Cylindrical And Spherical Capacitors, Series And Parallel Combinations

Here are the slides 1 to 10 on the topic “Cylindrical and Spherical Capacitors, Series and Parallel Combinations - Example on Cylindrical Capacitor”:

Cylindrical and Spherical Capacitors

Capacitors come in various shapes and sizes
Two common types are cylindrical and spherical capacitors
In this lecture, we will focus on cylindrical capacitors
We will also explore series and parallel combinations of capacitors
Let’s begin!

Cylindrical Capacitors

Cylindrical capacitors consist of two coaxial cylinders
The inner cylinder acts as the positive plate (+Q)
The outer cylinder acts as the negative plate (-Q)
The space between the plates is filled with a dielectric material
The capacitance of a cylindrical capacitor depends on its dimensions

Capacitance of Cylindrical Capacitor

The capacitance of a cylindrical capacitor is given by the formula: C = (2πε₀L) / ln(b/a) Where:

C is the capacitance
ε₀ is the permittivity of free space (8.85 × 10⁻¹² F/m)
L is the length of the cylinder
a is the radius of the inner cylinder
b is the radius of the outer cylinder

Capacitance Example

Let’s consider an example of a cylindrical capacitor
The inner cylinder has a radius of 0.02 m
The outer cylinder has a radius of 0.04 m
The length of the cylinder is 0.1 m
Calculate the capacitance of this cylindrical capacitor Solution: C = (2πε₀L) / ln(b/a) = (2π × 8.85 × 10⁻¹² × 0.1) / ln(0.04/0.02) ≈ 2.23 × 10⁻⁷ F

Series Combination of Capacitors

In a series combination, capacitors are connected in a daisy-chained fashion
The total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of individual capacitances (C₁, C₂, C₃, …):

1/C_total = 1/C₁ + 1/C₂ + 1/C₃ + …

The voltage across each capacitor remains the same
The total charge stored is divided among the capacitors

Series Combination Example

Let’s consider an example of two cylindrical capacitors connected in series
Capacitor 1 has a capacitance of 3 μF
Capacitor 2 has a capacitance of 5 μF
Calculate the total capacitance when they are connected in series Solution:

1/C_total = 1/C₁ + 1/C₂ = 1/3 × 10⁻⁶ + 1/5 × 10⁻⁶ = 5/15 × 10⁻⁶ + 3/15 × 10⁻⁶ = 8/15 × 10⁻⁶ C_total = 15/8 × 10⁻⁶ = 1.875 μF

Parallel Combination of Capacitors

In a parallel combination, capacitors are connected side by side
The total capacitance (C_total) is the sum of individual capacitances (C₁, C₂, C₃, …): C_total = C₁ + C₂ + C₃ + …
The voltage across each capacitor remains the same
The total charge stored is the sum of charges stored by individual capacitors

Parallel Combination Example

Let’s consider an example of three cylindrical capacitors connected in parallel
Capacitor 1 has a capacitance of 1 μF
Capacitor 2 has a capacitance of 2 μF
Capacitor 3 has a capacitance of 3 μF
Calculate the total capacitance when they are connected in parallel Solution: C_total = C₁ + C₂ + C₃ = 1 × 10⁻⁶ + 2 × 10⁻⁶ + 3 × 10⁻⁶ = 6 × 10⁻⁶ = 6 μF

Example: Capacitor with Dielectric

Let’s consider a cylindrical capacitor with a dielectric material between the plates
The dielectric material has a relative permittivity (εᵣ) of 4
The capacitance of the capacitor is 10 μF without the dielectric
Calculate the new capacitance with the dielectric Solution: C = (2πε₀L) / ln(b/a) C’ = C × εᵣ = 10 × 10⁻⁶ × 4 = 40 × 10⁻⁶ = 40 μF

Summary

Cylindrical capacitors consist of two coaxial cylinders
The capacitance of a cylindrical capacitor depends on its dimensions
Capacitors can be connected in series or parallel combinations
In a series combination, the total capacitance is the reciprocal of the sum of reciprocals of individual capacitances
In a parallel combination, the total capacitance is the sum of individual capacitances

Cylindrical Capacitors with Dielectric

When a dielectric material is placed between the plates of a cylindrical capacitor, the capacitance increases
The dielectric material increases the electric field strength and the stored charge for the same potential difference
The capacitance with dielectric is given by:
```
        C' = κ × C
```
where κ is the relative permittivity of the dielectric material
The presence of a dielectric also decreases the electric field between the plates

Capacitance Example with Dielectric

Consider a cylindrical capacitor with a relative permittivity (κ) of 5
The capacitance without the dielectric is 2 μF
Calculate the new capacitance with the dielectric Solution: C’ = κ × C = 5 × 2 × 10⁻⁶ = 10 × 10⁻⁶ = 10 μF

Energy Stored in a Capacitor

When a capacitor is charged, it stores energy in the electric field
The energy stored in a capacitor is given by the formula: E = ½ × C × V² where E is the energy stored, C is the capacitance, and V is the potential difference across the capacitor
The energy stored is directly proportional to the square of the potential difference

Example: Energy Stored in a Capacitor

Consider a cylindrical capacitor with a capacitance of 10 μF
The potential difference across the capacitor is 100 V
Calculate the energy stored in the capacitor Solution: E = ½ × C × V² = ½ × 10 × 10⁻⁶ × (100)² = ½ × 10 × 10⁻⁶ × 10000 = 5000 × 10⁻⁶ = 5 mJ

