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

In summary, the conversation discusses the calculation of the electric field on the axis of a circular plate with a given charge distribution. After approximating the right hand side of the equation, the field is found to be proportional to 1/z², similar to the field of a point charge. However, there was an error in the algebra that was corrected. The final result is that the electric field is approximately equal to sigma_0 * a^2 / 12 * epsilon_0 * z^2.
  • #1
loops496
25
3
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.
 
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  • #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}
 
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  • #3
Oh my algebra was off on that one! thanks.
 

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

1. What is the formula for the electric field of a circular plate with non-uniform charge density?

The formula for the electric field of a circular plate with non-uniform charge density is E = σ/(2ε0)(1 - r/R), where σ is the surface charge density, ε0 is the permittivity of free space, r is the distance from the center of the plate, and R is the radius of the plate.

2. How does the electric field vary with distance from the center of the plate?

The electric field varies inversely with the distance from the center of the plate. This means that as the distance increases, the electric field decreases.

3. Can the electric field be negative for a circular plate with non-uniform charge density?

Yes, the electric field can be negative for a circular plate with non-uniform charge density. This occurs when the distance from the center of the plate is greater than the radius of the plate, resulting in a negative value for the electric field.

4. How does the surface charge density affect the electric field of a circular plate with non-uniform charge density?

The surface charge density directly affects the magnitude of the electric field. As the surface charge density increases, the electric field also increases. This is because a higher surface charge density means there is more charge per unit area, resulting in a stronger electric field.

5. Is the electric field of a circular plate with non-uniform charge density affected by the shape of the plate?

No, the electric field of a circular plate with non-uniform charge density is not affected by the shape of the plate as long as the charge distribution remains uniform along the surface of the plate. The electric field only depends on the distance from the center of the plate and the surface charge density.

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