# How is the derivative of a potential with respect to an outward normal equal to the grad of the potential

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.

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