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If we first consider Gauss' Law $$\oint_s \boldsymbol{E\cdot} \,d\boldsymbol{A} = Q_{enclosed\\in\ surface}$$ We know from physics that $\boldsymbol{E}=-\nabla V$, but I want to know is it mathematically equivalent to say $-\partial V/\partial n = \nabla V$ -- and if so, how? Here, $n$ is the outward normal from the enclosed surface.

user27119
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    $\frac{\partial V}{\partial n}$ is a scalar, while $\nabla V$ is a vector. The correct relationship is $\frac{\partial V}{\partial n} = \nabla V \cdot n$, where $\cdot$ denotes the dot-product. – Ben Grossmann Dec 28 '17 at 16:26
  • @Omnomnomnom is that then saying that $n$ is really a unit vector/versor? – user27119 Dec 28 '17 at 16:46
  • Yes, $n$ is the unit vector which points outward from the enclosed surface – Ben Grossmann Dec 28 '17 at 16:54
  • @Omnomnomnom but surely in this case $\boldsymbol{E}$ is my normal vector? Because, $\boldsymbol{E} = \nabla V = (\partial _{x}V,\partial _{y}V,\partial _{z}V)$ – user27119 Dec 28 '17 at 17:00
  • What makes you think that the direction of $E$ is necessarily normal to the surface? – Ben Grossmann Dec 28 '17 at 17:05
  • @Omnomnomnom Nothing, which is the reason for my confusion. See http://mathworld.wolfram.com/NormalVector.html -- Specifically Eq. (3) – user27119 Dec 28 '17 at 17:10
  • Right, so the point is that $E$ is not your normal vector. – Ben Grossmann Dec 28 '17 at 17:11
  • In that context, the function $f$ is used to describe the surface. There is no connection between $\nabla f$ as it's used in that context and our $\nabla V$, the vector field to be integrated. – Ben Grossmann Dec 28 '17 at 17:12
  • @Omnomnomnom RIGHT. Yes you are correct. I was just pattern matching and not thinking there -- thanks! – user27119 Dec 28 '17 at 17:15

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