Show by experiment that the force of charge0 against charge1+charge2 equals the force of charge0 against charge1 plus charge0 against charge2. Induce an additive-homomorphic property F(r,q0,q1+q2) = F(r,q0,q1)+F(r,q0,q2). Then, exponent 1 follows.<p>I figured this out by listing a bunch of mathematical properties. I couldn't see how the author jumps from zero-preserving to multiply-charges, and I still don't know how, but we can call it out of scope lol<p><pre><code> r : distance between p and q
q0 : charge 0
q1 : charge 1
F : coulomb force function
charge-commutative: F(r,q0,q1) = F(r,q1,q0)
zero-preserving: 0 = F(r,q0,0)
additive-homomorphic: F(r,q0,q1+q2) = F(r,q0,q1)+F(r,q0,q2)
homogenous-degree-1: F(r,q0,n*q1) = n*F(r,q0,q1)
multiplicative-separability: F(r,q0,q1) = K*R(r)*Q(q0,q1)
multiply-charges: F(r,q0,q1) = K*R(r)*(q0*q1)^a
Given F(r,q0,q1) = K*R(r)*(q0*q1)^a, charge-commutative, zero-preserving, additive-homomorphic.
Induction using additive-homomorphic proves homogenous-degree-1. (For example, F(r,q0,2*q1) = F(r,q0,q1+q1) = 2*F(r,q0,q1))
Equational proof follows from homogenous-degree-1:
K*R(r)*(q0*n*q1)^a = n*K*R(r)*(q0*q1)^a
(q0*n*q1)^a = n*(q0*q1)^a
n^a*(q0*q1)^a = n*(q0*q1)^a
n^a = n
n = 0 or a = 1
n≠0, therefore a=1.</code></pre>