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.
