electric field, mathematical problem

[agolkar]

hi :) it's me again. This is my second question in a few days, w/o being able to help anyone else, hope to be not too much pretentious :biggrin:

my doubt this time regards the mathematical passage underneath the formula (23) of this speech:

http://scarface.ngi.it/elet.jpg

disregard the text part because i don't think it's helpful in an english forum :biggrin: it just introduces the electric field and its properties.
The fact is I cannot understand the derivation done from the E formula, i thought that the thing to do was just derivating r (that, considering an orthogonal system with the origin in the (x0,y0,z0) point, is (x^2+y^2+z^2)^1/2) because i had to consider the rest as costant so out of the derivation but it isn't it.. any help? thanks :smile:
alex

[speeding electron]

Could you explain your problem a little more? The vector r goes from the co-ordinates of the point charge to the the point (x,y,z) where you want to find the electric field. It has magnitude |r| = \sqrt{\left( x - x_0 \right) + \left( y - y_0 \right) + \left( z - z_0 \right)} . So if we divide r by this number, we have the vector of unit length pointing from the point charge to (x,y,z). We multiply this by the magnitude of the field, \frac{q}{4 \pi \epsilon_0 r^2} so that the field intensity is \frac{q \vec{r}}{4 \pi \epsilon_0 \r^2 \r } = \frac{q \vec{r}}{4 \pi \epsilon_0 r^3} = \frac{q \vec{r}}{4 \pi \epsilon_0} \frac{1}{\left[ \left( x - x_0 \right)^2 + \left( y - y_0 \right)^2 + \left( z - z_0 \right) \right]^{ \frac{3}{2} }} .

[agolkar]

why you pass from r^2 to r^3 ? :uhh:

[Romperstomper]

Multiply the magnitude of the field by the vector per unit length, which is \frac{\vec{r}}{r} . You get a r^3 at the bottom and the \vec{r} at the top.

So, you get E = \frac{q \vec{r}}{4 \pi \epsilon_0 r^3}

Then, you just substitute in for r :smile: