Capacitance and Capacitors Solved Examples (Class 12 Physics)
Capacitance numericals explore capacitors, energy storage, parallel and series combinations, and dielectrics. These problems are crucial for understanding
TL;DR: Capacitance numericals explore capacitors, energy storage, parallel and series combinations, and dielectrics. These problems are crucial for understan…
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Capacitance numericals explore capacitors, energy storage, parallel and series combinations, and dielectrics. These problems are crucial for understanding
Capacitance and Capacitors — Solved Numerical Examples (Step by Step)
Example 1: A parallel plate capacitor has plate area 0.1 m² and separation 0.01 m. Calculate its capacitance. (ε₀ = 8.85 × 10^-12 F/m)
Solution: Area A = 0.1 m², Separation d = 0.01 m, ε₀ = 8.85 × 10^-12 F/m. Capacitance C = (ε₀ × A) / d = (8.85 × 10^-12 × 0.1) / 0.01 = (8.85 × 10^-13) / 0.01 = 8.85 × 10^-11 F = 88.5 pF.
Example 2: A 10 μF capacitor is charged to 100 V. Calculate the charge stored and energy stored.
Solution: Capacitance C = 10 × 10^-6 F, Voltage V = 100 V. Charge Q = C × V = 10 × 10^-6 × 100 = 10^-3 C = 1 mC. Energy U = 0.5 × C × V² = 0.5 × 10 × 10^-6 × 100² = 0.5 × 10 × 10^-6 × 10^4 = 0.05 J.
Example 3: Two capacitors of 6 μF and 3 μF are connected in series across a 100 V supply. Calculate equivalent capacitance, total charge, and voltage across each.
Solution: C₁ = 6 μF, C₂ = 3 μF. Equivalent Ceq = (C₁ × C₂)/(C₁ + C₂) = (6 × 3)/(6 + 3) = 18/9 = 2 μF. Total charge Q = Ceq × V = 2 × 10^-6 × 100 = 2 × 10^-4 C. Voltage across C₁: V₁ = Q/C₁ = (2 × 10^-4)/(6 × 10^-6) = 33.33 V. Voltage across C₂: V₂ = Q/C₂ = (2 × 10^-4)/(3 × 10^-6) = 66.67 V.
Example 4: A capacitor with dielectric constant 5 has area 0.05 m² and separation 0.005 m. Calculate capacitance. (ε₀ = 8.85 × 10^-12 F/m)
Solution: Dielectric constant K = 5, Area A = 0.05 m², Separation d = 0.005 m. Capacitance C = (K × ε₀ × A) / d = (5 × 8.85 × 10^-12 × 0.05) / 0.005 = (2.2125 × 10^-12) / 0.005 = 4.425 × 10^-10 F = 442.5 pF.
Example 5: Two identical capacitors of 10 μF are connected in parallel and charged to 50 V. Calculate total capacitance, total charge, and energy stored.
Solution: C₁ = C₂ = 10 μF. Equivalent Ceq = C₁ + C₂ = 20 μF. Voltage across both = 50 V (parallel connection). Total charge Q = Ceq × V = 20 × 10^-6 × 50 = 10^-3 C = 1 mC. Energy U = 0.5 × Ceq × V² = 0.5 × 20 × 10^-6 × 2500 = 0.025 J.
Tips
- Capacitance C = Q/V for any capacitor.
- Series capacitors: 1/Ceq = 1/C₁ + 1/C₂. Parallel capacitors: Ceq = C₁ + C₂.
- Energy stored: U = 0.5CV² = 0.5QV = Q²/(2C).
Frequently Asked Questions
What is a dielectric and how does it affect capacitance?
A dielectric is an insulating material placed between capacitor plates. It increases capacitance by a factor K (dielectric constant) by reducing the effective electric field between plates.
Why are capacitors used in electronic circuits?
Capacitors store electrical energy, filter signals, block DC while passing AC, and smooth voltage in power supplies. They are essential for timing circuits, filters, and energy storage.
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