Spherical Capacitors

Spherical capacitors consist of two concentric spheres
The inner sphere acts as the positive plate (+Q)
The outer sphere acts as the negative plate (-Q)
The space between the plates is filled with a dielectric material
The capacitance of a spherical capacitor depends on its dimensions

Capacitance of Spherical Capacitor

The capacitance of a spherical capacitor is given by the formula: C = (4πε₀ab) / (b-a) Where:

C is the capacitance
ε₀ is the permittivity of free space (8.85 × 10⁻¹² F/m)
a is the radius of the inner sphere
b is the radius of the outer sphere

Capacitance Example: Spherical Capacitor

Let’s consider an example of a spherical capacitor
The inner sphere has a radius of 0.02 m
The outer sphere has a radius of 0.04 m
Calculate the capacitance of this spherical capacitor Solution: C = (4πε₀ab) / (b-a) = (4π × 8.85 × 10⁻¹² × 0.02 × 0.04) / (0.04-0.02) = (4π × 8.85 × 10⁻¹² × 0.02 × 0.04) / 0.02 ≈ 8.85 × 10⁻¹⁰ F

Series Combination of Cylindrical Capacitors

Cylindrical capacitors can be connected in a series combination
The total capacitance (C_total) for series combination is given by: 1/C_total = 1/C₁ + 1/C₂ + 1/C₃ + …
The voltage across each capacitor is not the same
The total voltage across the series combination is the sum of individual voltages

Parallel Combination of Cylindrical Capacitors

Cylindrical capacitors can be connected in a parallel combination
The total capacitance (C_total) for parallel combination is given by: C_total = C₁ + C₂ + C₃ + …
The voltage across each capacitor remains the same
The total charge stored is the sum of charges stored by individual capacitors

Summary

Cylindrical and spherical capacitors are common types of capacitors
Capacitance of cylindrical capacitor is given by (2πε₀L) / ln(b/a)
Capacitance of spherical capacitor is given by (4πε₀ab) / (b-a)
Capacitors can be connected in series or parallel combinations
Series combination: 1/C_total = 1/C₁ + 1/C₂ + 1/C₃ + …
Parallel combination: C_total = C₁ + C₂ + C₃ + …

Energy Stored in a Cylindrical Capacitor

The energy stored in a cylindrical capacitor can be calculated using the formula: E = ½ × C × V²
Where E is the energy stored, C is the capacitance, and V is the potential difference across the capacitor
The energy stored is directly proportional to the square of the potential difference

Example: Energy Stored in a Cylindrical Capacitor

Consider a cylindrical capacitor with a capacitance of 5 μF
The potential difference across the capacitor is 50 V
Calculate the energy stored in the capacitor Solution: E = ½ × C × V² = ½ × 5 × 10⁻⁶ × (50)² = ½ × 5 × 10⁻⁶ × 2500 = 6250 × 10⁻⁶ = 6.25 mJ

Electric Field in a Cylindrical Capacitor

The electric field between the plates of a cylindrical capacitor can be calculated using the formula: E = V / d
Where E is the electric field strength, V is the potential difference across the plates, and d is the separation between the plates

Example: Electric Field in a Cylindrical Capacitor

Consider a cylindrical capacitor with a potential difference of 100 V
The separation between the plates is 0.05 m
Calculate the electric field strength between the plates Solution: E = V / d = 100 / 0.05 = 2000 V/m

Dielectric Strength of a Capacitor

The dielectric strength of a capacitor is the maximum electric field that the dielectric material can withstand without breaking down
If the electric field exceeds the dielectric strength, the dielectric will become conductive and allow current to flow
Dielectric strength is measured in volts per meter (V/m)

Breakdown Voltage of a Capacitor

The breakdown voltage of a capacitor is the minimum potential difference that causes the dielectric to break down and start conducting
It is important to operate capacitors within their rated voltage to avoid breakdown and damage

Capacitive Reactance

Capacitive reactance, denoted as Xc, is the opposition to the flow of alternating current (AC) through a capacitor
It is similar to resistance in a direct current (DC) circuit
The capacitance reactance can be calculated using the formula: Xc = 1 / (2πfC)
Where Xc is the capacitive reactance, f is the frequency of the AC signal, and C is the capacitance

Example: Capacitive Reactance

Consider a cylindrical capacitor with a capacitance of 3 μF
The frequency of the AC signal is 50 Hz
Calculate the capacitive reactance Solution: Xc = 1 / (2πfC) = 1 / (2π × 50 × 3 × 10⁻⁶) ≈ 1.06 kΩ

Impedance of a Capacitor

Impedance, denoted as Z, is the total opposition to the flow of alternating current (AC) through a capacitor
It combines both the capacitive reactance (Xc) and any resistance (R) in the circuit
The impedance of a capacitor can be calculated using the formula: Z = √(R² + Xc²)
Where Z is the impedance, R is the resistance, and Xc is the capacitive reactance

Example: Impedance of a Capacitor

Consider a circuit with a cylindrical capacitor having a capacitance of 2 μF and a resistance of 1 kΩ
Calculate the impedance of the capacitor Solution: Xc = 1 / (2πfC) = 1 / (2π × 50 × 2 × 10⁻⁶) ≈ 1.59 kΩ Z = √(R² + Xc²) = √((1 × 10³)² + (1.59 × 10³)²) = √(1 × 10⁶ + 2.53 × 10⁶) = √(3.53 × 10⁶) ≈ 1.88 × 10³ Ω