Question 8 of 11intermediate🔧 ApplyNumerical3 marks

In a parallel plate capacitor with air between the plates, each plate has an area of 6×106 \times 10–3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Correct Answer

Capacitance C = 17.7 pF (or 1.77×101.77 \times 10⁻¹¹ F). Charge on each plate Q = 1.77×101.77 \times 10⁻⁹ C (or 1.77 nC).

Exercise: EXERCISES | Q: 2.8 | (Page 79)
For More Understanding

Explanation

This question tests the application of the parallel plate capacitor formula derived in Section 2.12 of the textbook. The formula C = ε₀A/d shows that capacitance depends only on geometry. The charge is then found using Q = CV, a fundamental relationship for capacitors.

Solution Steps

  1. Step 1: Identify given values — Area A = 6×106 \times 10⁻³ m², distance d = 3 mm = 3×103 \times 10⁻³ m, supply voltage V = 100 V. The dielectric is air (essentially vacuum).

  2. Step 2: Apply the capacitance formula for parallel plate capacitor: C = ε₀A/d, where ε₀ = 8.85×108.85 \times 10⁻¹² F/m.

  3. Step 3: Calculate capacitance: C = (8.85×108.85 \times 10⁻¹² × 6×106 \times 10⁻³) / (3×103 \times 10⁻³) = (8.85×108.85 \times 10⁻¹² × 6) / 3 = 17.7×1017.7 \times 10⁻¹² F = 17.7 pF.

  4. Step 4: Calculate charge using Q = CV: Q = 17.7×1017.7 \times 10⁻¹² × 100 = 1.77×101.77 \times 10⁻⁹ C (or 1.77 nC).

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