Capacitor — Capacitance and Dielectrics

Unit 22: Capacitor — Capacitance and Dielectrics

A capacitor is one of the three fundamental passive components of electric circuits, alongside the resistor and the inductor. While a resistor dissipates electrical energy as heat, a capacitor stores electrical energy in the form of an electric field. This chapter develops the concept of capacitance, derives the capacitance of a parallel plate capacitor using Gauss's law, analyses capacitor combinations, and examines how dielectric materials modify capacitor behaviour.

22.1 Capacitance and Capacitor

22.1(a) Uses of Capacitors in Simple Electrical Circuits

A capacitor is a device consisting of two conductors, generally called plates, separated by an insulating medium known as a dielectric. It can store electric charge and hence electrical energy. When connected across a potential difference, equal and opposite charges accumulate on the plates and an electric field is set up between them.

         ┌─────────────────┐
         │                 │
    ├────┤                 ├────┤
    │    │   (dielectric)  │    │
    ├────┤                 ├────┤
         │                 │
         └─────────────────┘
         
      Plate A (+Q)    Plate B (-Q)
      
            Circuit Symbol:
            
                  │  │
             ─────┤  ├─────
                  │  │

Figure 22.1: Capacitor structure and circuit symbol

Common uses of capacitors:

Energy storage: Used in camera flashes and similar rapid discharge applications.
Smoothing and filtering: Helps reduce voltage fluctuations in rectifier circuits.
Timing circuits: Used with resistors in RC circuits for delay and pulse timing.
Tuning circuits: Variable capacitors are used in radios and receivers.
Coupling and decoupling: Allows AC to pass while blocking DC in amplifier circuits.
Protection: Absorbs sudden voltage surges and protects components.

22.1(b) Definition of Capacitance

When the potential difference across a capacitor is changed, the charge stored on it also changes. Experiments show that, for a given capacitor, the charge stored is directly proportional to the potential difference. The constant ratio is called the capacitance.

C = ΔQ / ΔV

Capacitance is the ratio of the change in charge on a conductor to the corresponding change in its electric potential. It measures the ability of a system to store charge per unit potential difference.

For a capacitor charged from zero to a final value, the relation becomes:

C = Q / V

where Q is the charge on either plate and V is the potential difference between them. The SI unit of capacitance is the farad (F), defined as:

1 F = 1 C / 1 V

One farad is very large; practical capacitor values are often given in microfarads (μF = 10⁻⁶ F), nanofarads (nF = 10⁻⁹ F), or picofarads (pF = 10⁻¹² F).

22.1(c) Q–V Graph and Capacitance

For an ideal capacitor, charge is directly proportional to potential difference, so a graph of Q against V is a straight line through the origin. The slope of this graph gives the capacitance:

         Q
         ↑
         │        ╱
         │       ╱
         │      ╱
       Q │     ╱  ← slope = C
         │    ╱
         │   ╱
         │  ╱
         │ ╱
         │╱____________________→ V
         O                      V

Figure 22.2: Charge vs Potential Difference (slope = C)

slope = rise / run = ΔQ / ΔV = C

Conclusion: Capacitance is numerically equal to the gradient (slope) of the Q–V graph.

22.2 Parallel Plate Capacitor

22.2(a) Derivation of C = ε₀A/d Using Gauss's Law

A parallel plate capacitor consists of two large, flat conducting plates of area A separated by a small distance d. The region between them is usually vacuum or air.

         ┌───────────────────────────┐
         │  +  +  +  +  +  +  +  +   │
         │  +  +  +  +  +  +  +  +   │  (+Q)
         │  +  +  +  +  +  +  +  +   │
         │                           │
              ↓ E ↓ E ↓ E ↓ E ↓ E ↓
              (Electric field)
                  distance d
              ↑ E ↑ E ↑ E ↑ E ↑ E ↑
         │                           │
         │  -  -  -  -  -  -  -  -   │  (-Q)
         │  -  -  -  -  -  -  -  -   │
         │  -  -  -  -  -  -  -  -   │
         └───────────────────────────┘
         
              Area A

Figure 22.3: Parallel plate capacitor showing field lines

Step 1: Electric field between the plates

Using Gauss's law and considering an isolated charged sheet:

E_single sheet = σ / 2ε₀

In a parallel plate capacitor, the fields due to both charged plates add up between the plates:

E = σ/2ε₀ + σ/2ε₀ = σ/ε₀ = Q / (ε₀A)

Step 2: Potential difference between the plates

For uniform field:

V = E × d = Qd / (ε₀A)

Step 3: Capacitance formula

C = Q / V = Q / [Qd / (ε₀A)] = ε₀A / d

Parallel Plate Capacitor Formula:

C = ε₀A / d

where:

