- #1
raving_lunatic
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Homework Statement
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.
Homework 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.
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?