# Electric field, mathematical problem

1. Oct 4, 2004

### 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

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

http://scarface.ngi.it/elet.jpg [Broken]

disregard the text part because i don't think it's helpful in an english forum 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
alex

Last edited by a moderator: May 1, 2017
2. Oct 4, 2004

### 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 $$\frac{q \vec{r}}{4 \pi \epsilon_0 \r^2 \r } = \frac{q \vec{r}}{4 \pi \epsilon_0 r^3} = \frac{q \vec{r}}{4 \pi \epsilon_0} \frac{1}{\left[ \left( x - x_0 \right)^2 + \left( y - y_0 \right)^2 + \left( z - z_0 \right) \right]^{ \frac{3}{2} }}$$.

Last edited: Oct 4, 2004
3. Oct 5, 2004

### agolkar

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

4. Oct 5, 2004

### 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