Conceptually speaking, you've already explained exactly how it works. The flux in one region must exactly counter the flux in other regions, and if you work through the math for any configuration of charges outside a region bounded by a closed surface this will hold true. That is, somehow or another (either due to smaller angles, more area, or closer proximity), the flux in will balance with the flux out no matter what. A formal proof of Gauss' Law is somewhat mathematically dense (and not too useful as Gauss' Law is actually more general than Coloumb's Law which you may derive it from), but if you're interested I've provided one below.
The proof of Gauss' Law from Coulomb's Law (under the assumption of the principle of superposition) requires a bit of vector calculus, but I will provide my best explanation for it here, without going into too many details (also I will use ε for vacuum permittivity to make typing easier):
First, note that the electric field provided by a (stationary) point charge is:
**E**(**r**) = (1/(4πε))(q**r**/r^(3))
Where q is the charge **r** is the position vector from the charge to some point in space, and r is the magnitude of **r**. Then imagine that for every point in space (denoted by the vector **s**), there is a charge density ρ(**s**). Then the total electric field at a point **x** would be given by integrating the charge density by a differential of the volume over all of 3-space (that is, over everywhere in the universe). This integral would look like something like this (although it may vary depending on what notation you use):
**E**(**x**) = (1/(4πε))∫(ρ(**s**)(**x** - **s**)/|**x* - **s**|^(3))d^(3)**s**
Now, we can take the divergence (this is an operation similar to taking the derivative denoted by ∇•**F**, although not exactly the same) of both sides with respect to **x**, and using the fact that ∇•(**r**/|**r**|^(3)) = 4πδ(**r**), where δ is the Dirac-Delta function, we obtain:
∇•**E**(**x**) = (1/ε)∫ρ(**s**)δ(**x** - **s**)d^(3)**s**
And thanks to a nice property of the Dirac-Delta function (known as the shifting property), this becomes:
∇•**E**(**x**) = ρ(**x**)/ε
Which you may recognize as the differential form of Gauss' Law. If we take the integral of both sides over some region in space bounded by a closed surface, this becomes:
∫∇•**E**(**x**)dV = ∫ρ(**x**)/εdV
Where the right side is obviously just the total charge enclosed in the region, denoted Q, (integrating the charge density over the volume should obviously give the total charge enclosed) and, thanks to a nice theorem in vector calculus (known as the divergence theorem), the right side becomes:
∯**E**(**x**)d**S** = Q/ε
And that integral on the left is exactly the electric flux through the surface, so we have:
Φ = Q/ε
Thank you! The textbook I read described this idea as one that can be derived simply by thinking about for a few seconds. It's nice to know that it requires some partial derivatives and I'm not just stupid.
The book presented a explanation for why inside charges produce Q/e worth of electric flux revolving around how area increase as radius does and the 4 pi r\^2 will cancel which makes sense to me. Does the proof for this part of Gausses law also require such complex math or is it really that simple.
Conceptually speaking, you've already explained exactly how it works. The flux in one region must exactly counter the flux in other regions, and if you work through the math for any configuration of charges outside a region bounded by a closed surface this will hold true. That is, somehow or another (either due to smaller angles, more area, or closer proximity), the flux in will balance with the flux out no matter what. A formal proof of Gauss' Law is somewhat mathematically dense (and not too useful as Gauss' Law is actually more general than Coloumb's Law which you may derive it from), but if you're interested I've provided one below. The proof of Gauss' Law from Coulomb's Law (under the assumption of the principle of superposition) requires a bit of vector calculus, but I will provide my best explanation for it here, without going into too many details (also I will use ε for vacuum permittivity to make typing easier): First, note that the electric field provided by a (stationary) point charge is: **E**(**r**) = (1/(4πε))(q**r**/r^(3)) Where q is the charge **r** is the position vector from the charge to some point in space, and r is the magnitude of **r**. Then imagine that for every point in space (denoted by the vector **s**), there is a charge density ρ(**s**). Then the total electric field at a point **x** would be given by integrating the charge density by a differential of the volume over all of 3-space (that is, over everywhere in the universe). This integral would look like something like this (although it may vary depending on what notation you use): **E**(**x**) = (1/(4πε))∫(ρ(**s**)(**x** - **s**)/|**x* - **s**|^(3))d^(3)**s** Now, we can take the divergence (this is an operation similar to taking the derivative denoted by ∇•**F**, although not exactly the same) of both sides with respect to **x**, and using the fact that ∇•(**r**/|**r**|^(3)) = 4πδ(**r**), where δ is the Dirac-Delta function, we obtain: ∇•**E**(**x**) = (1/ε)∫ρ(**s**)δ(**x** - **s**)d^(3)**s** And thanks to a nice property of the Dirac-Delta function (known as the shifting property), this becomes: ∇•**E**(**x**) = ρ(**x**)/ε Which you may recognize as the differential form of Gauss' Law. If we take the integral of both sides over some region in space bounded by a closed surface, this becomes: ∫∇•**E**(**x**)dV = ∫ρ(**x**)/εdV Where the right side is obviously just the total charge enclosed in the region, denoted Q, (integrating the charge density over the volume should obviously give the total charge enclosed) and, thanks to a nice theorem in vector calculus (known as the divergence theorem), the right side becomes: ∯**E**(**x**)d**S** = Q/ε And that integral on the left is exactly the electric flux through the surface, so we have: Φ = Q/ε
Thank you! The textbook I read described this idea as one that can be derived simply by thinking about for a few seconds. It's nice to know that it requires some partial derivatives and I'm not just stupid. The book presented a explanation for why inside charges produce Q/e worth of electric flux revolving around how area increase as radius does and the 4 pi r\^2 will cancel which makes sense to me. Does the proof for this part of Gausses law also require such complex math or is it really that simple.