Consider the arrangement illustrated in the figure. Suppose a small...

Let $N$ be the normal reaction between the upper block of mass $m$ and the lower block of mass $M$.

For the upper block, the forces in the vertical direction are its weight $mg$ acting downwards, the normal reaction $N$ acting upwards, and the electric force $F_e = QE$ acting upwards (since the charge is positive and the electric field is vertically upward).
For vertical equilibrium of the upper block:
$N + QE = mg$
$\Rightarrow N = mg - QE$

The maximum static friction force between the two blocks is:
f_\max = \mu N = \mu(mg - QE)

Let $T$ be the tension in the string. When the maximum horizontal force $F$ is applied to the left on the upper block, it tends to slide to the left relative to the lower block. Thus, the friction force $f$ on the upper block acts to the right.
For the horizontal equilibrium of the upper block:
$F = T + f$

For the lower block, the string pulls it to the right with tension $T$, and the friction force $f$ exerted by the upper block acts to the left. Since the floor is smooth, there is no friction from the floor.
For the horizontal equilibrium of the lower block:
$T = f$

Substituting $T = f$ into the equation for the upper block, we get:
$F = f + f = 2f$

To find the maximum force $F$ without disturbing the equilibrium, the friction must be at its limiting value, f = f_\max.
F_\max = 2f_\max = 2\mu(mg - QE)