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Electron, point charge, and surface charge

  1. Jul 16, 2009 #1
    1. The problem statement, all variables and given/known data
    The [itex]z=0[/itex] plane is covered with the uniform surface charge density, [itex]\rho_s=10^{-9}[/itex], while at the same time the point charge [itex]q=10^{-9}[/itex] is located at (0,0,1). Where can an electron be place so if released, it will not move.


    2. Relevant equations

    [itex]\mathbf{E}(\mathbf{r})=\frac{1}{4\pi\varepsilon_0}\sum_{n=1}^N\frac{q_n(\mathbf{r}-\mathbf{r}_n)}{\left|\mathbf{r}-\mathbf{r}_n\right|^3}[/itex]
    [itex]\mathbf{F}=q\mathbf{E}[/itex]


    3. The attempt at a solution

    I assume that the forces will net to zero,

    [itex]\mathbf{F}=e\mathbf{E}_{\rho_s}+e\mathbf{E}_q=0[/itex]
    [itex]\therefore\mathbf{E}_{\rho_s}=-\mathbf{E}_q[/itex]

    I know that, due to symmetry, the surface charge only acts in the z-direction. Since the point charge is given to be at (0,0,1), I expect the solution to be of the form (0,0,m). I tried using the electric field of the surface charge to be
    [itex]\mathbf{E}_{\rho_s}=\frac{\rho_s}{2\varepsilon_0}[/itex]
    but I am not sure that that is acceptable since there is a charge in this problem and that answer is for just the plane alone.

    Not sure where to start really, other than the force equations, but after that I'm stuck as to how to go about solving this problem, any and all help is appreciated!
     
  2. jcsd
  3. Jul 16, 2009 #2

    gabbagabbahey

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    Gold Member

    Is the plane a conductor? Will the point charge induce some charge density on the plane?

    If not, why wouldn't the field due to the plane just be [itex]\mathbf{E}_{\rho_s}=\frac{\rho_s}{2\varepsilon_0}\hat{z}[/itex]?

    Well, you've already posted the field of the plane....what's the field of the point charge [itex]\textbf{E}_q[/itex]? Where does [itex]\textbf{E}_{\rho_s}=-\textbf{E}_{q}[/itex]?
     
  4. Jul 16, 2009 #3
    The electric field of the point charge is
    [itex] \mathbf{E}=\frac{1}{4\pi\varepsilon_0}\cdot\frac{q(0\mathbf{x}+0\mathbf{y}+(1-m)\mathbf{z})}{\left((0)^2+(0)^2+(1-m)^2\right)^\frac{3}{2}}[/itex]
    where I put the reference point at a position m, between 0 and 1 on the z-axis.
    This makes the force equation
    [itex]
    \frac{\rho_s}{2\varepsilon_0}=-\frac{1}{4\pi\varepsilon_0}\cdot\frac{q}{(1-m)^2}
    [/itex]

    Solving this for m,
    [itex]
    m=1-\sqrt{ \frac{2\varepsilon_0}{4\pi\varepsilon_0}\cdot\frac{q}{\rho_s}}
    [/itex]

    This gets me m=0.6011, which I am told is the answer. Thanks for your help! I think my problem was that I was over complicating the problem by assuming that the point charge would have an effect on the surface charge (and vice versa). I'll have to remember these things as I head into grad school this year!
     
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