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

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#