1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
    Follows that:
    The thing that puzzles me isn't the equation but this:

    2. Relevant equations

    The lenght of vector (norm):

    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


    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook