Capacitance

Why it matters

Capacitance is a fundamental concept in electromagnetics that plays a crucial role in the design and functioning of electronic circuits. It is essential for understanding how energy is stored in electric fields, which impacts everything from power supply design to signal processing in communication systems.

Key ideas

• Capacitance is the ability of a system to store electric charge per unit voltage. It is measured in farads (F).
• Capacitors are devices designed to store charge and energy in the electric field between a pair of conductors.
• The capacitance of a parallel plate capacitor is determined by the area of the plates, the distance between them, and the permittivity of the dielectric material between the plates.
• Dielectric materials increase the capacitance by reducing the electric field strength for the same charge on the plates.
• Energy stored in a capacitor is given by the work done to charge it, which is stored in the electric field.

Formulas

Assume a linear dielectric completely fills the gap and neglect fringing. The energy relation uses voltage-independent capacitance. At fixed charge, adding dielectric reduces E; at fixed voltage and plate spacing, E = V/d is unchanged while charge increases.

• Capacitance of a parallel plate capacitor: C = ε₀·εr·A / d
  • C = Capacitance (F)
  • ε₀ = Permittivity of free space (8.854 × 10⁻¹² F/m)
  • εr = Relative permittivity of the dielectric material (dimensionless)
  • A = Area of one plate (m²)
  • d = Distance between the plates (m)
• Energy stored in a capacitor: E = 1/2·C·V²
  • E = Energy (J)
  • C = Capacitance (F)
  • V = Voltage across the capacitor (V)

Worked example

Given: A parallel plate capacitor with plate area A = 0.02 m², plate separation d = 0.01 m, and dielectric material with εr = 4. The voltage across the capacitor is V = 100 V.

1. Calculate the capacitance using the formula: C = ε₀·εr·A / d C = (8.854 × 10⁻¹² F/m)·4·0.02 m² / 0.01 m C = 7.0832 × 10⁻¹¹ F

2. Calculate the energy stored using the formula: E = 1/2·C·V² E = 1/2·7.0832 × 10⁻¹¹ F·(100 V)² E = 3.5416 × 10⁻⁷ J

Final Answer: 3.5416 × 10⁻⁷ J

Common mistakes

• Confusing the units of capacitance (farads) with those of charge (coulombs).
• Forgetting to convert units, especially when dealing with microfarads or picofarads.
• Neglecting the effect of the dielectric material on capacitance.

For GATE EC

Questions on capacitance often involve calculating the capacitance of different configurations, understanding the effect of dielectrics, and energy storage. Practice problems involving series and parallel combinations of capacitors, as well as the impact of changing physical parameters on capacitance.

Quick check

1. What is the unit of capacitance?
2. How does increasing the plate area affect the capacitance of a parallel plate capacitor?
3. What role does a dielectric material play in a capacitor?

Answers: 1. Farad 2. Increases capacitance 3. Increases capacitance by reducing the electric field strength.