Cylindrical Capacitor Formula Calculator

\[ C = \frac{2 \pi \varepsilon_0 L}{\ln(b/a)} \]

Permittivity of Free Space (\(\varepsilon_0\)): F/m

Length (\(L\)):

Inner Radius (\(a\)):

Outer Radius (\(b\)):

1. What is the Cylindrical Capacitor Formula Calculator?

Definition: This calculator computes the capacitance (\(C\)) of a cylindrical capacitor using the formula \( C = \frac{2 \pi \varepsilon_0 L}{\ln(b/a)} \), where \( \varepsilon_0 \) is the permittivity of free space, \( L \) is the length of the cylinder, \( b \) is the outer radius, and \( a \) is the inner radius.

Purpose: It is used in physics and engineering to determine the capacitance of coaxial cylindrical capacitors, applicable in electronics (e.g., coaxial cables), high-voltage systems, and educational contexts.

2. How Does the Calculator Work?

The calculator uses the cylindrical capacitor formula:

Formula: \[ C = \frac{2 \pi \varepsilon_0 L}{\ln(b/a)} \] where:

\(C\): Capacitance (F, pF)
\(\varepsilon_0\): Permittivity of free space (F/m)
\(L\): Length (m, ft)
\(b, a\): Outer and inner radii (m, in)

Unit Conversions:

Length (\(L\)):
- 1 m = 1 m
- 1 ft = 0.3048 m
Radii (\(b, a\)):
- 1 m = 1 m
- 1 in = 0.0254 m
Capacitance (Output):
- 1 F = 1 F
- 1 pF = \( 10^{-12} \) F

The capacitance is calculated in farads (F) and can be converted to the selected output unit (F, pF). Results greater than 10,000 or less than 0.001 are displayed in scientific notation; otherwise, they are shown with 4 decimal places.

Steps:

Enter the permittivity (\(\varepsilon_0\)), length (\(L\)), inner radius (\(a\)), and outer radius (\(b\)) with their units (default: \(\varepsilon_0 = 8.8541878128 \times 10^{-12} \, \text{F/m}\), \(L = 1 \, \text{m}\), \(a = 0.01 \, \text{m}\), \(b = 0.02 \, \text{m}\)).
Convert inputs to SI units (m).
Validate that permittivity, length, and radii are greater than 0, and outer radius is greater than inner radius.
Calculate the capacitance in farads using the formula.
Convert the capacitance to the selected output unit.
Display the result, using scientific notation if the value is greater than 10,000 or less than 0.001, otherwise rounded to 4 decimal places.

3. Importance of Cylindrical Capacitor Calculation

Calculating the capacitance of a cylindrical capacitor is crucial for:

Electronics: Designing coaxial cables and capacitors, where capacitance affects signal transmission and energy storage (e.g., in RF applications).
Engineering: Developing high-voltage systems and sensors, where cylindrical capacitors are used for their uniform electric field and high capacitance per unit length.
Education: Teaching the principles of capacitance, electrostatics, and the effect of geometry on electric fields in physics.

4. Using the Calculator

Examples:

Example 1: Calculate the capacitance of a cylindrical capacitor with \( \varepsilon_0 = 8.8541878128 \times 10^{-12} \, \text{F/m} \), \( L = 1 \, \text{m} \), \( a = 0.01 \, \text{m} \), \( b = 0.02 \, \text{m} \), output in pF:
- Enter \( \varepsilon_0 = 8.8541878128 \times 10^{-12} \, \text{F/m} \), \( L = 1 \, \text{m} \), \( a = 0.01 \, \text{m} \), \( b = 0.02 \, \text{m} \).
- Ratio: \( \frac{b}{a} = \frac{0.02}{0.01} = 2 \).
- Denominator: \( \ln(b/a) = \ln(2) \approx 0.693147 \).
- Numerator: \( 2 \pi \varepsilon_0 L = 2 \pi \times 8.8541878128 \times 10^{-12} \times 1 \approx 5.56338 \times 10^{-11} \, \text{F} \).
- Capacitance in F: \( C = \frac{5.56338 \times 10^{-11}}{0.693147} \approx 8.026 \times 10^{-11} \, \text{F} \).
- Convert to output unit (pF): \( 8.026 \times 10^{-11} \times 10^{12} \approx 80.26 \, \text{pF} \).
- Result: \( \text{Capacitance} = 80.2600 \, \text{pF} \).
Example 2: Calculate the capacitance with \( \varepsilon_0 = 8.8541878128 \times 10^{-12} \, \text{F/m} \), \( L = 3.28084 \, \text{ft} \), \( a = 0.393701 \, \text{in} \), \( b = 0.787402 \, \text{in} \), output in F:
- Enter \( \varepsilon_0 = 8.8541878128 \times 10^{-12} \, \text{F/m} \), \( L = 3.28084 \, \text{ft} \), \( a = 0.393701 \, \text{in} \), \( b = 0.787402 \, \text{in} \).
- Convert: \( L = 3.28084 \times 0.3048 = 1 \, \text{m} \), \( a = 0.393701 \times 0.0254 = 0.01 \, \text{m} \), \( b = 0.787402 \times 0.0254 = 0.02 \, \text{m} \).
- Ratio: \( \frac{b}{a} = \frac{0.02}{0.01} = 2 \).
- Denominator: \( \ln(b/a) = \ln(2) \approx 0.693147 \).
- Numerator: \( 2 \pi \varepsilon_0 L = 2 \pi \times 8.8541878128 \times 10^{-12} \times 1 \approx 5.56338 \times 10^{-11} \, \text{F} \).
- Capacitance in F: \( C = \frac{5.56338 \times 10^{-11}}{0.693147} \approx 8.026 \times 10^{-11} \, \text{F} \).
- Output unit: F (no conversion needed).
- Result: \( \text{Capacitance} = 8.0260 \times 10^{-11} \, \text{F} \).

5. Frequently Asked Questions (FAQ)

Q: What is a cylindrical capacitor?
A: A cylindrical capacitor consists of two concentric cylindrical conductors separated by a dielectric (e.g., air or vacuum). The inner cylinder has radius \( a \), the outer cylinder has radius \( b \), and the capacitance is given by \( C = \frac{2 \pi \varepsilon_0 L}{\ln(b/a)} \), where \( L \) is the length. It is commonly used in coaxial cables and high-voltage applications.

Q: Why must the outer radius be greater than the inner radius?
A: The outer radius (\(b\)) must be greater than the inner radius (\(a\)) to form a physically meaningful capacitor with a gap between the cylinders where the electric field exists. If \( b \leq a \), the denominator \( \ln(b/a) \) would be zero or negative, leading to an undefined or negative capacitance, which is not possible.

Q: How does the length of the capacitor affect its capacitance?
A: The capacitance (\(C\)) of a cylindrical capacitor is directly proportional to its length (\(L\)), as seen in the formula \( C = \frac{2 \pi \varepsilon_0 L}{\ln(b/a)} \). A longer capacitor has a larger surface area for charge storage, increasing its capacitance, assuming the radii remain constant.