On proving that charge is linearly proportional to potential for a conductor In Mr. Purcell's Electricity and Magnetism, page 103, it is stated,

An isolated conductor carrying a charge $Q$ has a certain potential $\phi _{0}$, with zero potential at infinity. $Q$ is proportional to $\phi _{0}$. The constant of proportionality depends only on the size and shape of the conductor. We call this factor the capacitance of that conductor and 
  denote it by C. $$Q=C \phi _{0}$$

I understand that for a given charge $Q_{0}$ and its corresponding potential $\phi_{0}$ we could define a  $C_{0}$ as a function of the shape and size of the conductor such that $Q _{0} =C_{0}\phi_{0}$.
When we change the charge to $Q_{1}$, the potential will become $\phi_{1}$. How can we prove that it is the same constant $C_{0}$ that will link  $Q_{1}$ and  $\phi _{1}$ ? In other words is charge being linearly proportional to potential an experimental result or can we prove it? 
If one argues that it is the same constant because it depends only on the shape and size of the conductor, then they must also prove that this constant does satisfy $$Q=C _{0} \phi$$ 
for every given charge and its corresponding potential.
 A: TLDR: All the equations are linear relations, so their results have that too. 
Informally: 
Consider some solution to Maxwell’s equations. This relates the fields (hence potentials) and charges in some particular geometry of interest. 
Now multiply all the charges by some common constant. Since Maxwell’s equations have a linear relation between charge and electric field, the field at every point is multiplied by that same constant. 
Since the potential is just a spatial line integral (a linear operator) of the electric field, it to is multiplied by the same factor at every point. 
Hence there’s a linear relation between all-charges and potential. 
Formally:
For an initial configuration given by a 0 subscript, $\phi_0(\vec{x})$ is given by $\vec{E_0}(\vec{x})$ over all space, which in turn is given by the arrangement of charge $q_0(\vec{x})$ (and perhaps the current density).  
Now if charge everywhere where to increase by the same factor $k = q_1/q_0$, then the linearity of Maxwell's equations means that $\vec{E_1}(\vec{x}) = k \vec{E_0}(\vec{x})$ is a solution.  Since it's a solution, it's the only solution.  Then the potential comes from $\vec{E_1}$ in the same way, such that $\phi_1(\vec{x}) = k \phi_0(\vec{x})$.
A: Here's a  non-finished, non-rigorous attempt:
Let us say  we have a conductor having a surface $S$ at a potential $\phi_{0}$ and having a charge $Q_{0}$.
Supposing all the charges are located on the surface,  the potential can be defined at any point in space as:$$\phi (\textbf{x}) =\int _{S} \frac{ \sigma(\textbf{x}')}{\vert\lvert  {\textbf{x}-\textbf{x}'\lvert\lvert }} da'$$
$\phi (\textbf{x})$ is constant inside and on the surface of the conductor. Choosing an arbitrary $\textbf {x}_{0}$ within or on the surface of the conductor, we find:
$$\phi (\textbf{x}_{0})=\phi_{0} =\int _{S} \frac{ \sigma(\textbf{x}')}{\vert\lvert  {\textbf{x}_{0}-\textbf{x}'\lvert\lvert }} da'$$
where $da'$ is a surface element of $S$.
Since $\vert\lvert \textbf{x}_{0}-\textbf{x}'\lvert\lvert $ is a continuous function of $\textbf{x}'$, we can use the intermediate value theorem for integrals to rewrite the expression above as:$$\phi_{0} =\int _{S} \frac{ \sigma(\textbf{x}')}{\vert\lvert  {\textbf{x}_{0}-\textbf{x}'\lvert\lvert }} da'=C \int _{S} \sigma (\textbf{x}')da'=C \times Q_{0}$$
Where $C$ is a constant.
The flaw: We must prove that $C$ doesn't depend on the choice of $\textbf{x}_{0}$.
It'd be fantastic if someone could formalize this a little bit more.
A: we know charge and potential have linear relation , as charge is directly proportional to potential. so if you draw charge versus potential graph you'll get straight line  ( notice: the value you put on graph are experimental ) . now if you compare the equation y=mx  and Q=CV means C(capacitance) represent slope which comes constant . Means C doesn't depend on Q and V
