Unit 22: Capacitor

Capacitance and Capacitor

A capacitor is an electronic component that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material (dielectric) or vacuum. The ability of a capacitor to store charge per unit voltage is quantified by its capacitance.

The basic definition of capacitance is:

C = Q / V

where:

• C = capacitance (farads, F)
• Q = magnitude of charge on each plate (coulombs, C)
• V = potential difference between the plates (volts, V)

The SI unit of capacitance is the farad (F), defined as one coulomb per volt. Commonly used sub‑units are:

• 1 microfarad (µF) = 10⁻⁶ F
• 1 picofarad (pF) = 10⁻¹² F
• 1 nanofarad (nF) = 10⁻⁹ F

Example 1: A capacitor stores 5 µC of charge when a voltage of 10 V is applied. Its capacitance is:

C = Q / V = (5 × 10−6 C) / (10 V) = 0.5 × 10−6 F = 0.5 µF

Capacitors are characterised by their voltage rating, tolerance, and temperature coefficient, which determine their suitability for specific circuits.

Parallel Plate Capacitor

The simplest geometry for analysis is the parallel plate capacitor, consisting of two large, flat, conducting plates of area A separated by a distance d. When a voltage V is applied, a uniform electric field E exists between the plates (ignoring edge effects).

Capacitance Formula

Deriving from Gauss’s law, the capacitance of a vacuum‑filled parallel plate capacitor is:

C = ε₀ A / d

where:

• ε₀ = vacuum permittivity = 8.854 × 10⁻¹² F·m⁻¹
• A = overlapping area of one plate (m²)
• d = separation between plates (m)

Electric Field Between Plates

The magnitude of the uniform electric field is:

E = V / d

This relationship shows that for a given voltage, decreasing the plate separation increases the field strength.

Effect of a Dielectric

When a dielectric material of dielectric constant K (relative permittivity) fills the space between the plates, the capacitance increases by a factor K:

C = K ε₀ A / d

The dielectric reduces the effective electric field inside the capacitor to E' = E / K while allowing more charge to be stored for the same applied voltage.

Example 2: Calculate the capacitance of a parallel plate capacitor with plate area 0.02 m², separation 1 mm, and air (K≈1) as dielectric.

C = ε₀ A / d = (8.854×10−12 F/m) × (0.02 m²) / (0.001 m) = 1.77×10−10 F = 177 pF

If a dielectric with K = 4 is inserted, the capacitance becomes:

C' = K C = 4 × 177 pF = 708 pF

Combination of Capacitors

In practical circuits, capacitors are often connected together to achieve desired capacitance values. The rules for combining capacitors are analogous to those for resistors, but with series and parallel roles interchanged.

Parallel Combination

When capacitors are connected in parallel, the voltage across each capacitor is the same, and the total charge stored is the sum of individual charges. The equivalent capacitance C_eq is:

C_eq = C₁ + C₂ + C₃ + …

This results in an equivalent capacitance larger than any individual capacitor.

Series Combination

For capacitors in series, the charge on each capacitor is identical, while the voltage divides among them. The reciprocal of the equivalent capacitance equals the sum of reciprocals:

1 / C_eq = 1 / C₁ + 1 / C₂ + 1 / C₃ + …

Thus, the equivalent capacitance of a series network is always smaller than the smallest individual capacitor.

Mixed Networks

Complex networks can be reduced step‑by‑step by identifying series or parallel groups, computing their equivalent capacitance, and repeating until a single equivalent capacitance remains.

Example 3 (Parallel): Three capacitors of 2 µF, 4 µF, and 6 µF are connected in parallel across a 12 V supply.

Equivalent capacitance:

C_eq = 2 µF + 4 µF + 6 µF = 12 µF

Total charge stored:

Q_total = C_eq V = 12 µF × 12 V = 144 µC

Example 4 (Series): The same three capacitors are now connected in series across the same 12 V supply.

