Maxwell equation in differential and integral form proof
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Your confusion lies in failing to recognize that they are exactly the same equations. Take for example Gauss's law
∇⃗ ⋅E⃗ =ρϵ0
You can see that there ρ is the charge distribution, and in general can be a funcion of the position.
Now consider a volume V, you can just integrate the density to obtain the total charge in the volume, but you can also integrate the electric field gradient to get a measure of the electric field
∇⃗ ⋅E⃗ =ρϵ0/∫VdV
∫V∇⃗ ⋅E⃗ dV=∫Vρϵ0dV
Now you can use the Divergence's Theorem on the left side.
∇⃗ ⋅E⃗ =∭V∇⃗ ⋅E⃗ =∬SE⋅dS⃗
Here S is the surface of the volume and dS⃗ is the area element of the surface pointing perpendicularly to the surface out from the volume so the equation final form is
∬SE⋅dS⃗ =∭Vρϵ0dV
Which convention's aside is the same equation as shown in Wikipedia. You can do exactly the same to each of the Maxwell's equations using the
∇⃗ ⋅E⃗ =ρϵ0
You can see that there ρ is the charge distribution, and in general can be a funcion of the position.
Now consider a volume V, you can just integrate the density to obtain the total charge in the volume, but you can also integrate the electric field gradient to get a measure of the electric field
∇⃗ ⋅E⃗ =ρϵ0/∫VdV
∫V∇⃗ ⋅E⃗ dV=∫Vρϵ0dV
Now you can use the Divergence's Theorem on the left side.
∇⃗ ⋅E⃗ =∭V∇⃗ ⋅E⃗ =∬SE⋅dS⃗
Here S is the surface of the volume and dS⃗ is the area element of the surface pointing perpendicularly to the surface out from the volume so the equation final form is
∬SE⋅dS⃗ =∭Vρϵ0dV
Which convention's aside is the same equation as shown in Wikipedia. You can do exactly the same to each of the Maxwell's equations using the
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