How Does Gauss' Law Apply to Non-Symmetrical Shapes and Charge Distributions?

[Harrisonized]

Homework Statement

Gauss' Law states:

∫∫ E.dS = ∫∫∫ div(E) dV = Q_enc/ε₀

The proof is as follows (this is from Marsden's Vector Calculus 5e):

Let M be a elementary region in ℝ³. Then if (0,0,0) ∉ ∂M, we have:
∫∫_∂M r.n/r³ dS
= 4π if (0,0,0) ∈ M
= 0 if (0,0,0) ∉ M

Construct a sphere of radius δ with boundary ∂N such that the sphere is contained completely within M. Let W be the region between ∂N and ∂M. W has a divergence of 0.

Note that ∂N has a normal vector pointing inward toward the deleted point (0,0,0) and ∂M has a normal vector pointing outward away from the deleted point (0,0,0).

Since the total flux out of W must be 0 based on the divergence theorem, the flux out of ∂N and the flux out of ∂M must be equal. The flux out of ∂N is 4π, since

-∫∫_∂N r.n/r³ dS = ∫∫_∂N δ²/δ⁴ dS = 1/δ² ∫∫_∂N dS

∫∫_∂N dS = 4πδ², the surface area of a sphere. Take the limit as the sphere approaches 0.

Therefore, the flux out of ∂M containing the deleted point (0,0,0) is 4π.

Let E = Q/(4πε₀) r/r³. Then

∫∫ E.dS = Q/(4πε₀) ∫∫_∂M r.n/r³ dS = Q/ε₀

My questions are as follows:

1. Let M be a convex region of any finite size with a charge distribution. From the exterior of M, does the force follow Coulomb's law as though M were a point charge acting from the centroid? (I'm inclined to believe yes, because of the proof of Gauss' Theorem, but I can't be certain.)

2. Suppose I have a sphere with a conical "bite" on its side. Like this: (< The centroid is thus to the left of the sharp tip of the "cone". Suppose I put a test charge somewhere inside the cone, such that it's not touching the surface of the object. What force does it experience? (I'm inclined to believe that F = q₀Q/(4πε₀r²) toward the centroid.)

3. Suppose I have an off-center cavity near the surface of the sphere, such that it doesn't pass through the centroid of the shape. What force does a test charge placed in this cavity experience?

4. In my physics class, I'm told that Gauss' Law may only be used when there is a high degree of symmetry. However, the proof I mentioned above allows me to use Gauss' law for ANY shape, given that its centroid is contained within its boundary. Why is Gauss' Law used only for symmetrical regions? What allows me to use Gauss' Law for shapes that don't have centroids, such as infinite planes or infinite lines?

5. Can anyone tell me what's so special about a solid angle? I really don't understand proofs of Gauss' law that just mention the solid angle, such as this one:

http://faculty.cua.edu/sober/536/Gauss_solid_angle.pdf

I see in too many physics textbooks and it's driving me nuts because it just seems so... unsatisfactory.

 

[jbsteele]

Homework Equations Gauss' Law: ∫∫ E.dS = ∫∫∫ div(E) dV = Qenc/ε₀Coulomb's Law: F = q0Q/(4πε0r2)The Attempt at a Solution 1. Yes, the force follows Coulomb's Law as though M were a point charge acting from its centroid.2. The force would be q0Q/(4πε0r2) toward the centroid.3. The force would still be q0Q/(4πε0r2), but it would be directed away from the cavity.4. Gauss' Law can be used for shapes that don't have centroids because the law is based on the idea that the net electric flux out of a closed surface is zero. This means that the total electric flux into the surface must be equal to the total electric flux out of the surface, regardless of the shape of the surface.5. A solid angle is the three-dimensional analog of an angle. It measures the amount of the surface of a sphere that is subtended by a given area on the sphere's surface. It is useful for calculating the magnitude of a vector field in three-dimensional space, as it is related to the divergence of the field.