A parallel-plate capacitor with a capacitance of 5 is connected to...

Given, potential difference across the plates, $V = 6 \text{ V}$

Separation between the plates, $d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$

The electric field $E$ between the plates of a parallel-plate capacitor is given by the relation:

$E = \dfrac{V}{d}$

Substituting the given values:

$E = \dfrac{6}{2 \times 10^{-3}}$

$E = 3 \times 10^3 \text{ V m}^{-1}$

The given value of capacitance is extra information and is not required for this calculation.