# Experimental relationship between linear dependence and superposition

From Griffith's Introduction to Electrodynamics

The principle of superposition may seem obvious to you, but it did not have to be so simple: if the electromagnetic force were proportional to the square of the total source charge, for instance, the principle of superposition would not hold, since $\left(q_1+q_2\right)^2\ne q_1^2 + q_2^2$ (there would be cross terms to consider). Superposition is not a logical necessity, but an experimental fact.

I am confused as to why it is important that electromagnetic force must depend linearly on charge for the superposition principle to hold. Suppose it was the case that electromagnetic force depended on the square of charge. Then, defining, a new quantity "squarecharge" $k = q^2$, electromagnetic force would depend linearly on "squarecharge", and therefore does superposition not hold?

Put another way, every object with charge $q$ has an associated quantity $m = q^{1/3}$. Suppose that $m$ is more naturally measured than $q$, therefore it is used in place of $q$ in all formulas. Then, even though electromagnetic forces are proportional to the cube of $m$, superposition still holds.

• The author tells you in the bit you quote. Compare $A q_1^\alpha + A q_2^\alpha$ to $A(q_1 + q_2)^\alpha$ for any alpha not equal to one. Sep 20, 2013 at 23:58

Let's suppose we have a source particle and and a test particle a distance $r$ from each other. Upon measuring the force on the test particle, we find some value $F_\text{one source}$. By varying the distance, we discover it depends on the inverse distance squared:$$F_\text{one source}=\frac{C}{r^2},$$ where $C$ is just some constant. (No 'charge' appears here because we haven't introduced the concept of charge.)
Now let's bring an additional source particle that is identical to the original, and place it right on top of the original. Again, we measure the force, but now we call it $F_\text{two sources}$. How is this new force related to the original one? We have no idea, because I haven't told you anything about the particles or the nature of the forces.
Suppose now it is an experimental fact that the force doubles:$$F_\text{two sources}=2F_\text{one source}=\frac{2C}{r^2}=\frac{C}{r^2}+\frac{C}{r^2}.$$ Here, the possibility of superposition exists. I've made the superposition part (where you just add up the contribution due to each source alone) very explicit.
Now, suppose instead of the force doubling, the force happened to quadruple:$$F_\text{two sources}=4F_\text{one source} = \frac{4C}{r^2}\ne\frac{C}{r^2}+\frac{C}{r^2}.$$
Thus, if the force didn't double when you added another identical source particle, superposition would not be possible. It doesn't matter how you end up 'expanding' $C$: if you try to do $C=C'q$ or $C=C'q^2$, you can't get the math to work out (i.e., you can't get the net force on your test particle to be the sum of individual forces).
This second example (where the force quadrupled) is an example of what Griffiths means when he says "...proportional to the square of the source charge." One could say $F_\text{two sources}=(C+C)^2/r^2$ and get the right expression for the force, but you couldn't then go and write the force using superposition.