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Electric potential, hard problem

  1. Feb 11, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data

    There's a cylindrical conductor of radius d, whose linear charge density is [tex]\lambda[/tex] which is situated at a distance D from a infinite conductor plane.
    1)Calculate the potential in the region that starts from the plane and ends at infinity (hence the cylindrical conductor is inside the region).
    2)Determine the capacity of the system.
    2. Relevant equations
    None given. I guess [tex]\vec E=-\vec \nabla \phi[/tex].


    3. The attempt at a solution
    I realize I must find the potential in a 2 dimensional region, which seems really complicated to me.
    What I've done so far is [tex]\phi (r)=-\int _{\infty}^r \vec E \cdot d\vec s[/tex]. I calculate the E field to be [tex]E=\frac{2k\lambda}{r}[/tex] outside the cylinder (inside E is worth 0 since it's a conductor, hence the potential inside the cylindrical conductor is constant and is worth the value of the potential over the surface of the cylindrical conductor).
    But now, my big problem is that I have that [tex]\phi (r)=-\int _{\infty}^r \frac{2k\lambda dr}{r}=-2k\lambda \ln \left ( \frac{r}{\infty} \right )[/tex], which diverges... I don't know what I'm doing wrong. I feel really strange, how is that possible that I get all wrong? It seems so simple, however I keep getting impossible results.
     
  2. jcsd
  3. Feb 11, 2010 #2

    collinsmark

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    Is the cylinder infinite in length? It's not really clear from your problem statement.

    If the cylinder is infinitely long, you can't pick infinity as a reference for the potential. It will blow up, as you have shown. You can still do the problem though, but you'll just have to pick some other point as your reference. In other words, instead of saying, "the potential with respect to infinity is this...", say, "the potential with respect to point A is...", once you've decided on a suitable point A.

    On the other hand, if the cylinder is finite in length, you should be able to use infinity as a reference point. Of course the relevant equations will be a lot tougher to work with if the cylinder is finite in length.
     
    Last edited: Feb 11, 2010
  4. Feb 11, 2010 #3

    fluidistic

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    Thank you, it makes perfectly sense. Yes it is infinite. So basically if someone asks me what is the energy required to construct the configuration of charges (namely the charged cylinder), I have to say it's in fact impossible because the energy would be infinite.
    Ok I understand that I have to define the difference of potential between 2 points, in which infinity is discarded.
    I talked to a friend about this problem an hour ago, he told me to use the method of images. I'm not sure if it would work though. I never seen the demonstration nor the axioms that this method uses. Hence I have no idea if I can use it in this problem.
     
  5. Feb 11, 2010 #4

    collinsmark

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    Forgive me, I may have spoken too soon. What I said before about not being able to choose infinity as a reference point would be true if the infinite plane was insulating, but you might be able to get by with using infinity as a reference because the infinite plane is conductive. Please forgive me for the confusion. Either way though, you could still do with problem if you picked some finite point.

    Now back to the problem. There is a possible way to do this. Keep in mind the following things:

    (a) The electric field lines immediately outside a static conductor (not carrying current) are always perpendicular to the surface.
    (b) The electric field inside a static conductor is zero.
    (c) This means that the charge distribution on the surface of the conductor aligns itself to force the electric field inside the conductor to be zero.

    So the equipotential field lines created from the cylinder will always end up hitting the infinite plane conductor at right angles to the plane -- always.

    What other situation could exactly model the electric field produced by this, at least in the region above the plane? (Hint: think mirror. Hint: what if we replaced the conductor [mirror] with a cylinder of opposite charge, in exactly the same place where the image of the first cylinder was in the mirror -- so instead of an image of a cylinder in a mirror there is actually another cylinder of opposite charge [and no mirror]).

    [Edit] Before this edit, I said "sort of easy". I've now changed that to "possible."
     
    Last edited: Feb 11, 2010
  6. Feb 11, 2010 #5

    collinsmark

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    Let me put it another way.

    Consider placing the conducting plane such that it crosses the z-axis at z = 0, and runs along with both the x and y axes. Suppose the cylinder goes parallel to the y-axis at z = D.

    Your vector variable [tex] \vec r [/tex] is in reference to the y-axis itself, not the center of the cylinder!

    For the cylinder, create a new radius vector variable (I'm having a problem with getting my desired font for this variable, so I'm just going to use a capitol R -- I'd rather use a script r, but I can't seem to do it), called [tex] \vec R [/tex]. Define things such that

    [tex] \vec r = \vec R + \vec D, [/tex]

    where [tex] \vec D = (0, 0, D). [/tex]

    In summary, [tex] \vec r [/tex] moves around the y-axis and [tex] \vec R [/tex] moves around the center of the cylinder. Therefore,

    [tex] \vec R = \vec r - \vec D [/tex]

    Now go ahead and solve the potential integral, choosing a point on the y-axis as the reference.

    [tex] \phi(\vec R) - \phi(\vec D) = -\int _{D} ^{R} \frac{\lambda}{2 \pi \epsilon _{0}} \frac{1}{R'} dR' [/tex]

    Note that under the integral, [tex] R [/tex] and [tex] D [/tex] are scalars -- the absolute values of corresponding vectors. That is intentional. Here, [tex] R' [/tex] is a dummy scalar variable that gets integrated out. Note that the above integral is for a wire. Do whatever you think is necessary to properly represent a cylinder.

    Then make the substitution to get the equation for [tex] \phi(\vec r). [/tex] When you make the substitution, remember [tex] R = \left| \vec r - \vec D \right| [/tex].

    But you're not finished yet. You need to do the same thing again for the image cylinder -- the image created by the conductor. Then sum the results together.

    But remember to place the image on the other side of the conductor. That means your equation that relates [tex] \vec R [/tex], [tex] \vec D [/tex], and [tex] \vec r [/tex] will be a little different.

    Also remember to give the [tex] \lambda [/tex] a minus sign, since its negative the charge of the real cylinder.

    I chose the y-axis as my reference. But after you complete everything above, if you let r --> infinity, the potential doesn't blow up. So if you wish, at this point you can switch to using infinity as your reference by adding a constant, if you wanted to. ('Turns out to be trivial in the case, I believe. Quite trivial, in fact.)
     
    Last edited: Feb 12, 2010
  7. Mar 1, 2010 #6

    fluidistic

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    Sorry for being so late, but thanks a lot for your help!
     
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