# Vector identity and Electric Field Screening in Tinkham's

I'm going through Tinkham's "Introduction to Superconductivity" Second Edition. In Chapter 2, the author states that "we obtain

\begin{align} -\nabla \times \nabla \times {\bf E} = \nabla^2{\bf E} = {\bf E}/\lambda \end{align} after using a vector identity. These results apply only to time-varying electric fields since (1.3') would imply a current accelerating to infinity in response to a strictly dc electric field."

I have two questions regarding this passage:

1) What does the author mean by this last sentence? Why would a lack of a bound on the current for the dc case imply that the equation does not apply at all? Is it because we would leave a regime required for the London equations? If so, which assumption would we be breaking?

Possibly related to this:

2) By "vector identity", I suppose Tinkham is referring to \begin{align} {\bf A } \times {\bf B } \times {\bf C } = ({\bf A}\cdot {\bf C}) {\bf B} - ({\bf A}\cdot {\bf B}) {\bf C}, \end{align} which in the relevant equation becomes \begin{align} {\bf \nabla } \times {\bf \nabla } \times {\bf E } = {\bf \nabla} ({\bf \nabla}\cdot {\bf E}) - {\bf \nabla}^2 {\bf E}. \end{align}

So it seems to me that the author is assuming that $$\nabla \rho = 0$$, so that the first term of the rhs of the equation disappears. Is this the case? If so, what justifies this assumption?

A related question has been made here, but it deals with the magnetic version of this equation, presented shortly after in the book. In that case the discussion is different because $$\nabla \cdot {\bf h}$$ is always zero.