Self capacitance I have two spheres of metal, each of radius $1m$, these spheres are effectively an infinite distance apart.
I now connect a $1V$ power supply between these two spheres: (and wait an infinite amount of time, for the charges to reach equilibrium.)
   O----------------| 1V |----------------O
Sphere 1         Power Supply          Sphere 2



*

*What property of the metal from which the spheres are made determines the magnitude of the charge on the spheres?

*How can the magnitude of the charge on one of the spheres be calculated?

*If one of the spheres had its mass doubled what would be the charge on the spheres be then?


(I am hoping to discover a new equation which will allow me to solve a more in depth problem, which I dreamt up whilst sitting on the bus)
EDIT
I have now discovered what I think is the equation which I was looking for: $Q=4\pi \epsilon_0 R V$ for a single isolated sphere. I therefore (by intuition) think that the equivalent for the 2 sphere set-up is $Q=2\pi \epsilon_0 R V$
Therefore the charge I was looking for would be $Q=2\pi \epsilon_0 \times 1V \times 1m=5.56325...\times10^{-10}C$
 A: Let the distance between the two spheres be $d$ and let one of them have a charge of $Q$ and the other, $-Q$. The net field at a point at a distance $r$ from the centre of the positively charged sphere is the sum of fields due to both charges, and the direction of field is same for both direction, i.e. towards the negatively charged sphere. Thus the field is $\displaystyle E(r) = \dfrac {Q}{4\pi\varepsilon_0}\left(\dfrac 1{r^2} +\dfrac1{(d-r)^2}\right)$, and the potential difference between the two spheres is:
$$ \begin{align}V &= \int_R^{d-R}\dfrac {Q}{4\pi\varepsilon_0}\left(\dfrac 1{r^2} +\dfrac1{(d-r)^2}\right) \\[2ex]
&=\dfrac {Q}{4\pi\varepsilon_0R}\left(\dfrac {-1}{r} +\dfrac1{d-r}\right)_{R}^{d-R}\\[2ex]
&=\dfrac {2Q}{4\pi\varepsilon_0R}\left(\dfrac{d-2R}{d-R}\right)
\end{align}$$
$$\therefore Q = 2\pi\varepsilon_0RV\cdot\dfrac{d-R}{d-2R}$$
Taking the limit $d\to\infty$, we get:
$$Q=2\pi\varepsilon_0RV$$
Please correct me if i am wrong. Hope this helps. Cheers!
Note: A mention of thanks to @Anubhav Goel, for helping me realize my previous approach using direct expressions for potential of charged sphere wasn't that good, and for pointing out the mistake in my integral. So i have edited the answer to show another method using electric field, and corrected the integral.
