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Homework Help: Electrostatic potential of unit charge in vacuum

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm having hard time seeing that from this picture:
    wiugio.jpg
    Follows that:
    [tex]\phi(\vect{r})=\frac{q}{4\pi\varepsilon_0}\cdot\frac{1}{|\vec{x}|}[/tex]
    The thing that puzzles me isn't the equation but this:
    [tex]\frac{1}{|\vec{x}|}=\frac{1}{|\vec{r}-\vec{k}|}=\frac{1}{\sqrt{1+r^2-2r\cos\theta}}[/tex]

    2. Relevant equations

    The lenght of vector (norm):
    [tex]||\vec{a}||=\sqrt{a_1^2+a_2^2+a_3^2}[/tex]

    3. The attempt at a solution

    Now I know why is [tex]|\vec{x}|=|\vec{r}-\vec{k}|[/tex]

    That follows from simple subtraction of two vectors. And from triangle I see that [tex]\cos\theta=\frac{|\vec{k}|}{|\vec{r}|}[/tex], so I can get [tex]\vec{k}[/tex] from that, but how do I get that thing under the square? I read a bit in Jacson, that I would need to transform that into polar coordinate system, but how?

    Can someone give me detailed expansion of that?
     
  2. jcsd
  3. Jan 16, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi dingo_d! :smile:

    Either …

    i] it's a standard trig formula for any triangle … a2 = b2 + c2 - 2bccosA

    ii] just expand (r - k).(r - k) = r2 + … ? :smile:
     
  4. Jan 16, 2010 #3
    Wow, really it's trig formula for triangle! the k is unit vector so k^2=1! I see it now :D Thanks!
     
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