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Force on a point charge due to constant sphere surface charge density.

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data
    The surface of a sphere of radius a is charged with a constant surface density [tex]\sigma[/tex]. What is the total charge Q' on the sphere? Find the force produced by this charge distribution on a point charge q located on the z axis for z > a and for z < a.


    2. Relevant equations



    3. The attempt at a solution
    For z>a, I found [tex]\vec{F}[/tex] = [tex]\frac{\normalsizeq\sigma\hat{z}}{\epsilon_{0}z^{2}}[/tex]

    Now, during my calculations, there was one point where I evaluated the integral for [tex]\normalsize\vartheta[/tex] to be [tex]\normalsize\frac{1}{z}\left(\frac{z-r^{'}}{\left(z^{2}-r^{'}^{2}-2zr^{'}\right)^{1/2}+\frac{z+r^{'}}{\left(z^{2}+r^{'}^{2}+2zr^{'}\right)}[/tex]

    My thinking is that when I simplify the fractions, I obtain |z-r| and |z+r| in the denominator if I want the positive square root, which is the z > a case. So would the z<a case simply mean the square roots would have negative signs out, ie -|z-r| and |z+r| (z+r is stil positive).

    Edit: My latex syntax is messed up. But basically I got (z-r') / (z^2+r'^2-2zr')^(1/2) + (z+r')/(z^2+r'^2+2zr')^(1/2)
     
  2. jcsd
  3. Sep 28, 2009 #2

    gabbagabbahey

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    Hmmm... isn't there a [itex]q[/itex] missing from this expression?:wink:...and [itex]\sigma[/itex] is a charge density, not a charge....so the units on this aren't quite right!

    What did you get for the total charge on the sphere?

    (I fixed your [itex]\LaTeX[/itex]...just click on the images to see the code I used)

    First, aren't you integrating over [itex]r'[/itex]?...It shouldn't be present in your final result!

    Second,

    [tex]|z\pm a|=\left\{\begin{array}{lr}z\pm a &, z\geq \mp a\\a\pm z &, z<\mp a\end{array}\right.[/tex]

    Third, if you've learned Gauss's Law, a clever choice of Gaussian Surface would have allowed you to avoid integrating entirely!:wink:
     
  4. Sep 28, 2009 #3
    Yes, I did forget a q in my answer for z > a.

    The sphere charge density is on the surface, so I would be integrating over the surface area of the sphere (r is constant), correct?

    We're learning Gauss' Law in the next chapter, which is why the tedious integral was carried out here.


    Thanks for confirming my thoughts on where the z < a would come into play in the integral :)
     
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