What would E field addition be if Gauss's law was different

1. Feb 17, 2013

Curl

What if Gauss' Law was
div(E)=q2/eps0

Then if we had two point charges, how could we calculate the resultant E field at an arbitrarily point? Obviously superposition would not work anymore, so how could it be done mathematically?

This is essentially a math question

2. Feb 17, 2013

WannabeNewton

Did you stop to check if it was first dimensionally correct before proceeding? The units for the real gauss's law pertain to a physically interpretable / measurable quantity.

3. Feb 17, 2013

Curl

4. Feb 17, 2013

Jorriss

I'm trying to understand your question but I don't quite follow? The right side seems to have units of charge squared divided by farads per meter, but the lefts units are completely different. What is this supposed to mean?

5. Feb 18, 2013

Staff Emeritus
Physics does.

You have received your answer: what you wrote does not have the correct units, so there is no answer to your question. It would be like asking "how long would it take a ball to fall 7 gallons"

6. Feb 19, 2013

Curl

I guess another way to ask this is to say if a particle of special type of matter had the property that it created a vector field given by
F=(c*m^2/r°r)*r/|r| where m is the mass of the particle,

then if I have two particles, each of mass m, one at the origin and one at x=1, then what would be the resultant field F at an arbitrary point r0 ???

7. Feb 19, 2013

cwilkins

However, if I'm understanding you correctly, I think you might be getting at the concept of linearity and the superposition of fields. Is your question what the resulting field would be if it varied as the square of the sum of all the masses/charges in some way? If so, you likely could not use Gauss's law at all, since it's a sum which assumes that the total field can be produced by summing each up piece of field, which in turn assumes a sum over each charge/mass. If the field produced by each infinitesimal source cannot be added linearly to produce the net overall field (that is, it is not independent of the other charges/masses present), then this sum does not give you the total flux, and therefore cannot give you the total mass/charge. In other words, by summing the field, you are no longer summing the total charge/mass, since they are not linearly related.

To answer your original question, if the divergence of the electric field were proportional to the square of charge, then nothing would really change other than we'd probably treat the fundamental charge as its squared counterpart (and some units would need to be added or modified). It would still be linear in superposition, though.

(Note: This is intuitive reasoning. Correctness not guaranteed.)

Last edited: Feb 19, 2013
8. Mar 8, 2013

Curl

I don't quite follow this part, how did I explicitly state it? I stated that the masses are at different locations.

9. Mar 9, 2013

cwilkins

You gave the expression for the force in your post. The values you gave could have been plugged into it.