If the aforementioned capacitor is connected across a 6.0 V...

The capacitance of the parallel-plate capacitor formed by the coated surfaces is given by:

$C = \dfrac{K \varepsilon_0 A}{d}$

Using the standard parameters associated with this problem context (area $A = 100 \text{ cm}^2$, thickness $d = 1 \text{ mm}$, and dielectric constant $K = 8$):

$C = \dfrac{8 \times 8.854 \times 10^{-12} \times 100 \times 10^{-4}}{10^{-3}}$

$C = 7.083 \times 10^{-10} \text{ F}$

When the capacitor is connected across a $6.0 \text{ V}$ battery, the net charge appearing on one of the coated surfaces is:

$Q = CV$

$Q = (7.083 \times 10^{-10}) \times 6.0$

$Q = 4.2498 \times 10^{-9} \text{ C}$

$Q \approx 4.2 \text{ nC}$

Answer: $4.2 \text{ nC}$