electromagnetism - How to derive the expression for the electric field in terms of the potential?

How can I derive that

$$\vec{E}=-\vec{\nabla}\phi-\frac{\partial \vec{A}}{\partial t}$$ where $\phi$ is the scalar potential and $\vec{A}$ the vector potential?

Answer

$\def\vA{{\vec{A}}}$ $\def\vB{{\vec{B}}}$ $\def\vD{{\vec{D}}}$ $\def\vE{{\vec{E}}}$ $\def\vH{{\vec{H}}}$ $\def\vS{{\vec{S}}}$ $\def\eps{\varepsilon}$ $\def\rot{\operatorname{rot}}$ $\def\div{\operatorname{div}}$ $\def\grad{\operatorname{grad}}$

Faraday's law:

$$\rot(\vE)+\dot\vB=0$$ Source-less B-field: $$\div(\vB)=0$$ From this in a simply connected domain there follows the existence of a vector potential $\vA$ with $$\vB = \rot\vA$$ Therewith, Faraday's law reads $$\rot(\vE+\dot{\vA})=0$$ The curl-freeness of the vector field $\vE+\dot{\vA}$ (in a simply connected domain) implies the existence of a scalar potential with $$\vE+\dot{\vA} = -\grad\varphi$$ and that is your formula. You see $\varphi$ is just defined in the way you wrote it down.

The purpose of introducing $\vA$ is to solve $\div\vB=0$ and the purpose of introducing $\varphi$ is to solve Faraday's law. In most cases the only equation which remains to be solved is Ampre's law

$$\rot\vH = \vS + \dot\vD$$ which reads with our new independent variables $\varphi$ and $\vA$ as $$\rot(\mu\rot \vA) = (\kappa + \eps\partial_t)(-\grad\varphi - \dot\vA).$$ Maybe, one has also a pre-defined space-charge density $\rho$. That would imply the equation $$\rho = \div \vD = \div (\eps(-\grad\varphi - \dot\vA))$$ For constant $\eps$ you have $$-\frac{\rho}{\eps} = \Delta\varphi + \div \dot\vA$$ Now, you have some degrees of freedom in the choice of $\vA$. If $\vA$ is a vector potential for $\vB$ then for any smooth scalar function $\varphi'$ also $\vA':=\vA+\grad\varphi'$ is a vector potential for $\vB$ since $\rot\grad=0$. One possible choice is $\div\vA = 0$. Therewith, the equation for the space charge reads just $$-\frac{\rho}{\eps} = \Delta\varphi$$ which we know from electro-statics. The nice thing of $\div\vA=0$ is that the above equation decouples from the magnetics. (But only if $\rho$ assumed to be predefined.) So one can solve the problem staggered.

The condition $\div\vA=0$ can always be satisfied. If we initially have a vector potential $\vA'$ with $\div\vA'\neq 0$ then we define $\vA := \vA'+\grad\varphi'$ such that

$$0 = \div \vA = \div(\vA' + \grad\varphi')= \div\vA' + \Delta \varphi'$$ To find the function $\varphi'$ for the modification of $\vA'$ we just have to solve the poisson equation $$\Delta\varphi' = -\div\vA'$$ for $\varphi'$. The coice $\div A=0$ is the so called Coulomb gauging.

A field is determined by its curl and sources. Under the assumption that we have already the required boundary conditions for $\varphi$ and $\vA$ then the fixation of the divergence for the vector potential (i.e., the gauging) determines the potentials uniquely. Beside the Coulomb gauging there are other forms of useful gauging (e.g., Lorentz-gauging).