# Homework Help: Electric potential due to two point charges near origin

1. Jan 19, 2014

### raving_lunatic

1. The problem statement, all variables and given/known data
A charge + q is ﬁxed at each of the two points (±a,0,0). Show that the potential V at a point (x,y,z) near to the origin may be expressed as

V = [q/4∏*ε*a] [[2+(2x2-y2-z2)/a2]
where x, y and z are small compared with a.

2. Relevant equations

V(x) = q/4∏i*ε*r , where r is the distance between the point charge and x and V is the potential at x.

Superposition, i.e. V due to multiple point charges = the sum of V due to each point charge

(1+x)^n = 1+nx+... valid for small x.

3. The attempt at a solution

This seemed like an incredibly straightforward problem, which is probably why it's quite frustrating. I tried adding together the potentials due to each point charge, approximated using the binomial expansion:

V(q1) = q/4∏ε[(x+a)2 + y2 +z2]^-1/2

V(q2) = q/4∏εa[(a-x)2 + y2 +z2]^-1/2

Taking a out of the brackets as a common factor gives:

V(q1) = q/4∏εa [(x/a + 1)2+ (y2+z2)/a2)]^-1/2

and a similar expression for V(q2) but with (1-x/a)2.

I then expanded the squares and used (2x/a + (x2+y2+z2)/a2) as a small parameter to perform the binomial expansion; this gave

V(q1) = q/4∏εa [1 - x/a - (x2+y2+z2/2a2)]

V(q2) = q/4∏εa [1 + x/a - (x2+y2+z2/2a2)]

Which doesn't lead to the desired result when added; where did I go wrong?

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2. Jan 19, 2014

### raving_lunatic

I think I've worked it out; need to keep the terms to the second order. Sorry.