A parallel-plate capacitor has plates of area 0.040 m² separated by a vacuum gap of 2.0 mm. Calculate its capacitance. (Permittivity of free space ε0 = 8.85 × 10⁻¹² F/m)
Answer and explanation
Correct answer: C. 177 pF
- Step 1: Capacitance formula: C = ε0 A / d. Step 2. Convert gap to meters: d = 2.0 × 10⁻³ m. Step 3. Compute: C = (8.85 × 10⁻¹² × 0.040) / (2.0 × 10⁻³) = 3.54 × 10⁻¹³ / 2.0 × 10⁻³ = 1.77 × 10⁻¹⁰ F = 177 pF. Answer: 177 pF. Why not the others: 354 pF does not divide by 2; 88.5 pF halves the plate area; 1.77 nF is an order-of-magnitude error.
Practise more Electricity, Magnetism and Modern Physics.
Free questions with an explanation for every one, and full timed Unified Tertiary Matriculation Examination mock tests.
More Electricity, Magnetism and Modern Physics questions
- Calculate the electrostatic force of attraction between two point charges of +2.0 μC and -4.0 μC separated by a distance of 0.20 m in a vacu…
- Determine the electric field intensity at a distance of 0.30 m from an isolated positive point charge of 5.0 nC in air. (Take k = 9.0 × 10⁹…
- Two capacitors of capacitances 4.0 μF and 6.0 μF are connected in series across a direct current voltage supply. What is the equivalent capa…
- A 10 μF capacitor is charged to a potential difference of 200 V. Calculate the total electrostatic energy stored in the capacitor.
- A chemical cell of electromotive force (emf) 1.50 V and internal resistance 0.50 Ω is connected across an external resistor of 2.50 Ω. Calcu…
- A cylindrical resistance wire of length 2.0 m and cross-sectional area 5.0 × 10⁻⁷ m² has an electrical resistance of 4.0 Ω. Calculate the el…