Exercise 5.5.6: One static field, two sets of potentials

Pair 1: $B=\nabla \times 0=0$ and $E=-\nabla (-E_{0}x)=E_0{a}_x$. Pair 2: $A_2$ does not depend on position, so $B=\nabla \times A_2=0$, and $E=-\partial A_2/\partial t=E_0{a}_x$.

We need $A_2=A_1+\nabla \psi$ and $V_2=V_1-\partial \psi /\partial t$. The first gives $\partial \psi /\partial x=-E_{0}t$, $\partial \psi /\partial y=\partial \psi /\partial z=0$, so $\psi =-E_{0}xt+h(t)$. The second gives $0=-E_{0}x+E_{0}x-h^{\prime }(t)$, so $h$ is a constant, which can be dropped:

$\nabla \cdot A_1=0$ with $\partial V_1/\partial t=0$, and $\nabla \cdot A_2=\partial (-E_{0}t)/\partial x=0$ with $V_2=0$: both pairs satisfy $\nabla \cdot A+{\mu }_0{\epsilon }_0\partial V/\partial t=0$. For $\psi$: ${\nabla }^2\psi =0$ (linear in $x$) and ${\partial }^2\psi /\partial t^2=0$ (linear in $t$), so ${\nabla }^2\psi -c^{-2}{\partial }^2\psi /\partial t^2=0$.

In the first description $V_1(0)-V_1(1\mathrm{m})=E_0\times 1\mathrm{m}=1\mathrm{kV}$, the familiar result. In the second $V_2=0$ everywhere, and the whole field is carried by $-\partial A_2/\partial t$. The line integral ${\int }_0^1{E}_{x}dx=1\mathrm{kV}$ is the same in both, because it is built from $E$ alone.

Units: $E_{0}xt$ is $(V/m)\mathrm{m}\mathrm{s}=\mathrm{V}\mathrm{s}=\mathrm{Wb}$, so $\nabla \psi$ is in $Wb/m$ like $A$ and $\partial \psi /\partial t$ in $\mathrm{V}$ like $V$.