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Electric Boundary Value Problem

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data

    A pair of infinite, parallel planes are equipotential surfaces. The plane at z = 0 has an electric potential of 0 and the plane at z = b also has a potential of zero. The electric field at b is 0 at time t at which there is a constant, positive charge density between the planes of Rho_naught.

    (i) Use Poisson's eqn. to write down the appropriate diifferential equation for 0 <= z <= b.

    (ii) Solve the equation to get the potential in terms of arbitrary constants.

    (iii) Use the resulting potential to find the vector electric field in terms of arbitrary constants.

    (iv) Use the boundary conditions to find the constants.

    2. Relevant equations

    Poisson's eqn = Laplacian (Psi) = Del ^2 Psi = d2Psi/dx2 + d2Psi/dy2 + d2Psi/dz2 = -Rho_naught/epsilon_naught

    E = - Del(Psi)

    3. The attempt at a solution

    Honestly I am at a loss. I don't understand how the E field can be zero everywhere at b if there is a charge density right next to it.

    I have no idea how time comes into this.

  2. jcsd
  3. Dec 1, 2008 #2
  4. Dec 1, 2008 #3


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    Do you understand how the plane at z=0 can have a voltage of V=0 when there is a charge next to it? The mention of time is confusing to me as well; I would just ignore that bit.

    Do you know the method of separation of variables? Do you know what is a Green function?
  5. Dec 2, 2008 #4
    I can understand that you could define the voltage as being zero there. But if you are next to a charge you are experiencing an E field unless there is a field equal and opposite.

    I do know separation of variables, I don't know Green's function but we aren't 'supposed' to know that for this apparently.
  6. Dec 2, 2008 #5
    Ok, I'm thinking about it some more. There is a pair of planes, parallel and infinite in area. So basically the symmetry is of the kind that only one dimension matters, since it is cartesian, it is z in this case.

    [tex]Lap(\Psi) = \nabla^{2}\Psi = \frac{d^{2}\Psi}{dz^{2}}=-\frac{\rho_{0}}{\epsilon_{0}}[/tex]







    [tex]\vec{E} = -\nabla\Psi=-\nabla[-\frac{\rho_{0}}{2\epsilon_{0}}z^{2}+zc_{1}+c_{2}][/tex]

    [tex]\vec{E} =-\frac{d}{dz}[-\frac{\rho_{0}}{2\epsilon_{0}}z^{2}+zc_{1}+c_{2}]\hat{z}[/tex]

    [tex]\vec{E} =-[-\frac{\rho_{0}}{\epsilon_{0}}z+c_{1}]\hat{z}[/tex]


    At z=b, E = 0;

    [tex]0 =-\frac{\rho_{0}}{\epsilon_{0}}b+c_{1}[/tex]


    At z=0, Psi = 0;


    [tex]0 =-\frac{\rho_{0}}{2\epsilon_{0}}0^{2}+0c_{1}+c_{2}[/tex]


    Does this seem like a plausible solution?

    Note: I still don't get how E could be zero at the surface.
  7. Dec 2, 2008 #6
    Your solution looks okay to me, but I think there is something wrong with the problem. I don't think you can have both planes having zero potential, and the electric field being zero at one plane. You can only satisfy two of these conditions. For example, your solution does not give psi = 0 at z = b. However, it does satisfy the two boundary conditions you chose to impose.

    As to how the field can be zero, what can cause an opposing field to cancel out the field due to this volume charge?
  8. Dec 2, 2008 #7
    For the last question, another plane charge configuration could do it. If there was a plane charge configuration of positive charge with equal effectiveness opposite the point that you are measuring at, it would cancel out the E field. But this would also double the efield on the other side, meaning that the potentials couldn't be the same going from one plate to the other. This question is all a convoluted mess to me. By the way, this is a math methods of physics class, not a physics class, so we aren't supposed to know what is happening, but I obviously would like to know as it helps in solving the problem!
  9. Dec 2, 2008 #8


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    Oh, sorry Prologue. I didn't catch that. The electric FIELD is zero. I have no idea how this could be guarunteed for an arbitrary charge density between the planes. This problem seems strange to me.

    Oh, and due to the presence of the charge density, which I'm assuming is arbitrary, the problem is not x,y symmetric, and you cannot reduce Poisson's equation to a 1-D equation like that.
  10. Dec 2, 2008 #9
    That's true turin, I didn't give it all that much thought. I was taking constant to mean it is the same everywhere, while you are right that it could just mean that it is not changing but be different in different points. I am betting that it is meant to mean the same everywhere.
  11. Dec 2, 2008 #10
    1. I think treating it as a constant charge density is fine with the problem stated the way it is. When you say [tex]\rho_{0}[/tex] with a subscript, it is usually meant to be constant. The problem does say "constant, positive charge density" :wink: Being "different at different points" is not constant!

    Without this symmetry, you'd have to do variable separation, as turin says.

    2. You are (almost) right about the surface charge also. There are surface charges at both the planes (assume metal plates for example, these are good equipotential surfaces), these can guarantee that the field is zero at z = b and that the potential is zero at z = 0. You are right about the fact that this will mean that there is a potential difference between the plates. As I said before, you cannot guarantee that the plates are at the same potential, if you want to guarantee that the field is zero. Mathematically, you have two constants c1 and c2, but three boundary conditions. In general you cannot satisfy all these boundary conditions.
  12. Dec 2, 2008 #11
    This was a problem on a previous test, and it could have been a rewrite of a rewrite of a etc. My professor has been teaching this class for probably 30 years, so it is completely possible that something is off in the problem due to a mistake in a rewrite of the problem.

    Thanks for the help everyone.
  13. Dec 3, 2008 #12


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    "constant" does not mean that it has to be the same at every spatial point. There is a word for that condition: "uniform". "constant" may mean "uniform" as a special case, but it doesn't have to. I still think this problem is strange.
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