Thread: Integrating vectors (NOT vector fields)

1. Integrating vectors (NOT vector fields)

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 $\displaystyle h$ 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:

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

What did I do wrong?

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

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

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

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

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

3. Dear zzzoak,