# Symmetry arguments in Gauss law and Superposition [duplicate]

Right now I am working on electrostatics. One of the techniques used is to use coulombs law and superposition in a situation where one can sum the point charges and corresponding fields to get the total field. I am having trouble understanding how the E field can be completely described by charge only. Charge is a number and E is a 3-vector at a certain point in space. The dimension of charge is 1 and the dimension of E is 3. Does that mean the E field really only has 1 degree of freedom? ( 2 degrees of freedom for each point charge) That is the E field only depends on distance. For a set of charges (p1,p2,...), the superposition principle states that E is a function of (r1,r2,...) (p1,p2,r1,r2...). Is there a way to show from Gauss's law and superposition (sources are independent) the empirical fact of Coulombs law? del * E = p/e ,E1 + E2 = E(p1/e + p2/e) => E(r1,r2) E(p1,p2,r1,r2). Coulombs Law is an axiom or http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html#c1?

In class we used a del*E = p/e and "symmetry" to derive Coulombs law. What is this symmetry? Is this symmetry important for a point charge and electrostatics as a whole? Basically because rotating the charge in the theta or phi direction does not change the point charge, the field (for a point charge). Does this imply that E must point in the r direction and only depend on r? p(r,theta,phi) where dp/dtheta = 0 and dp/dphi = 0 => E(r)r direction? How? Gauss' Law and Symmetry

• To find the electric field due to a charged particle you would need to know both the charge and the location of the particle. Commented Jan 29, 2022 at 17:31
• The electric field is used to describe the force on a test charge q: $\mathbf{F} = q\mathbf{E}$. So for a point charge, the field is given by $\mathbf{E} = \mathbf{F}/q$, where $\mathbf{E}$ and $\mathbf{F}$ are evaluated at the position of the test charge, given the validty of Coulombs Law. No you cannot derive Coulombs Law from Gauss' Law, since Gauss' Law is derived from Coulombs Law. Commented Jan 29, 2022 at 17:35
• So E would be a function of (p1,r1,p2,r2) . B Brekke?
– dfdf
Commented Jan 29, 2022 at 17:38
• E has two degrees of freedom for each point charge?
– dfdf
Commented Jan 29, 2022 at 17:42
• @Samuel the electric field exists regardless of the existance of the test charge , the field is more fundamental than the force. Commented Jan 29, 2022 at 17:45