F(a)?

    Thanks for your help!
    Last edited: May 18, 2006
  14. May 18, 2006 #13


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    *This* is your lambda! Notice the units: coulomb per meter. What is the meaning of this quantity? Consider the following question: if you take a piece of the line of charge which is, let's say, 80 cm long. How much charge does it contain? The answer is: multiply the length by lambda! So here, you would multiply 0.80 meter with your lambda and get an answer in Coulomb, which represents the amount of charge contained in a section 0.80 m long.

    So, the amount of charge contained in a small piece of length dx is equal to dx times lambda. Since this is a tiny charge, we use the symbol dq instead of q. So dq= lambda dx
    It's the value you gave above. And no, it won't disappear at the end. It will be needed to get your numerical answer at the end.

    Hope this makes sense.

    It is kind of crazy to have you do a problem like this without having explained those concepts!

  15. May 18, 2006 #14
    Hi Patrick, thank you very much. All that makes perfect sense now. You are very kind to explain it all to me! :smile:

    I have to say, that integration at the end is rather complicated.

    I've heard for integrating that format you use

    [tex]\ln{|{x + \sqrt{y^2 + x^2}\over y}|}[/tex]

    This is very complicated to calculate, especially with infinites involed. Might you suggest an easier approach to integrate such a formula?

    Thanks! :biggrin:
    Last edited: May 18, 2006
  16. May 18, 2006 #15


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    [tex] \int {dx \over {\sqr{ x^2 + y^2} }} = \int {dx \over y} {1 \over {\sqrt{ 1 + (x^2/y^2)}}} = \int { dz \over {\sqrt{ 1 + z^2}}} [/tex]

    where z is defined to be x/y (and the limits are -infinity to +infinity in all the integrals).

    This last integral is a standard one. If you have a table of integral, it will surely be listed.

  17. May 18, 2006 #16
    My tables book doesn't have anything like that, I'm sorry. I have a special one for the Irish education system.

    I know this is slightly cheeky, but do you think you could perform the integration on one of the numbers (say the point charge at 2m), I'm really having trouble with it. :frown:

    Thanks a million!
  18. May 19, 2006 #17
    Patrick, or anyone else who might be able to help.

    After many attempts I got the following:

    [tex][\sinh^{-1} ({\frac{x}{2}})]^{+\infty}_{-\infty}[/tex]

    But I can't go on from there. I have to get rid of the infinite somehow, but I can't see how. Can you point me in the right direction? Thank you so much!
    Last edited: May 19, 2006
  19. May 19, 2006 #18


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    Hi.. sorry I replied so late. Got a phone call yesterday night and by the time I was done you had logged out. Also, it was a bit more tricky than I initially thought (for the reason I mention below) so I had to give it some thought.

    The website http://integrals.wolfram.com/index.jsp is useful to do integrals.

    This is a tiny bit more tricky that I was expecting because I had in mind the calculation of the *electric field* at a point, which involves a simpler integral which gives an arctan.

    But we still can do it. going back to the form in terms of x and y (that will make things a bit simpler), we get
    [tex] \int { dx \over {\sqrt{ x^2 + y^2} } } = ln(|x + {\sqrt{ x^2 + y^2}}|) [/tex]

    which I think you gave earlier in the thread! (sorry if I went in a different direction! It's because I still had in mind the result for the calculation of the electric field which involves an arctan). Now, this integral still diverges at infinity!!

    The trick is that we will calculate the difference of potential between the two points before evaluating at x= plus or minus inifnity!

    So the difference of potential between y = 4 and y =2 meters is (I assume that we want the former minus the second one) given by

    [tex] { \lambda \over 4 \pi \epsilon_0} ln {| x + {\sqrt{x^2+16}} |\over| x + {\sqrt{x^2 + 4}}|} [/tex]

    Now we need to evaluate this between x= + infinity and x=-infinity. As x goes to infinity, the argument of the ln goes to 1 so we get ln(1)=0. The case minus infinity is more interesting. Notice that as x goes to minus infinity, we get

    [tex] x + {\sqrt{x^2 + C}} \rightarrow x + |x| {\sqrt{1 + {C \over x^2}}} \approx x +|x| + {C \over 2 |x|} [/tex]
    which, as x goes to minus infinity, goes to
    [tex] {C \over 2 |x| } [/tex]

    Therefore, as x goes to minus infinity, the expression
    [tex]ln {| x + {\sqrt{x^2+16}}| \over| x + {\sqrt{x^2 + 4}}|} [/tex]
    [tex] ln ( {16 \over 2 |x|} {2 |x| \over 4}) = ln({16 \over 4})[/tex]

    Finally plugging this back in the equation for the potential at 4 minus the potential at 2 meters, we get
    [tex] - { \lambda \over 4 \pi \epsilon_0} ln({ 16 \over 4}) = - 2{ \lambda \over 4 \pi \epsilon_0} ln({ 4 \over 2}) [/tex]

    The sign is correct (I took the potential at 4m minus the potential at 2 meters and you expect the potential at 4 meters to be smaller than at 2 meters).

    Notice that you can see the general rule right away: for the difference of potential between a point at r_1 and at r_2, we get

    [tex] V(r_2) - V(r_1) = - 2{ \lambda \over 4 \pi \epsilon_0} ln({ r_2 \over r_1}) [/tex]
    (the sign is correct..if r_2 > r_1, the difference of potential is negative).

    Notice that we can only calculate the difference of potential between two points. Often people will choose to set V=0 at r=1 meter , in which case the equation becomes

    [tex] V(r) = - 2{ \lambda \over 4 \pi \epsilon_0} ln(r ) [/tex]
    with the choice of ground at r=1 meter!

    You may ask: why did we get infinities in the first place? It is because the formula kq/r for a point charge assumes a ground (surgace of zero potential) at r=infinity. This choice is *not* possible for an infinite line of charge.

    Phew...that was quite a post :surprised

    I hope it makes sense. Don't hesitate to ask questions if anything is not completely clear.

    Last edited: May 19, 2006
  20. May 19, 2006 #19
    Patrick, thanks a million! This is a huge help. I know the general formula now, but more importantly I understand how that was derived. In the exam, I'll probably use this formula but explain how I got it (this way, I don't have to do that horrible integration). You've been a huge help, thanks a million!
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