ε₀ = 8.85 × 10⁻¹² F m⁻¹ (permittivity of free space)
A = area of overlap of plates (m²)
d = separation between plates (m)

22.2(b) Effect of Plate Area and Separation

Effect of plate area (A):

C ∝ A (at constant d)
Larger plate area → more space for charge → larger capacitance

Effect of separation (d):

C ∝ 1/d (at constant A)
Smaller separation → stronger electric field → larger capacitance

22.2(c) Effect of a Dielectric

When a dielectric material fills the space between the plates:

C = εᵣ ε₀ A / d

where εᵣ is the relative permittivity (dielectric constant). Since εᵣ > 1 for all dielectrics, inserting a dielectric always increases the capacitance.

Typical dielectric constants:

Air ≈ 1.0006
Paper ≈ 2 to 3.5
Glass ≈ 4 to 10
Mica ≈ 3 to 6
Water ≈ 80

22.3 Combination of Capacitors

22.3(a) Capacitors in Series

When capacitors are connected in series, the same charge Q flows through each capacitor. The total potential difference is the sum of individual potential differences.

         ┌────────────────────────────┐
         │                            │
      ┌──┤  C₁   C₂   C₃          ├──┐
      │  │ ┌─┐  ┌─┐  ┌─┐         │  │
   ───┤  └─│ │──│ │──│ │─────────┤  ├───
      │    └─┘  └─┘  └─┘         │  │
      │                          │  │
      │     ─── V₁ ───   ─ V₂ ─  │  │
      │           ─ V₃ ─          │  │
      │         ─── V_total ───   │  │
      │                          │  │
      └──────────────────────────┴──┘
              (V₁ + V₂ + V₃ = V)

Figure 22.4: Capacitors in series (same charge Q on each)

1/Cₛ = 1/C₁ + 1/C₂ + 1/C₃ + ...

Key point: Equivalent capacitance in series is always smaller than the smallest individual capacitor.

Example 1 (Series):

Two capacitors of 4 μF and 6 μF connected in series. Find equivalent capacitance.

1/Cₛ = 1/4 + 1/6 = (3+2)/12 = 5/12
Cₛ = 12/5 = 2.4 μF

22.3(b) Capacitors in Parallel

When capacitors are connected in parallel, the same potential difference V appears across each capacitor. The total charge is the sum of individual charges.

         ┌─────────────────────────────┐
         │                             │
         ├──┬──────────────┬───────────┤
         │  │              │           │
         │ ┌─┐            ┌─┐        ┌─┐
    ┌────┤ │ │C₁         │ │C₂      │ │C₃
    │    │ └─┘            └─┘        └─┘
    │    │  │              │           │
    │    └──┼──────────────┼───────────┘
    │       │              │
    │       ─ V ─   ─ V ─   ─ V ─
    │       (same voltage across all)
    │
    │    Q = Q₁ + Q₂ + Q₃
    │
    └─────────────────────────────────

Figure 22.5: Capacitors in parallel (same voltage V across each)

Cₚ = C₁ + C₂ + C₃ + ...

Key point: Equivalent capacitance in parallel is always greater than the largest individual capacitor.

Example 2 (Parallel):

Two capacitors 5 μF and 8 μF connected in parallel. Find equivalent capacitance.

Cₚ = 5 + 8 = 13 μF

22.4 Energy Stored in a Charged Capacitor

22.4(a) Derivation of E = ½QV and E = ½CV²

To charge a capacitor, work must be done against the electric field. This work is stored as electrostatic potential energy.

During charging, the potential difference rises from 0 to V while the charge rises from 0 to Q. The Q–V graph is a straight line through the origin, and the energy stored equals the area under the graph:

         Q
         │
         │                  ╱ ● (Q, V)
       Q ├─ ─ ─ ─ ─ ─ ─ ─ ╱ │
         │                ╱��██│
         │              ╱█████│
         │            ╱███████│  ← Area = Energy
         │          ╱█████████│     = ½QV
         │        ╱███████████│
         │      ╱█────────────│
         │_____╱──────────────┼──→ V
         O                    V

Figure 22.6: Energy stored = area under Q–V graph = ½QV

E = ½ Q V

Using Q = CV:

E = ½ C V²

Or from V = Q/C:

E = Q² / 2C

⚠️ Important Note:

The factor ½ appears because the potential difference is not constant during charging—it increases gradually from 0 to V. The area under the Q–V graph is triangular (not rectangular), hence the factor ½.

Example 3:

A 10 μF capacitor is charged to 20 V. Find the energy stored.

