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

The charge on a capacitor is given by the relation $Q = CV$.

Given:
Capacitance, $C = 5 \mu\text{F}$
Potential difference, $V = 6 \text{ V}$

Substituting the values:
$Q = 5 \mu\text{F} \times 6 \text{ V} = 30 \mu\text{C}$

The separation between the plates ($d = 2 \text{ mm}$) is not required for this calculation. The charge on the positive plate is $30 \mu\text{C}$.