Capacitance of two conducting spheres

[AxiomOfChoice]

Homework Statement

We're supposed to find the capacitance of a system of two conducting spheres, one of radius r_1 and charge Q, the other of radius r_2 and charge -Q, separated by a distance L (this is the distance between their centers) that's very large compared to either r_1 or r_2.

Homework Equations

We define the capacitance by C = Q/V, where V is the potential difference between the spheres.

The Attempt at a Solution

Really, my only question, as of right now, is what approximations or assumptions we can make based on the L >> r_1,r_2 assumption. Is it just that the charge distribution on eiter sphere is unaffected by the presence of the other sphere? Such that we can assume the potential is just the superposition
\dfrac{Q}{4 \pi \epsilon_0 R_1} - \dfrac{Q}{4 \pi \epsilon_0 R_2},
where R_1 is the distance from the center of the first sphere and R_2 is the distance from the center of the other sphere?

 

[AxiomOfChoice]

Here's a new attempt at a solution:

(1) The potential due to a sphere is kQ/r, where r is the distance from the center of the sphere.

(2) If we consider the two spheres mentioned above, set the origin at the center of the left-most, and let the x-axis join the centers of the spheres, then the potential at any point along this axis is given by the linear superposition of the potentials of sphere 1 and sphere 2:
<br /> V = V_1 + V_2 = \frac{kQ}{x} - \frac{kQ}{L-x}.<br />

(3) This means the potential difference between the spheres is given by \Delta V = V(r_1) - V(r_2), which is just
<br /> \frac{kQ}{r_1} - \frac{kQ}{L-r_1} + \frac{kQ}{r_2} - \frac{kQ}{L-r_2}.<br />

I guess we can reduce this, and put Q/\Delta V to get the capacitance.

 

[AxiomOfChoice]

...anyone? Does it look like I'm heading in the right direction here?