E = ½CV² = ½ × 10×10⁻⁶ × (20)² = ½ × 10×10⁻⁶ × 400
E = 2 × 10⁻³ J = 2 mJ

22.5 Loss of Energy When Joining Capacitors with Like Terminals

When two charged capacitors are connected with their like terminals (positive to positive, negative to negative), a sudden charge redistribution occurs, resulting in a loss of energy. This energy is dissipated as heat due to the current pulse during connection.

22.5(a) Initial and Final States

         BEFORE CONNECTION          AFTER CONNECTION
         
         C₁        C₂                    C₁ ∥ C₂
         ┌─┐       ┌─┐                    ┌─────┐
    ───┤ │───  ───┤ │───            ────┤       ├────
         │ │       │ │               │    │   ∥   │
         │ │       │ │               │    │       │
    ───┤ │───  ───┤ │───            ────┤       ├────
         └─┘       └─┘                    └─────┘
         
        V₁, Q₁    V₂, Q₂          Common V, Q_total
        Q₁ = C₁V₁  Q₂ = C₂V₂      Q_total = Q₁ + Q₂

Figure 22.7: Connection of capacitors: before and after

22.5(b) Mathematical Analysis

Initial State:

C₁ at voltage V₁ has charge Q₁ = C₁V₁
C₂ at voltage V₂ has charge Q₂ = C₂V₂
Initial energy: E₁ = ½C₁V₁² + ½C₂V₂²

Final State (connected in parallel):

The two capacitors reach a common final voltage V_f. By conservation of charge:

Q₁ + Q₂ = (C₁ + C₂)V_f

V_f = (C₁V₁ + C₂V₂) / (C₁ + C₂)

Final energy stored:

E₂ = ½(C₁ + C₂)V_f²

E₂ = ½(C₁ + C₂) × [(C₁V₁ + C₂V₂)/(C₁ + C₂)]²

E₂ = (C₁V₁ + C₂V₂)² / [2(C₁ + C₂)]

Energy Loss:

ΔE = E₁ - E₂

ΔE = C₁C₂(V₁ - V₂)² / [2(C₁ + C₂)]

⚠️ Important:

ΔE is always positive (energy is lost)
ΔE depends on the difference in initial voltages (V₁ - V₂)
If V₁ = V₂, then ΔE = 0 (no energy loss)
The energy is dissipated as heat, electromagnetic radiation, and mechanical vibrations during the transient current

Example 4 (Energy Loss):

A 4 μF capacitor charged to 100 V is connected to a 2 μF capacitor charged to 40 V (positive terminals connected). Find:

(i) Common final voltage

(ii) Initial total energy

(iii) Final total energy

(iv) Energy lost

Solution:

(i) V_f = (C₁V₁ + C₂V₂) / (C₁ + C₂)
V_f = (4×100 + 2×40) / (4 + 2)
V_f = (400 + 80) / 6 = 480/6 = 80 V

(ii) E₁ = ½C₁V₁² + ½C₂V₂²
E₁ = ½(4×10⁻⁶)(100)² + ½(2×10⁻⁶)(40)²
E₁ = 0.02 + 0.0016 = 0.0216 J = 21.6 mJ

(iii) E₂ = ½(C₁ + C₂)V_f²
E₂ = ½(6×10⁻⁶)(80)²
E₂ = ½ × 6×10⁻⁶ × 6400 = 0.0192 J = 19.2 mJ

(iv) ΔE = E₁ - E₂ = 21.6 - 19.2 = 2.4 mJ

OR: ΔE = C₁C₂(V₁ - V₂)² / [2(C₁ + C₂)]
ΔE = (4×2×10⁻¹²)(100-40)² / [2(6×10⁻⁶)]
ΔE = (8×10⁻¹²)(3600) / (12×10⁻⁶)
ΔE = 2.4 × 10⁻³ J = 2.4 mJ ✓

22.6 Common Potential When Joining Capacitors

22.6(a) General Formula for Common Potential

When multiple capacitors charged to different voltages are connected in parallel, they reach a common potential given by:

V_common = (C₁V₁ + C₂V₂ + C₃V₃ + ...) / (C₁ + C₂ + C₃ + ...)

V_common = ΣCᵢVᵢ / ΣCᵢ

This is derived from conservation of charge: total charge before = total charge after.

22.6(b) Physical Interpretation

Understanding common potential:

The common potential is a weighted average of initial voltages, weighted by capacitances
Capacitors with larger capacitance have more weight in determining the final voltage
The common potential always lies between the highest and lowest initial voltages
Charge flows from higher potential to lower potential until equilibrium

         Before:                    After:
         
         V₁ > V₂                    Common V
            ↑                          ↑
         ┌──┐        ┌─────────┐   
         │C₁│        │    +    │
         └──┘        │   C₁∥C₂ │
            ↓        │    -    │
         ┌──┐        └─────────┘
         │C₂│           (Equilibrium)
         └──┘
         
         Charge flows from C₁ → C₂

Figure 22.8: Charge redistribution to reach common potential

22.6(c) Special Cases

Case 1: Identical Capacitances

If C₁ = C₂ = ... = C, then:

V_common = (V₁ + V₂ + V₃ + ...) / n

The common potential is simply the arithmetic mean of the initial voltages.

