On what basis can we say that the surface charge density of a conductors is proportional to the total charge in it Suppose we've two conductors which are separated by a distance. We put charge $+Q\ $ on one and a charge $-Q\ $ on the other. Let the conductor with $+Q$ charge be called $+$  conductor and the conductor with $-Q$ charge be called $-$ conductor.
Then potential difference between the conductors is
$$V=V_{+}-V_{-}=-\int_{-}^{+}\vec{E}\cdot d\vec{l}$$ where $V_{+}$ and $V_{-}$ are the potentials of positive and negative conductors respectively.
The electric field at $\ \vec{r}\ $ due to $\ \pm\ $ conductor is
$$\vec{E}_\pm (\vec{r})=\frac{1}{4\pi\epsilon_0} \int \displaystyle{\frac{\sigma_\pm}{{r'_{\pm}}^2}}\  da'_\pm\ \hat{r'}_\pm$$
where $\sigma_\pm$ is the surface charge density of the $\pm$ conductor.
In some text books it's written that the surface charge density on a conductor is proportional to the total charge in it.
$$\sigma_\pm =(\pm Q) k_\pm$$
Then the resultant electric field at $\ \vec{r}\ $ is
\begin{align}
\vec{E}(\vec{r})
& = \frac{1}{4\pi\epsilon_0} \left[ \left(\int \displaystyle{\frac{\sigma_+}{{r'_{+}}^2}}\  da'_+\ \hat{r'}_{+}\right) +\left(\int \displaystyle{\frac{\sigma_-}{{r'_{-}}^2}}\  da'_-\ \hat{r'}_{-}\right)  \right]
\\ & = \frac{Q}{4\pi\epsilon_0} \left[ \left(\int \displaystyle{\frac{k_+}{{r'_{+}}^2}}\  da'_+\ \hat{r'}_{+}\right) -\left(\int \displaystyle{\frac{k_-}{{r'_{-}}^2}}\  da'_-\ \hat{r'}_{-}\right)  \right]
\end{align}
Since $\vec{E}$ proportional to $Q$, so also is V.
$$V=\frac{Q}{C}$$
The constant of proportionality $C$ is called capacitance.
My question is, on what basis can we say that the surface charge density of a conductors is proportional to the total charge in it. How do we know that doubling $\pm Q$ simply doubles $\sigma_\pm$? May be the charge moves around into a completely different configuration, quadrupling $\sigma_\pm$ in some places and halving it in others, just so the total charge on the conductor is doubled.
 A: The proof is that the charges in the conductor must move to a configuration of minimum electrostatic potential, but that configuration is specified by the relative amount of charge at various places, not the total amount.  This is because the field is linearly proportional to the charges, and the field-charge interaction energy for a given field is also linearly proportional to the charges, so this means that all you need to get a potential energy is the total charge, and the relative field configuration and the relative charge configuration (by "relative", I mean ratios from place to place).  
So if we have total charge Q, we get some given spatial distribution of those charges to minimize the electrostatic potential.  Now we change the total charge to kQ.  If we start with the same relative distribution of charge, we just multiply the surface density everywere by k (all the charges are at the surface, that's easy to show with Gauss' law).  So the field has the same geometry, it is also simply multiplied by k.  But a field that is simply multiplied by k will have the same relative charge distribution that minimizes its potential energy.  The potential energy itself changes by the factor $k^2$, but the relative charge configuration must still work to minimize it because it is only the relative field configuration and relative charge configuration that determine if it is minimized.
