A parallel plate capacitor acquires a charge of 40\ when connected...

The capacitance of a capacitor is given by the formula $C = \dfrac{Q}{V}$.

Substituting the given values:

$C = \dfrac{40\ \mu\text{C}}{20\ \text{V}} = 2\ \mu\text{F}$

The capacitance of a parallel plate capacitor depends only on its physical dimensions (area of the plates and the distance between them) and the dielectric medium between the plates. It is independent of the voltage applied across it or the charge stored in it.

Therefore, when the same capacitor is connected across a $60\ \text{V}$ battery, its capacitance remains unchanged at $2\ \mu\text{F}$.

Answer: $2\ \mu\text{F}$