I've been trying to understand the solution to a question based on the Gauss Divergence Theorem. The question says,
If $\phi$ is harmonic in $V$, then $\iint_{S} \frac{\partial \phi}{\partial n} dS = 0$. (We need to prove this)
The solution goes like this, and I've added my comments too-
$\iint_{S} \frac{\partial \phi}{\partial n} dS = \iint_{S} (\frac{\partial \phi}{\partial n}n).n dS$. This makes sense because $n$ is the unit normal.
$= \iint_{S} (\nabla \phi).n dS$ I have trouble here. How can we write that $\frac{\partial \phi}{\partial n}n = \nabla \phi$??
$= \iiint_{V} div(\nabla \phi)dV$, by Divergence theorem, this step is clear
$= \iiint_{V} \nabla^{2} \phi dV = 0$, since $\phi$ is harmonic in $V$, this step is clear too..
Can anybody please explain the second step to me? Firstly, if n is the normal, what do we even, which is a vector, what do we mean by taking a partial derivative with respect to it? Even if we do, won't that result in a vector rather than a scalar? How would that result in the gradient of the vector? I tried writing that $\frac{\partial \phi}{\partial n} = \frac{\partial \phi}{\partial x}\frac{\partial x}{\partial n} + \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial n} + \frac{\partial \phi}{\partial z}\frac{\partial z}{\partial n}$, but was unable to proceed. Help would be appreciated.