Capacitance of two spheres with equal radius

[bdvjofrni]

I'm considering two identical spherical conductor each of radius $a$ and separated by a distance $d$, and trying to figure out the capacitance of this configuration.
My thoughts are that since capacitance is
$$C=\frac {Q}{V}$$
and that the spherical conductors are equipotential surfaces, $V$ would just be
$$V=V_1-V_2=(\frac {kq}{a} - \frac {kq}{a+d})- (- \frac {kq}{a} + \frac {kq}{a+d}) = 2kq (\frac{1}{a} - \frac{1}{a+d})$$
and thus the capacitance would just be
$$C=\frac{1}{2k(\frac{1}{a}-\frac{1}{a+d})}=2πεa \frac {a+d}{d}$$

And then I read Wikipedia, and what I did turns out to be incorrect. Wikipedia says that the correct formula is
$$C=2πεa \sum\limits_{n=1}^∞ {\frac {\sinh (\ln (D+\sqrt {D^2-1}))}{\sinh (n \ln (D+\sqrt {D^2-1}))}}$$

I only know some very basic physics and this expression looks scary to me, but when $a \ll d$ they give pretty much the same result.
If I graph $C$ vs $d$ with a fixed radius $a=1$, it looks like:

The blue line is the graph of the correct formula while red one is mine. It seems that when the separation $d$ is 10 times as big as radius $a$, the formula I derived yields a result very close to the correct one.
So my question is, what did I fail to take into account when I considered this that led to my incorrect conclusion, and why is it that the formula I get approximately equals the correct formula when the separation gets large? (physically, not mathematically; I can see that both formulas approach horizontal asymptote $2πεa$.
(I don't expect to really understand the correct formula on Wikipedia; that seems scary. Just wanted to know what I failed to take into account)
Thanks!

 

[Ibix]

I think you are writing down the solution for an isolated charged sphere, writing down the solution for another, and adding them together. So you are ignoring the effect of the electric field of each sphere on the charge distribution on the other one.

That's why your solution is about right at large separations where the interaction is weak, but fails at small separations where you can't neglect it.