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Electric field of a circular plate with non uniform charge density.

  1. Aug 24, 2014 #1
    Hey!

    I need to calculate the electric field on the axis of a circular plate of radius a with the following charge distribution:
    [tex]\sigma_0 \frac{r^2}{a^2} \delta (z), \; r\leq a[/tex][tex]0, \; r>a[/tex]
    where [itex]\sigma_0[/itex] is a constant.
    I've already calculated the potential and taken its gradient to get the field, which is only in z and its given by,
    [tex]E_z = \frac{\sigma_0}{2 \epsilon_0 a^2} \left[\frac{4}{3}z \sqrt{a^2+z^2}- 2z \sqrt{z^2} - \frac{z(a^2 - 2z^2)}{3 \sqrt{a^2+z^2}}\right][/tex]

    Now I have to take the approximation [itex]z>>a[/itex] and describe the field. Intuitively it should be the like the field of a point charge, but i can not get the [itex]1/r^2[/itex] dependence after the approximation.
    Is there something wrong?

    Thanks in advance.
     
  2. jcsd
  3. Aug 24, 2014 #2
    You just need to approximate your right hand side. Taking just the square brackets, it is equal to:
    \begin{align}
    &\frac{4}{3}z^2\sqrt{1+\frac{a^2}{z^2}}-2z^2-\frac{2z^2\left(\frac{a^2}{2z^2}-1\right)}{3\sqrt{\frac{a^2}{z^2}+1}}\simeq \frac{4}{3}z^2\left(1+\frac{a^2}{2z^2}\right)-2z^2-\frac{2}{3}z^2\left(1-\frac{a^2}{2z^2}\right)\left(1-\frac{a^2}{2z^2}\right) \\
    &=-\frac{2}{3}z^2+\frac{2}{3}a^2+\frac{2}{3}z^2-\frac{2}{3}a^2+\frac{a^4}{6z^2}=\frac{a^4}{6z^2},
    \end{align}
    and so your field is:
    \begin{align}
    E_z\simeq \frac{\sigma_0a^2}{12\epsilon_0z^2}.
    \end{align}
     
  4. Aug 25, 2014 #3
    Oh my algebra was off on that one! thanks.
     
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