Electrostatic potential of unit charge in vacuum

[dingo_d]

Homework Statement

I'm having hard time seeing that from this picture:

Follows that:
$$\phi(\vect{r})=\frac{q}{4\pi\varepsilon_0}\cdot\frac{1}{|\vec{x}|}$$
The thing that puzzles me isn't the equation but this:
$$\frac{1}{|\vec{x}|}=\frac{1}{|\vec{r}-\vec{k}|}=\frac{1}{\sqrt{1+r^2-2r\cos\theta}}$$

Homework Equations

The length of vector (norm):
$$||\vec{a}||=\sqrt{a_1^2+a_2^2+a_3^2}$$

The Attempt at a Solution

Now I know why is $$|\vec{x}|=|\vec{r}-\vec{k}|$$

That follows from simple subtraction of two vectors. And from triangle I see that $$\cos\theta=\frac{|\vec{k}|}{|\vec{r}|}$$, so I can get $$\vec{k}$$ 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?

 

[tiny-tim]

… how do I get that thing under the square? …

Hi dingo_d!

Either …

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

ii] just expand (r - k).(r - k) = r² + … ?

 

[dingo_d]

Wow, really it's trig formula for triangle! the k is unit vector so k^2=1! I see it now :D Thanks!