Electromagnetism: Can anyone find the mistake?

[*Alice*]

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

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 [tex] E= \frac {sigma} {2*E0} * [1 - \frac {x} {(x^2 + a^2)^{1/2}}][/tex]

this is what I did:

charge on each ring:

[tex] 2*pi*r*sigma*dr = A*sigma=Q [/tex]

Electric field on each ring:

[tex] 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}} [/tex]

Integrate over ring:

[tex] \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}}] [/tex]

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

 

[siddharth]

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

 

[*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?

 

[siddharth]

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

 

[*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...

Oh dear!

Anyways - thank you so much!