A

Step 1: Calculate the capacitance (C) of the parallel plate capacitor using the formula:
$$ C = \frac{\varepsilon_0 A}{d} $$
where:
- $\varepsilon_0 = 8.85 \times 10^{-12} \: \text{F/m}$ (permittivity of free space)
- $A = 100 \: \text{cm}^2 = 100 \times 10^{-4} \: \text{m}^2 = 1 \times 10^{-2} \: \text{m}^2$
- $d = 5 \: \text{mm} = 5 \times 10^{-3} \: \text{m}$

Substituting the values:
$$ C = \frac{(8.85 \times 10^{-12}) (1 \times 10^{-2})}{5 \times 10^{-3}} = \frac{8.85 \times 10^{-14}}{5 \times 10^{-3}} = 1.77 \times 10^{-11} \: \text{F} $$

Step 2: Calculate the charge (Q) on the capacitor using the formula:
$$ Q = C \cdot V $$
where $V = 5000 \: \text{V}$
$$ Q = (1.77 \times 10^{-11}) \cdot (5000) = 8.85 \times 10^{-8} \: \text{C} $$

Step 3: Calculate the electrostatic force (F) between the plates using:
$$ F = \frac{Q^2}{2 \varepsilon_0 A} $$
Substituting the values:
$$ F = \frac{(8.85 \times 10^{-8})^2}{2 (8.85 \times 10^{-12}) (1 \times 10^{-2})} = \frac{7.83 \times 10^{-15}}{1.77 \times 10^{-13}} = 0.0442 \: \text{N} $$

Step 4: Relate the force to the additional mass (m) that needs to be added:
$$ F = m g $$
where $g = 9.81 \: \text{m/s}^2$
$$ m = \frac{F}{g} = \frac{0.0442}{9.81} = 0.0045 \: \text{kg} = 4.5 \: ext{grams} $$
Thus, the additional mass needed to maintain balance is approximately 4.5 grams.