Calculating Capacitance
It is the ability of an object to store an electric charge. It only depends on the capacitor's geometry and the dielectric material choice between the plates. In most cases, it does not depend on the potential difference or the charge stored on the plates.
Capacitance equation
How to calculate the capacitance, then? You need to use the following formula:
\( C = \frac{s}{\varepsilon A} \)
where:
- C — Capacitance, measured in farads (symbol: F);
- ε — Dielectric permittivity (a measure of resistance) in farads per meter;
- A — Area where the plates overlap; and
- s — Separation distance between the plates.
The permittivity depends on the dielectric material used. By default, this capacitance calculator uses the value of 8.854 p F / m 8.854 pF/m, which is the permittivity of the vacuum. If you wish to change this value, click on the field.
Example
Calculating Capacitance
Capacitance is the ability of a component or circuit to store an electrical charge. It is a measure of the charge that a capacitor can store per unit voltage. Capacitance is a fundamental property of capacitors, and it plays an important role in electrical circuits.
The general approach to calculating capacitance includes:
- Identifying the surface area of the plates (A) and the distance between them (d).
- Knowing the permittivity of the material between the plates (ε).
- Applying the formula for capacitance to calculate the result.
Capacitance Formula
The general formula for capacitance is:
\[ C = \frac{\varepsilon A}{d} \]Where:
- C is the capacitance (in farads, F).
- A is the area of the plates (in square meters, m²).
- \(\varepsilon\) is the permittivity of the material between the plates (in farads per meter, F/m).
- d is the distance between the plates (in meters, m).
Example:
If a capacitor has plates with an area of \( 0.01 \, \text{m}^2 \) and a distance between the plates of \( 0.001 \, \text{m} \), and the permittivity of the material between the plates is \( 8.85 \times 10^{-12} \, \text{F/m} \), the capacitance is:
- Step 1: Apply the formula: \( C = \frac{8.85 \times 10^{-12} \times 0.01}{0.001} \).
- Step 2: Calculate the result: \( C = 8.85 \times 10^{-12} \, \text{F} \).
Capacitance with Different Materials
The capacitance of a capacitor depends on the material between the plates. The permittivity (\(\varepsilon\)) value varies for different materials, which affects the capacitance. Materials with higher permittivity allow the capacitor to store more charge for the same voltage.
Example:
If the dielectric material between the plates is air, its permittivity is \( 8.85 \times 10^{-12} \, \text{F/m} \). However, if the material is a dielectric such as barium titanate, the permittivity may be much higher, which would increase the capacitance.
Real-life Applications of Capacitance
Calculating capacitance is crucial for various practical applications, such as:
- Designing capacitors for power supply circuits, energy storage, and filtering applications.
- Choosing appropriate capacitors for timing circuits in devices such as clocks and radios.
- Creating and tuning circuits for signal processing and communications equipment.
Common Units of Capacitance
SI Unit: The standard unit of capacitance is the farad (F). In practice, capacitors often have values in microfarads (µF), nanofarads (nF), or picofarads (pF).
Example: A capacitor with a capacitance of 1 µF can store \( 1 \times 10^{-6} \) coulombs of charge for each volt of applied potential difference.
Common Operations with Capacitance
Series Connection: When capacitors are connected in series, the total capacitance is found using the formula: \[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \]
Parallel Connection: When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances: \[ C_{total} = C_1 + C_2 + \cdots \]
Negative Capacitance: A phenomenon where the capacitor seems to behave as if it has a negative capacitance, often observed in certain special materials and experimental setups.
| Problem Type | Description | Steps to Solve | Example |
|---|---|---|---|
| Calculating Capacitance from Plate Area | Finding the capacitance of a parallel plate capacitor when the area of the plates and the distance between them are known. |
|
If a capacitor has plates with an area of \( 0.01 \, \text{m}^2 \), a distance between the plates of \( 0.001 \, \text{m} \), and the permittivity of the material is \( 8.85 \times 10^{-12} \, \text{F/m} \), the capacitance is \( C = \frac{8.85 \times 10^{-12} \times 0.01}{0.001} = 8.85 \times 10^{-12} \, \text{F} \). |
| Capacitance with Different Materials | Finding the capacitance when the dielectric material between the plates changes. |
|
If the dielectric material between the plates is barium titanate with a permittivity of \( 120 \times 10^{-12} \, \text{F/m} \), the capacitance will increase compared to air as the permittivity is higher. |
| Capacitance in Series Connection | Finding the total capacitance when capacitors are connected in series. |
|
If two capacitors with capacitances of \( 5 \, \mu \text{F} \) and \( 10 \, \mu \text{F} \) are connected in series, the total capacitance is \( \frac{1}{C_{total}} = \frac{1}{5} + \frac{1}{10} \), so \( C_{total} = 3.33 \, \mu \text{F} \). |
| Capacitance in Parallel Connection | Finding the total capacitance when capacitors are connected in parallel. |
|
If two capacitors with capacitances of \( 5 \, \mu \text{F} \) and \( 10 \, \mu \text{F} \) are connected in parallel, the total capacitance is \( C_{total} = 5 + 10 = 15 \, \mu \text{F} \). |
