For a parallel plate capacitor having plate area $1.13 × 10^3\, m^2$ and separation between plates 0.5 cm, the capacitance is

2.0 μF

The correct answer is Option (2) → 2.0 μF

Given:

Plate area, $A = 1.13 \times 10^3 \, \text{m}^2$

Separation between plates, $d = 0.5 \, \text{cm} = 0.005 \, \text{m}$

Permittivity of free space, $\varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}$

Formula for capacitance of a parallel plate capacitor:

$C = \frac{\varepsilon_0 A}{d}$

Substituting the values:

$C = \frac{8.854 \times 10^{-12} \times 1.13 \times 10^3}{0.005}$

$C = \frac{1.000502 \times 10^{-8}}{0.005}$

$C = 2.001 \times 10^{-6} \, \text{F}$

$C = 2.001 \, \mu\text{F}$