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Electric field, mathematical problem

  1. Oct 4, 2004 #1
    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:


    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:
  2. jcsd
  3. Oct 4, 2004 #2
    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 [tex] |r| = \sqrt{\left( x - x_0 \right) + \left( y - y_0 \right) + \left( z - z_0 \right)} [/tex]. 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, [tex] \frac{q}{4 \pi \epsilon_0 r^2} [/tex] so that the field intensity is [tex] \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} }} [/tex].
    Last edited: Oct 4, 2004
  4. Oct 5, 2004 #3
    why you pass from r^2 to r^3 ? :uhh:
  5. Oct 5, 2004 #4
    Multiply the magnitude of the field by the vector per unit length, which is [tex]\frac{\vec{r}}{r} [/tex]. You get a [tex] r^3[/tex] at the bottom and the [tex]\vec{r}[/tex] at the top.

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

    Then, you just substitute in for r :smile:
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