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.

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    $\begingroup$ $\frac{\partial V}{\partial n}$ 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. $\endgroup$ – Omnomnomnom Dec 28 '17 at 16:26
  • $\begingroup$ @Omnomnomnom is that then saying that $n$ is really a unit vector/versor? $\endgroup$ – QuantumPenguin Dec 28 '17 at 16:46
  • $\begingroup$ Yes, $n$ is the unit vector which points outward from the enclosed surface $\endgroup$ – Omnomnomnom Dec 28 '17 at 16:54
  • $\begingroup$ @Omnomnomnom but surely in this case $\boldsymbol{E}$ is my normal vector? Because, $\boldsymbol{E} = \nabla V = (\partial _{x}V,\partial _{y}V,\partial _{z}V)$ $\endgroup$ – QuantumPenguin Dec 28 '17 at 17:00
  • $\begingroup$ What makes you think that the direction of $E$ is necessarily normal to the surface? $\endgroup$ – Omnomnomnom Dec 28 '17 at 17:05

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