A parallel plate capacitor separated by an air gap is made of 0.8m^2 tin plates and 20 mm apart. It is…

1. Analyze the question

We are asked to calculate the charge ($Q$) stored on the plates of a parallel plate capacitor. We are given:

• Area of plates ($A$) = $0.8 \, \text{m}^2$
• Distance between plates ($d$) = $20 \, \text{mm} = 0.02 \, \text{m}$
• Voltage ($V$) = $120 \, \text{V}$
• Permittivity of free space ($\varepsilon_o$) = $8.85 \times 10^{-12} \, \text{Fm}^{-1}$

2. Explain the concept

To find the charge, we use two fundamental formulas for capacitors:

1. Capacitance ($C$): For a parallel plate capacitor, $C = \frac{\varepsilon_o A}{d}$
2. Charge ($Q$): The relationship between charge, capacitance, and voltage is $Q = C \times V$

By combining these, we get: $Q = \frac{\varepsilon_o A V}{d}$

3. Show the solution

Step 1: Convert units to SI units

• $d = 20 \, \text{mm} = 20 \times 10^{-3} \, \text{m} = 0.02 \, \text{m}$

Step 2: Calculate the Capacitance ($C$)

Step 3: Calculate the Charge ($Q$)

To express this in nanoCoulombs (nC), where $1 \, \text{nC} = 10^{-9} \, \text{C}$:

4. Identify the answer

Comparing our result to the options provided, the correct value is 42.5 nC. This matches option B.

5. Tip

When dealing with capacitors, always ensure your distance ($d$) is in meters and your area ($A$) is in square meters before plugging them into the formula. A common mistake in JAMB exams is forgetting to convert millimeters (mm) to meters (m), which leads to a calculation error by a factor of 1000!