TITLE: How many closed measure zero sets are needed to cover the real line?
QUESTION [9 upvotes]: This question assumes familiarity with combinatorial cardinal characteristics of the continuum.
Let $\mathcal{E}$ be the $\sigma$-ideal generated by closed measure zero subsets of the real line. It is known that
$$\operatorname{cov}(\mathcal{N})\cdot\operatorname{cov}(\mathcal{M})\le 
\operatorname{cov}(\mathcal{E})\le 
\operatorname{cov}(\mathcal{N})\cdot \mathfrak{d},$$
and that the first inequality is consistently strict
(Bartoszynski-Shelah).
Is the second inequality consistently strict?
I conjecture that the answer is positive, and known, but did not find it in the cited paper (or elsewhere, but I didn't search thoroughly).
Update: This problem is solved below by Ashutosh, and the solution suggests a follow-up question.

REPLY [7 votes]: Since we can cover the reals by at most $\mathfrak{r}$ (reaping number) closed null sets, we can do a countable support iteration of rational perfect set forcing over a model of CH to get a model of $\omega_1 = \mathfrak{r} < \mathfrak{d} = \omega_2$. 
To see that $\operatorname{cov}(\mathcal{E}) \leq \mathfrak{r}$, just note that for every infinite $A \subseteq \omega$, $i \in \{0, 1\}$, the set $N_{A, i} = \{x \in 2^{\omega} : (\forall n \in A) (x(n) = i)\}$ is closed null.
$\mathfrak{d} = \omega_2$ in this model because Miller real is not dominated by any ground model real. Also the ground model p-points are preserved so $\mathfrak{r} = \mathfrak{u} = \omega_1$. The proof of this fact can be found in chapter 23 of Halbeisen's Combinatorial set theory.