# Coulomb's Law

## Coulomb's Law Revision

**Coulomb’s Law**

We should recall from GCSE level that charges interact. **Like charges** **repel** and **opposite charges attract**. This section takes this idea a step further allowing us to calculate the** force of attraction** or **repulsion** using an equation called **Coulomb’s Law**.

**Coulombs Law**

All **charged particles** produce an **electric field** around them. As with any **force field**, the **electric field **will exert a **force** on any object within the field.

The force can be an **attractive force** if between two **opposite charges** or a** repulsive force** if between two **like charges**.

**Coulomb’s Law** describes how two charged particles interact with each other if they are within the **electric field **of one another. It states that the **electrostatic force** is **proportional **to the **product of the two charges** and** inversely proportional** to the square of the distance between the charges. We consider the charge to be at the centre of a charged sphere.

E \propto Q_1 Q_1 and E \propto \dfrac{1}{r^2}

This is represented in the equation for **Coulomb’s Law**:

F = \dfrac{Q_1 Q_2}{4 \pi \epsilon_0 r^2}

- F is the
**force**between the two charges in Newtons \left(\text{N}\right). - Q_1 and Q_2 are the
**charges**in Coulombs \left(\text{C}\right). - \epsilon_0 is the permittivity of free space, 8.85 \times 10^{-12} \: \text{Fm}^{-1}. This is given on the formula sheet.
- r is the
**distance**between the centre of the two charges in metres \left(\text{m}\right).

When using Coulomb’s Law, we treat air as a vacuum, hence why we use the permittivity of free space \left( \epsilon_0\right).

**Example**: Two protons are 2 \: \text{mm} apart. Calculate the electrostatic force of repulsion between the protons given that the charge of a proton is 1.60 \times 10^{-19} \: \text{C}.

**[3 marks]**

F = \dfrac{Q_1 Q_2}{4 \pi \epsilon_0 r^2}

F = \dfrac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{4 \pi 8.85 \times 10^{-12} \times 0.002^2}

F = 5.75 \times 10^{-23} \: \text{N}

**Electric Fields**

By convention, the direction of **electric field lines** are always drawn **away from the positive charge** and **towards the negative charge**.

As the field lines are not evenly distributed at all points within the field, **radial** fields like those above are considered **non-uniform**.

An example of a **uniform electric field **can be found between positively charged and negatively charged plates. Remember, the field lines will always flow from** positive to negative**:

At all points between the plates, the **electric field** lines are equally spaced. Therefore, the** electric field strength** is **equal** at all points**.**

## Coulomb's Law Example Questions

**Question 1:** State the definition of Coulomb’s Law.

**[2 marks]**

The **electrostatic force is proportional to the product of the two charges** and **inversely proportional to the square of the distance between the charges**.

OR

F = \dfrac{Q_1 Q_2}{4 \pi \epsilon_0 r^2}

Where F is the force between the two charges, Q_1 and Q_2 are the charges, \epsilon_0 is the permittivity of free space \left(8.85 \times 10^{-12} \: \text{Fm}^{-1}\right) and r is the distance between the centre of the two charges.

**Question 2**: Two electrons are 1 \: \text{mm} apart. Calculate the electrostatic force of repulsion between the electrons given that the charge of an electron is -1.60 \times 10^{-19} \: \text{C}.

**[3 marks]**

**Question 3:** The diagram below represents a uniform electric field. How would this diagram change if the field was no longer uniform?

**[1 mark]**

The field lines would vary in distance apart.

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