Integrating vectors (NOT vector fields)

**rebghb** · October 16th 2010, 03:37 AM

This is a problem in electromagnetics applying coulomb's law, in brief, we have a disc(wire) with charges on it. We want to calculate the electric field at a point $h$ on the axis of the disc. The book had a rather annoying way of solving the problem, so I wanted to go with something simpler and went with the following:

$(1) \, d\vec{E} = \int \frac{1}{4\pi \epsilon R^2} \vec{R} dq$ . R is a unit vector
$(2) \, \vec{R} = \frac{1}{\sqrt{r^2 + h^2}} (-r\cos \theta \vec{i} - r\sin \theta \vec{j} + h\vec{k})$ .
$(3) \, dq = \rho_s dA = \rho_s rdrd\theta$ .
when we put all of that together, we get:
$\vec{E} = \int\int _s \frac{\rho_s}{4\pi \epsilon (r^2 + h^2)^{3/2} } (-r\cos \theta \vec{i} - r\sin \theta \vec{j} + h\vec{k}) rdrd\theta$ .
But unless we take away the 2 $r$ s from the vector $(-r\cos \theta \vec{i} - r\sin \theta \vec{j} + h\vec{k})$ we won't get to the right answer which is $\frac{\rho_s}{2\epsilon}( \, \frac{h}{\sqrt{a^2 + h^2}} - 1 \,) \, \vec{k}$ .
Note: integral limits $0 < r < a$ and $0 < \theta < 2\pi$ .

What did I do wrong?

**zzzoak** · October 17th 2010, 02:22 PM

It is not clear without a picture. It takes much time.
Using the symmetry along the z axis

$ \displaystyle { dE=\frac{1}{4 \pi \epsilon _0} \; \frac{dq}{R^2} \; cos {\alpha}} } $

$ dq=\rho \; 2 \pi r dr $

where
$ \displaystyle { cos {\alpha}= \frac {h}{\sqrt{r^2+h^2}} } $
is the projection of electric field to z axis
and
$ R^2=r^2+h^2 $

$ \displaystyle { E=\frac{\rho \; h}{2 \epsilon _0} \int_0^a \frac{r \; dr} {(r^2+h^2)^{3/2}}=\frac{\rho}{2 \epsilon _0} (1-\frac{h}{\sqrt{a^2+h^2}}) } $

**rebghb** · October 17th 2010, 09:11 PM

Dear zzzoak,

Thanks for the reply

I know it takes a lot of time... My professor used your way but I didn't like it, I was testing a new way. Now my question is, why doesn't my way work?

Is there something we must know when integrating vectors? (integrating vectors as in sum of a very large number of vectors).

Best,

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