Reciprocal sum:

1/C_eq = 1/2 + 1/4 + 1/6 = (6 + 3 + 2)/12 µF⁻¹ = 11/12 µF⁻¹

Hence:

C_eq = 12/11 µF ≈ 1.09 µF

Charge on each capacitor (same for series):

Q = C_eq V ≈ 1.09 µF × 12 V ≈ 13.1 µC

Energy of a Charged Capacitor

Energy is stored in the electric field between the plates. When a capacitor is charged, work done by the voltage source appears as electrostatic potential energy.

Energy Formulas

The energy U stored in a capacitor can be expressed in three equivalent forms:

U = ½ C V² = ½ Q V = Q² / (2 C)

Derivation: The infinitesimal work done to move a small charge dq against the instantaneous voltage v = q/C is dW = v dq = (q/C) dq. Integrating from 0 to Q gives:

U = ∫₀ᴠ (q/C) dq = (1/C) ∫₀ᴠ q dq = (1/C) (Q²/2) = Q²/(2C)

Using Q = CV yields the other forms.

Energy Density

For a parallel plate capacitor, the energy per unit volume (energy density) in the field is:

u = ½ ε₀ E²

where E = V/d is the electric field magnitude. This expression shows that the energy resides in the field itself, not in the plates.

Example 5: A 10 µF capacitor is charged to 100 V. Compute the stored energy.

U = ½ C V² = 0.5 × (10×10⁻⁶ F) × (100 V)² = 0.5 × 10×10⁻⁶ × 10⁴ = 0.5 × 0.1 = 0.05 J

The same result using Q = CV = (10×10⁻⁶)(100) = 1×10⁻³ C gives:

U = Q²/(2C) = (1×10⁻³)² / (2 × 10×10⁻⁶) = 1×10⁻⁶ / (2×10⁻⁵) = 0.05 J

Effect of Dielectric

Inserting a dielectric material between the plates of a capacitor influences its behaviour in several interconnected ways.

Increase in Capacitance

As shown earlier, the capacitance is multiplied by the dielectric constant K:

C_with_dielectric = K C_without_dielectric

Typical values: K ≈ 1 for vacuum/air, K ≈ 3.9 for silicon dioxide, K ≈ 80 for water (static), and K ≈ 1000 for certain ferroelectric ceramics.

Polarization

Dielectric molecules contain bound charges that tend to align with the applied electric field. This alignment creates an internal polarization P (dipole moment per unit volume) that opposes the field, reducing the net field inside the material:

E_net = E₀ – P/ε₀

where E₀ is the field that would exist in vacuum.

Electric Displacement

The electric displacement field D accounts for both free and bound charges:

D = ε₀ E + P = ε E

with ε = K ε₀ being the absolute permittivity of the dielectric. In a capacitor, D remains continuous across the dielectric interface, while E is reduced by factor K.

Practical Implications

• Dielectrics allow capacitors to achieve larger capacitance in a compact size.
• They increase the voltage rating because the material can withstand higher electric fields before breakdown.
• Different dielectrics exhibit varying losses (dielectric loss tangent) and temperature stability, influencing selection for AC vs. DC applications.

Example 6: A parallel plate capacitor with air (K=1) has capacitance 200 pF. If a mica sheet (K≈5) completely fills the gap, what is the new capacitance?

C_new = K C_air = 5 × 200 pF = 1000 pF = 1 nF

If the same capacitor is charged to 50 V, the stored energy increases from:

U_air = ½ × 200 pF × (50 V)² = 0.5 × 200×10⁻¹² × 2500 = 0.25 µJ

to

U_mica = ½ × 1 nF × (50 V)² = 0.5 × 1×10⁻⁹ × 2500 = 1.25 µJ

Thus, inserting the dielectric raises the stored energy by a factor of five for the same voltage.

---

In summary, this chapter has laid out the quantitative foundation of capacitors—from the basic definition of capacitance, through geometry‑specific formulas, combination rules, energy storage, and dielectric effects. Mastery of these concepts enables students to analyse and design circuits involving capacitors for filtering, timing, energy storage, and signal processing applications.