A parallel plate capacitor separated by an air ... - JAMB Physics 2023 Question

The charge (\(Q\)) on each plate of the capacitor can be calculated using the formula \(Q = CV\), where:

- \(C\) is the capacitance,
- \(V\) is the voltage.

The capacitance (\(C\)) for a parallel plate capacitor with air as the dielectric is given by \(C = \frac{\varepsilon_0 A}{d}\), where:

- \(\varepsilon_0\) is the vacuum permittivity (\(8.85 \times 10^{-12} \, \text{F/m}\)),
- \(A\) is the area of the plates (\(0.8 \, \text{m}^2\)),
- \(d\) is the separation between the plates (\(0.02 \, \text{m}\)).

Let's substitute the values:

\[
C = \frac{(8.85 \times 10^{-12} \, \text{F/m})(0.8 \, \text{m}^2)}{0.02 \, \text{m}} = 3.54 \times 10^{-10} \, \text{F}
\]

Now, calculate the charge (\(Q\)):

\[
Q = CV = (3.54 \times 10^{-10} \, \text{F})(120 \, \text{V}) = 4.25 \times 10^{-8} \, \text{C}
\]

Convert the charge to nanocoulombs:

\[
4.25 \times 10^{-8} \, \text{C} = 42.5 \, \text{nC}
\]

Therefore, the correct answer is: \(42.5 \, \text{nC}\).