# Electromagnetism: Can anyone find the mistake?

1. Jan 22, 2006

### *Alice*

given: the electric field at a point on the axis a distance x from the plane of a ring is $$E = \frac {q*x} {4*pi*E0*(x^2+r^2)^{3/2}}$$

where E0
is the permeability coefficient

The charged ring is replaced by a circular sheet of charge of radius a a surface charge density sigma. The ring can be divided into infinitessimally small rings of radius r and thicknes dr. Show that the electric field is given by $$E= \frac {sigma} {2*E0} * [1 - \frac {x} {(x^2 + a^2)^{1/2}}]$$

this is what I did:

charge on each ring:

$$2*pi*r*sigma*dr = A*sigma=Q$$

Electric field on each ring:

$$E = \frac {2*pi*sigma*dr*x*r} {4*pi*E0*(x^2 + r^2)^{3/2}} = \frac {sigma*dr*x*r} {2*E0*(x^2 + r^2)^{3/2}}$$

Integrate over ring:

$$\frac {sigma} {2*E0} * \int_{0}^{a} \frac {r} {(x^2 + r^2)^{3/2}} dx = \frac {sigma} {2*E0} * [-1/2*\frac{1} {(x^2+a^2)^{0.5}}] (from 0 to a) = \frac {sigma} {4*E0}* [1 - \frac {x} {(x^2+a^2)^{.5}}]$$

why is that factor 4 here (it's supposed to be 2)? Help's very much appreciated!
LaTeX

Last edited: Jan 22, 2006
2. Jan 22, 2006

### siddharth

Looks like you missed a 'r' in the numerator in when you calculated Electric field on each ring:

3. Jan 22, 2006

### *Alice*

Yes, sorry...missed to write that one in one line. However, I had it back in the integration the line below, so that it didn't affect the answer. It's now edited.

Does anyone have any idea about that factor 4?

Last edited: Jan 22, 2006
4. Jan 22, 2006

### siddharth

The factor is supposed to be '2'. The derivative of $r^2$ is $2r$. So if you take that into account, you will not get '4'.

5. Jan 22, 2006

### *Alice*

That's exactly what I did and that caused all the trouble:

substitute: u= x^2 + a^2

so then you have to multiply by (1/2)...oh yeah....I see! I didn't multiply by two when I did the integration......:yuck:

Oh dear!

Anyways - thank you so much!!!