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?



From here on how shall I proceed? I can't factor out $.\vec{dr}$ as all paths may not necessarily be same.

  • 1
    $\begingroup$ "However superposition principle of electric potential doesn't seem to be following from it." That is incorrect. Potential is defined as the line integral of field, i.e. $V_a - V_b \equiv \int_a^b \vec{E} \cdot d\vec{x}$. Superposition of $V$ follows superposition of $E$ by the linearity of integration. $\endgroup$
    – DanielSank
    Dec 23, 2017 at 5:20

1 Answer 1


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.

  • $\begingroup$ What if for first charge path is $AB$ and for second charge path is $CD$. In that case we can't factor out $.\vec{dr}$. So how shall we proceed? $\endgroup$
    – Joe
    Dec 23, 2017 at 5:53
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    $\begingroup$ @Joe The electric field is a conservative field, and that means the value of the integral is independent of the path taken. The proof of this is simply that the curl of a static electric field is zero. Anyhow this guarantees that our result holds. $\endgroup$ Dec 23, 2017 at 6:04
  • $\begingroup$ One little confusion to clear: Our integral is path independent. But it is dependent on end points. In my example the end points are different. For first charge, it is $A$ and $B$; and for second charge it is $C$ and $D$. In such cases how is factoring out $.\vec{dr}$ be justifiable? $\endgroup$
    – Joe
    Dec 23, 2017 at 6:13
  • $\begingroup$ @Joe It's not obvious to me how you start with $\int_A^B (\mathbf E_1 + \mathbf E_2) \cdot d\mathbf r$ and divide it into two integrals with different start and end points. If you have different start and end points you are calculating the potential difference between different pairs of points. $\endgroup$ Dec 23, 2017 at 6:16
  • $\begingroup$ Since there are two (or more) different charges, I generally mean the end points of their path may not necessarily be same. Then in this case how can we say total work done by the two charges IS the difference in "the sum of their final potentials" and "the sum of their initial potentials". $\endgroup$
    – Joe
    Dec 23, 2017 at 6:44

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