# Electric Potential

## Electric Potential Revision

**Electric Potential**

Just like in a **gravitational field**, moving a charge in an** electric field **changes the amount of **electric potential energy** the charge has. In this section we look at the idea of** electric potential** and calculate the **work done** moving a charged particle in an **electric field**.

**Electric Potential**

To move like charges together, or opposite charges further apart, **work** must be done. This is because we are moving the charges in the opposite direction to overcome the **electrostatic forces** of attraction or repulsion. In doing so, the** potential energy** of the charge being moved is increased.

As seen above, the **electrostatic force** for two like charges is **repulsive**. Therefore, moving the like charges together requires work to be done and increases the **electric potential**. Likewise, if a pair of opposite charges are separated against their **attractive electrostatic force**, work is done and the** electric potential increases**.

In both cases, at **infinity **(a point outside the electric field of both point charges) the** electric potential** is **zero**.

The **electric potential** can therefore be defined as the **work done per unit of charge** in moving a charge from infinity to a specific point within the electric field of the charge.

**Electric Potential **at a specific point can be calculated using the equation:

V = \dfrac{Q}{4 \pi \epsilon_0 r}

- V is the
**electric potential**at the point you are calculating in volts \left(\text{V}\right). - Q is the
**charge**of the point charge causing the electric field in Coulombs \left(\text{C}\right). - \epsilon_0 is the
**permittivity of free space**\left(8.85 \times 10^{-12} \: \text{Fm}^{-1}\right). - r is the
**distance**from the centre of the point charge in metres \left(\text{m}\right).

**Example:** A Van de Graaf generator has a charge of 5.0 \times 10^{-6} \: \text{C}. Calculate the electric potential at a distance of 0.5 \: \text{m} away from the centre of the generator.

**[2 marks]**

V = \dfrac{Q}{4 \pi \epsilon_0 r}

V = \dfrac{5 \times 10^{-6}}{\left(4 \pi \times 8.85 \times 10^{-12} \times 0.5 \right)}

V = 90000 \: \text{V} \: \text{or} \: 90 \: \text{kV}

**Electric Potential Difference**

As with Gravitational Potential, we can calculate the **electrical potential difference** between two points in an **electric field**. This is known as the **electrical potential difference** \left( \Delta V \right).

\Delta V = V_f – V_i

where V_f is the **final electric potential** and V_i is the **initial electric potential**, both in joules per coulomb \left(\text{JC}^{-1}\right).

**Example:** An electron is moved from a point in an electric field where the electric potential is +1.9 \: \text{JC}^{-1} to another point where the electric potential is +0.2 \: \text{JC}^{-1} . Calculate the potential difference between the two points in the electric field.

**[1 mark]**

\Delta V = V_f - V_i

\Delta V = 0.2 - 1.9

\Delta V = -1.7 \: \text{JC}^{-1}

**Work Done on a Charged Particle**

**Work** needs to be done to move a charge through an** electric field**. As the charge moves though the field, it’s electric potential is changing. This is described by the following equation:

\Delta W = q \Delta V

- \Delta W is the
**change in work done**in joules \left(\text{J}\right). - q is the
**charge**of the moving charged particle in Coulombs \left(\text{C}\right). - \Delta V is the
**change in electric potential**in volts \left(\text{V}\right).

Therefore we can see that the** change in work done** is equal to the **change in electric potential energy**.

**Change in** **electric potential energy** can be calculated using the following equation:

\text{Change in electric potential energy} = \dfrac{Qq}{4 \pi \epsilon_0 \Delta r}

where Q is the point charge, q is the moving charge, \epsilon_0 is the permittivity of free space \left(8.85 \times 10^{-12}\right) and \Delta r is the change in distance from the centre of the charge \left(r_2 – r_1\right).

In the above example, the **change in electric potential energy** and the **work done** are positive as we are moving against the electrostatic attraction so we need to do work. If the moving charge was moving towards the point charge, the **work done** would be negative.

If a **moving charge** moves perpendicular to the field lines, no work is done as there would be no \Delta r and therefore no change in **electric potential**.

## Electric Potential Example Questions

**Question 1: **A positive charge is moved away from a negative charge. Describe why the moving charge gains electric potential.

**[2 marks]**

As the **electrostatic force between opposite charges is attractive**, moving the charges apart** requires work to be done** and therefore, electric potential increases.

**Question 2: **Calculate the electric potential at a distance of 0.1 \: \text{m} away from the centre of a point charge of 1 \times 10^{-7} \: \text{C}.

**[2 marks]**

**Question 3:** An alpha particle is moved from a point in an electric field where the electric potential is +0.9 \: \text{JC}^{-1} to another point where the electric potential is +2.2 \: \text{JC}^{-1} . Calculate the potential difference between the two points in the electric field.

**[1 mark]**