Proof of superposition principle for electric potential Superposition principle of electric force is empirical. However superposition principle of electric potential doesn't seem to be following from it. Is there a formal proof of superposition principle for electric potential?

Edit
$V_1+V_2+...+V_n=\int\vec{F_1}.\vec{dr_1}+\int\vec{F_2}.\vec{dr_2}+....+\int\vec{F_n}.\vec{dr_n}$

From here on how shall I proceed? I can't factor out $.\vec{dr}$ as all paths may not necessarily be same.
 A: The superposition of the electric force is a consequence of the suporposition of electric fields. That is if we have two electric fields $\mathbf E_1$ and $\mathbf E_2$ then we add them to get:
$$ \mathbf E = \mathbf E_1 + \mathbf E_2 $$
This is equivalent to the superposition of the force because the force on a charge $q$ is:
$$\begin{align}
 \mathbf F &= q\mathbf E \\
&= q(\mathbf E_1 + \mathbf E_2) \\
&= q\mathbf E_1 + q\mathbf E_2 \\
&= \mathbf F_1 + \mathbf F_2 
\end{align}$$
The field and the force are vectors and they add just like all vectors do. Ultimately this is because the electromagnetic field is described by Maxwell's equations and these are linear. This isn't true of all fields. For example general relativity is a nonlinear theory and in GR gravitational fields do not simply add (though in everyday life the deviations from nonlinearity are negligible).
Anyhow, the potential difference between two points $A$ and $B$ is just the integral of the electric field along a line between those two points:
$$ V_{AB} = \int_A^B \mathbf E \cdot d\mathbf r $$
And since $\mathbf E = \mathbf E_1 + \mathbf E_2$ we can write this as:
$$\begin{align}
V_{AB} &= \int_A^B (\mathbf E_1 + \mathbf E_2) \cdot d\mathbf r \\
&= \int_A^B \mathbf E_1 \cdot d\mathbf r + \int_A^B \mathbf E_2 \cdot d\mathbf r \\
&= V_1 + V_2
\end{align}$$
So the superposition of the potentials follows immediately from the superposition of the field.
