# Find E, circular disk. Radial component of E question.

1. Aug 21, 2009

### jfierro

Hi, I'm not sure this falls under the homework category, but I posted this here because I think this is more of an insight question than a question for help on how to solve the excercise (which I have).

A circular disk of radius a has a uniform charge $$\rho_s \frac{C}{m^2}$$. If the disk is on the plane z = 0 with its axis along the z axis,

a) Demonstrate that at point (0, 0, h)

$$\textbf{E} = \frac{\rho_s}{2\epsilon_0}\left\{ 1 - \frac{h}{\sqrt{h^2 + a^2}} \right\} a_z$$

This is a visual representation, if lame, of the problem:

http://img191.imageshack.us/img191/1258/chargeddisk.gif [Broken]

Well, using Coulomb's I get to:

$$\textbf{E} = \frac{\rho_s}{ 4\pi\epsilon_0 } \int_0^a\int_0^{2\pi} \frac{ -r a_r + h a_z } { \left( r^2 + h^2 \right)^\frac{3}{2} }rd\theta dr$$

Now, I recall reading in the book that the individual contributions to the E field along the $$a_r$$ (radial) direction sum up to zero, so I just discarded that vectorial component in the integral above and got easily to the demonstration. However, my question is:

Why is it that the integration of the differential charges along the radial direction $$a_r$$ is 0? I tried integrating:

$$\int_0^a \frac{ -r \textbf{a_r} } { \left( r^2 + h^2 \right)^\frac{3}{2} }rdr$$

Lets say the disk had radius 1, and we want to know the magnitude of the radial component at point (0, 0, 1), so that the integral becomes:

$$-\int_0^1 \frac{ r^2 \textbf{a_r} } { \left( r^2 + 1 \right)^\frac{3}{2} }dr$$

But that, according to mathematica, is:

$$\frac{1}{\sqrt{2}} - ArcSinh[1]$$

(Command used: Integrate[-(x^2)/((x^2 + 1^2)^(3/2)), {x, 0, 1}])

Which is not 0

Last edited by a moderator: May 4, 2017
2. Aug 21, 2009

### jfierro

Sorry for the double post, I created a new thread while fixing some tex formatting issues... If a moderator reads this please delete the other thread.

3. Aug 22, 2009

### Born2bwire

There is a \phi dependence that you are missing. Ignore the math and just visualize the electric field vector. The field vector from the charge at \phi=0 and some arbitrary radius on the disc is going to point up along the z-axis and away from the source point at the observation point on the z-axis. Now look at the field vector from the charge at \phi=\pi at the same radius. It will have the same magnitude and angle with the z-axis, the only difference is that it will be pointing in the opposite \rho direction. The \rho component of the electric field from a charge at \phi will be cancelled out by the charge at \phi+\pi. So you need to double check your integrals because it is easy to demonstrate this to yourself conceptually.

4. Aug 22, 2009

### jfierro

Thanks for your response, I guess it is easy to visualize it conceptually, the same goes for an infinite plane at any point perpendicular to it. However, There is something inside of me asking for some math to back it up :). So, if we add the \phi dependence, the integral for the radial component becomes (I'll use \rho and \phi since it seems it's more standard than r and \theta ):

Integrating on \phi:

Which will be zero only if:

Could you help me find out what I am missing math-wise?

5. Aug 22, 2009

### Born2bwire

The problem is that cylindrical coordinates is inadequate for the integral. What happens if you try to sum the vectors x-hat and -x-hat when represented in cylindrical coordinates? You do not get the correct answer by just doing a naive addition. Convert it to cartesian coordinates for a no-brainer integration. In this case:

$$\vec{E} = \frac{\rho_s}{ 4\pi\epsilon_0 } \int_0^a\int_0^{2\pi} \frac{ \rho\cos\phi \hat{x} + \rho\sin\phi\hat{y} } { \left( \rho^2 + h^2 \right)^\frac{3}{2} }\rho d\phi d\rho$$

It is obvious now that the integration over \phi will be zero.

Last edited: Aug 22, 2009
6. Aug 22, 2009

### jfierro

It is indeed very clear now that the contributions along the rho axis sum up to zero. Thanks :).