Question

How is solid angle related to electric field according to Gauss's Law?

Answer 1

There is a "proof" of Gauss' Law that draws from the idea of the solid angle for a simplification.

Explanation:

By definition, the electric flux through a closed surface `sigma` ( not limited to a sphere) enclosing a single point charge `q` is:

`Phi = oint_sigma \ mathbf E * mathbf hat n \ d sigma`

If we use a coordinate system with the point charge at the origin, then Coulomb's Law states that the field at `mathbf r`, a point on `sigma`, is:

`mathbf E = q/(4 pi varepsilon_o r^2) mathbf hat r`

So the integral looks like this:

`Phi = oint_sigma \ q/(4 pi varepsilon_o r^2) mathbf hat r \ * mathbf hat n \ d sigma = q/(4 pi varepsilon_o ) \ oint_sigma \ (mathbf hat r \ * mathbf hat n \ d sigma)/r^2`

If `dA` is the projection of `d sigma` onto a sphere of radius r centred on the charge (and touching `sigma` at `mathbf r`), then:

• `dA = mathbf hat r \ * mathbf hat n \ d sigma`

The solid angle, by definition, is:

• `d Omega = (dA)/r^2 = (mathbf hat r \ * mathbf hat n \ d sigma)/r^2`

So this simplifies the integration to:

`Phi = q/(4 pi varepsilon_o) \ int_Omega \ \ d Omega = q/varepsilon_o`

This can then be extended using superposition, repeating the process for all other charges within `sigma`, so that:

`Phi = sum_i \ oint_sigma \ mathbf E_i * mathbf hat n \ d sigma = sum q_(enc)/varepsilon_o`