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Cylinder potential problem

  1. Mar 15, 2008 #1
    1. The problem statement, all variables and given/known data

    An infinite grounded conducting cylinder without charge is placed inside a uniform electric field E perpendicular to the surface of the cylinder. Find the electric potential outside the cylinder.

    2. Relevant equations
    3. The attempt at a solution

    I know I have to use the general solution for the Laplace equation in cylindrical coordinates and use the boundary conditions to determine the necessary coefficients. However, I have only one condition (the potential on the surface is equal to V), and after elementary calculations I get: [tex]V(r,\phi)=a_{0} + b_{0}lnr + c_{0}r cos\phi + d_{0}\frac{1}{r}cos\phi - E_{0}rcos\phi[/tex]. The boundary condition gives me two equations, but I have 4 coefficients, so I need another boundary condition. I think something has to be assumed about the potential for [tex]r \rightarrow \infty[/tex], but I don't see any sensible assumptions (we can't just assume that it vanishes there?).
  2. jcsd
  3. Mar 15, 2008 #2
    However, you do know what the electric field is at infinity, and this will give you another boundary condition. You also need to think about where your solution needs values, is the potential infinite or finite valued at the origin?
    Last edited: Mar 15, 2008
  4. Mar 15, 2008 #3
    As [tex]r \rightarrow \infty[/tex] we have [tex]\vec{E} \rightarrow \vec{E_{0}}[/tex], an this implies [tex]c_{0}=0[/tex], i.e. no part with r. V=const inside the cylinder, so at the origin V must not be singular. However, V has two singular parts with opposite signs at the origin [tex]ln(r=0) = -\infty, \frac{1}{r=0} = +\infty[/tex], so we cannot eliminate any of these parts. What can be done?
    Last edited: Mar 15, 2008
  5. Mar 15, 2008 #4
    It is better to use this one as the general solution,

    [tex]V(r,\phi)=a_{0} + b_{0}lnr + c_{0}r cos\phi + d_{0}\frac{1}{r}cos\phi[/tex]

    As [tex]r\rightarrow\infty[/tex], we know the solution must be [tex]-E_{0}r cos\phi[/tex] so [tex]c_{0} = -E_0[/tex].

    The only other term that will be able to satisfy the boundary condition for all values of r is [tex]d_{0}\frac{1}{r}cos\phi[/tex].

    At [tex]r=R, V=0[/tex] and as such, we can then work out the value for [tex]d_0[/tex] with [tex]a_0, b_0 = 0[/tex].

    This satisfies the boundary conditions, and by the uniqueness theorem, we know that this is the only solution that will satisfy these boundary conditions.
  6. Mar 15, 2008 #5
    >>The only other term that will be able to satisfy the boundary condition for all values of r is [tex]d_{0}\frac{1}{r}cos\phi[/tex]

    Why? The [tex]lnr[/tex] term generates an electric field of magnitude [tex]\frac{1}{r}[/tex], so it also vanishes at infinity, just as [tex]d_{0}\frac{1}{r}cos\phi[/tex], which generates a field [tex]\frac{1}{r^2}[/tex], which vanishes at infinity.What's the vital difference between these two terms?
  7. Mar 15, 2008 #6
    The ln(r) term doesn't satisfy the boundary conditions for all phi, given that the current solution has some angular dependence.
  8. Mar 17, 2008 #7
    I'm not convinced. The term ln(r) generates a field [tex]\frac{1}{r}[/tex] which vanishes at infinity, so it doesn't matter if it has any angular dependence or not, since it vanishes anyway - the sole boundary condition for infinity is that [tex]E \rightarrow Ercos\phi[/tex], and electric field [tex]\frac{1}{r}[/tex] has no impact there, with or without angular component If it wasn't ln(r) but something like [tex]ln(r)f(\phi)[/tex], it would be the same.
  9. Mar 17, 2008 #8

    Ben Niehoff

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    Gold Member

    Nope, you can't do this. [itex]\ln r[/itex] and [itex]1/r[/itex] do not approach infinity at the same rate, so they cannot cancel each other. To verify, do it rigorously by combining them into a single fraction and applying L'Hopital's rule:

    [tex]\ln r + \frac1{r} = \frac{r \ln r + 1}r[/tex]

    Clearly, the numerator is of higher order than the denominator, and so this limit will be infinite. You can work out the details yourself if you like.

    But the upshot is that the coefficients of your [itex]\ln r[/itex] and [itex]r^{-n}[/itex] terms must all independently vanish. This leaves you with only half as many coefficients to solve for.

    Edit: Whoops, this only applies INSIDE the cylinder. Outside the cylinder, you don't care if the potential is finite at the origin or not. Hmm...
    Last edited: Mar 17, 2008
  10. Mar 18, 2008 #9
    You don't need to do all of this. Once a solution has been found that satisfies the boundary conditions, you know that this is the only solution by the uniqueness theorem. This means that we can use trial solutions to find a solution to a problem using Laplace's equation.
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