Case 2: One Capacitor Much Larger

If C₁ ≫ C₂, then:

V_common ≈ V₁

The final potential is dominated by the larger capacitor.

Example 5 (Three Capacitors):

Three capacitors: C₁ = 3 μF at V₁ = 60 V, C₂ = 2 μF at V₂ = 40 V, C₃ = 5 μF at V₃ = 20 V are connected in parallel. Find the common potential and total charge redistribution.

Solution:

V_common = (C₁V₁ + C₂V₂ + C₃V₃) / (C₁ + C₂ + C₃)
V_common = (3×60 + 2×40 + 5×20) / (3 + 2 + 5)
V_common = (180 + 80 + 100) / 10
V_common = 360 / 10 = 36 V

Charge Redistribution:

Q_initial = C₁V₁ + C₂V₂ + C₃V₃
Q_initial = 3×60 + 2×40 + 5×20 = 180 + 80 + 100 = 360 μC

Q_final = (C₁ + C₂ + C₃) × V_common
Q_final = 10 × 36 = 360 μC ✓

Charge on each:
Q₁' = 3 × 36 = 108 μC
Q₂' = 2 × 36 = 72 μC
Q₃' = 5 × 36 = 180 μC

22.7 Effect of Dielectric

22.7(a) Dielectric and Polarization

A dielectric is an electrical insulator. When placed in an electric field, it becomes polarized. The positive and negative charges within molecules shift slightly, producing induced dipoles aligned with the applied field.

         Without Field          With Electric Field
         
         (O)  (O)  (O)          (+ -)  (+ -)  (+ -)
         (O)  (O)  (O)          (+ -)  (+ -)  (+ -)
         (O)  (O)  (O)          (+ -)  (+ -)  (+ -)
         
         Random orientation     Aligned with E→

Figure 22.9: Polarization of dielectric molecules in electric field

22.7(b) Effect of Inserting Dielectric

When a dielectric is inserted between capacitor plates:

Bound charges appear on dielectric surfaces
Induced field opposes the original field
Net field is reduced by factor εᵣ
Voltage across capacitor decreases (if isolated)
Charge remains constant (if isolated)
Capacitance increases by factor εᵣ

For isolated capacitor:
E = E₀ / εᵣ (field reduced)
V = V₀ / εᵣ (voltage reduced)
Q = constant
C = εᵣ C₀ (capacitance increased)

C = εᵣ ε₀ A / d

Key Formulas and Summary

Fundamental Relations:

C = Q/V
C = ε₀A/d (vacuum)
C = εᵣε₀A/d (with dielectric)

Combinations:

Series: 1/Cₛ = 1/C₁ + 1/C₂ + ...
Parallel: Cₚ = C₁ + C₂ + ...

Energy:

E = ½QV = ½CV² = Q²/2C
Energy loss on joining:
ΔE = C₁C₂(V₁ - V₂)² / [2(C₁ + C₂)]
Common potential:
V_common = (C₁V₁ + C₂V₂ + ...) / (C₁ + C₂ + ...)

Multiple Choice Questions

1. The SI unit of capacitance is:

(a) Volt

(b) Coulomb

(c) Farad ✓

(d) Ohm

2. Capacitance C = ε₀A/d is directly proportional to:

(a) Distance d

(b) Area A ✓

(c) Charge Q

(d) Voltage V

3. In series combination of capacitors:

(a) Voltage is same

(b) Charge is same ✓

(c) Both are same

(d) None are same

4. For parallel combination, equivalent capacitance is:

(a) Sum of capacitances ✓

(b) Reciprocal of sum

(c) Less than smallest C

(d) Always equal

5. When joining two charged capacitors with like terminals, energy loss occurs because:

(a) Charge is dissipated

(b) Voltage difference causes transient current dissipation ✓

(c) Capacitance changes

(d) Dielectric breaks down

6. Energy stored in capacitor is given by:

(a) QV

(b) ½QV ✓

(c) Q²V

(d) CV

7. Inserting a dielectric between capacitor plates:

(a) Decreases capacitance

(b) Increases capacitance ✓

(c) No change

(d) Depends on charge

8. When two identical capacitors at different voltages are connected in parallel, the common potential is:

(a) Sum of voltages

(b) Average of voltages ✓

(c) Higher voltage only

(d) Lower voltage only

⚡ Capacitor — Capacitance and Dielectrics