I have a uniformly charged disk, radius $R$, with surface charge density $\sigma$. I want to find the electric field along the axis through the centre of the disk (which I've called the $x$ axis).

I am aware that one method is to consider small rings and integrate from $0$ to $R$, but that is not how I approached the problem and so I would like to know how to do it my way.

I considered a small section on the disk (I don't know how to formulate its charge), but it had position vector $r \hat{r}$ in cylindrical coordinates. Then I got stuck on how to express the charge of that small section. I think it is $dq = \sigma dz dr$ (again in cylindrical coordinates), but I am not sure.

Any help is appreciated!


If a small piece of your disk is located at $$ x=r\cos\theta\, ,\qquad y=r\sin\theta,\qquad z=0 $$ then the density would be $dq=\sigma dA=\sigma r dr d\theta$ where $\theta$ is the polar angle on the disk since $dA=r dr d\theta$ is the area of a small piece of your disk, with radius $r$ and arclength $rd\theta$.

For a point located on the $\hat z$ axis at $Z\ne 0$, this small amount of charge will produce the infinitesimal field $$ d\vec E=\frac{dq}{4\pi\epsilon_0} \frac{Z\hat z- r\cos\theta\hat x - r\sin\theta\hat y}{(r^2+Z^2)^{3/2}}\, . $$ The full field is obtained by integrating the infinitesimal fields.

  • $\begingroup$ What would be the coordinates of the point on the axis then in polar coordinates? $\endgroup$ – PhysicsMathsLove Jun 8 '17 at 15:09
  • $\begingroup$ also why is the angular opening $rd\theta$ and not just $\theta$? $\endgroup$ – PhysicsMathsLove Jun 8 '17 at 15:15
  • $\begingroup$ @PhysicsMathsLove updated and expanded my answer. $\endgroup$ – ZeroTheHero Jun 8 '17 at 15:18
  • $\begingroup$ Thanks. But I still don't understand why the $dq = \sigma r dr d\theta$ rather than just $dq = \sigma dr d\theta$ or something else :( $\endgroup$ – PhysicsMathsLove Jun 8 '17 at 15:20
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    $\begingroup$ The surface element in polar is $r dr d\theta$. As you have written it $dr d\theta$ does not have units of surface since $\theta$ is an angle and thus dimensionless. $\endgroup$ – ZeroTheHero Jun 8 '17 at 15:23

$$dq=\sigma dA$$

Thus, your question reduces to writing the surface element on a horizontal disk in polar coordinates.

If I move an infinitesimal distance $dr$ along $\mathbf{\hat r}$, I have traced an infinitesimal line segment.

I want to make an infinitesimal rectangle, so what should I choose for the other side? I can move in the direction of $\mathbf{\hat \theta}$.

So if I trace an infinitesimal angle $d\theta$, what length will I have traced? If the angle is small enough and in radians, the arc length (which will be like a straight line segment if the angle subtending it is small enough) is given by $dl = r d\theta$ (you can verify it by considering a triangle with two equal sides. Write the length of the other side as a function of the angle which is facing it. If the angle shrinks toward zero, the formula can be seen.).

So I have an infinitesimal rectangle with sides $dr$ and $rd\theta$. My surface element will be $$dA = dr \cdot rd\theta = rdrd\theta$$

  • $\begingroup$ Another great answer, thank you!! Hopefully I can do this if it comes up in my EM exam tomorrow :D $\endgroup$ – PhysicsMathsLove Jun 8 '17 at 21:29
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    $\begingroup$ Thank you @PhysicsMathsLove . In famous coordinate systems (Cartesian, Cylindrical, and Spherical), for surfaces that are specified by fixing one of the three coordinates and varying the other two (like this disk here), you can easily find the surface element using this method of making infinitesimal rectangle (the sides lie along unit vectors). But if the surface is more complex like a tilted plane in Cartesian coordinates, it is not that easy. Currently, I don't know how to find the surface element in the tilted plane. $\endgroup$ – AHB Jun 9 '17 at 8:19

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