# Capacitance

## Capacitance Revision

**Capacitance**

A **capacitor** is an electrical component which is capable of **storing **and **releasing energy**. The **capacitor** is capable of storing **energy** before releasing the energy to supply another component or device.

**Capacitance**

Each capacitor has a **capacitance** which represents the **amount of energy** the capacitor can store. The greater the **capacitance** of a capacitor, the more **energy **the **capacitor **can store when fully charged.

The most common type of **capacitor** is the** parallel plate capacitor** shown below. This diagram also shows the circuit symbol for the **capacitor**. **Capacitance** is measured with the units Farads, although many capacitors will have a capacitance stated in microFarads \left( \mu \text{F}\right) or nanoFarads \left(\text{nF}\right):

- 1 \: \mu \text{F} = 1 \times 10^{-6} \: \text{F}
- 1 \: \text{nF} = 1 \times 10^{-9} \: \text{F}

**Capacitance** is the **amount of charge stored per unit of potential difference**, giving the equation:

C = \dfrac{Q}{V}

- C is the
**capacitance**in Farads \left(\text{F}\right). - Q is the amount of
**charge**stored in the capacitor in coulombs \left(\text{C}\right). - V is the
**potential difference**across the plates of the capacitor \left(\text{V}\right).

**Example:** A parallel plate capacitor has a capacitance of 2.5 \: \text{nF}. The capacitor is connected to a 500 \: \text{V} power supply. What is the charge on the plates of the capacitor?

**[2 marks]**

C = \dfrac{Q}{V} rearranges to Q = CV

Q = 2.5 \times 10^{-9} \times 500

Q = 1.25 \times 10^{-6} \: \text{or} \: 1.25 \: \mu \text{C}

**Capacitance** may also be calculated using the equation:

C = \dfrac{A \epsilon_0 \epsilon_r}{d}

- C is the
**capacitance**of the capacitor in Farads \left(\text{F}\right). - A is the
**area**of the plates of the capacitor in metres squared \left(\text{m}^2\right). - \epsilon_0 is the
**permittivity of free space**\left(8.85 \times 10^{-12} \: \text{Fm}^{-1}\right). - \epsilon_r is the
**relative permittivity**in Farads per metre \left(\text{Fm}^{-1}\right). - d is the
**separation**between the plates of the capacitor in metres \left(\text{m}\right).

**Parallel Plate Capacitors**

As seen in the previous diagram, most **capacitors** make use of a **dielectric** material. A **dielectric** material is made of **polar molecules**.

A single **dielectric/polar** molecule has a **positive pole** and a** negative pole** as seen here.

When no charge is applied to the plates of the **capacitor**, the dielectric molecules are free to be in any position and direction. Therefore, the **dielectric molecules** in a material will not be aligned but will be pointing in random directions.

However, when a **charge** is put across the **parallel plates**, each plate takes on an opposite charge and all the **dielectric molecules align** accordingly. This is because the positive pole is attracted to negative plate and the negative pole is attracted to the positive plate.

The **dielectric** material has a **relative permittivity**. The** permittivity** of a material quantifies how easy it is for the material to generate an **electric field**.

The **relative permittivity** \left(\epsilon_r\right) of a material (also known as the **dielectric constant**) is given by the equation:

\epsilon_r = \dfrac{\epsilon}{\epsilon_0}

where \epsilon_r is the **relative permittivity**, \epsilon is the **permittivity of the material** you are measuring and \epsilon_0 is the** permittivity of free space **\left(8.85 \times 10^{-12}\right). All are measured in \text{Fm}^{-1}.

**Example:** Calculate the permittivity of a material with a relative permittivity of 9 \times 10^{10}.

**[2 marks]**

\epsilon_r = \dfrac{\epsilon}{\epsilon_0}

\epsilon = \epsilon_r \times \epsilon_0

\epsilon = 9 \times 10^{10} \times 8.85 \times 10^{-12}

\epsilon = 0.80 \: \text{Fm}^{-1}

When a charge is passed through to the plates of the **capacitor**, an **electric field **is produced between the plates. At the same time, the **dielectric molecules** align themselves and create their own** electric field**. From the diagram below, we can see that when this occurs, the** electric fields **oppose each other.

The better the **dielectric molecules** align, the greater the opposing **electric field **would be and the larger the **permittivity of the dielectric material**. This consequently causes a decrease in the **potential difference **between the plates and increases the **capacitance**.

## Capacitance Example Questions

**Question 1:** A parallel plate capacitor has a capacitance of 10 \: \text{nF}. The capacitor is connected to a 250 \: \text{V} power supply. What is the charge on the plates of the capacitor?

**[2 marks]**

**Question 2:** What is meant by the permittivity of a material?

**[1 mark]**

A quantity which determines **how easy it is for an electric field to be produced in the material**.

**Question 3:** Water has a relative permittivity of 78.4. Calculate the permittivity of water.

**[2 marks]**