1.5 mJ of work is done when a charge of 30 μC is moved between two points, M and N, in an electric field - AQA - A-Level Physics - Question 15 - 2019 - Paper 2

The potential difference (V) can be calculated using the formula:

$V = \frac{W}{Q}$

where:

• $W$ is the work done (in joules)
• $Q$ is the charge (in coulombs)

In this case, we have:

• $W = 1.5 \, \text{mJ} = 1.5 \times 10^{-3} \, \text{J}$
• $Q = 30 \, \mu\text{C} = 30 \times 10^{-6} \, \text{C}$

Substituting values into the formula:

$V = \frac{1.5 \times 10^{-3}}{30 \times 10^{-6}} = \frac{1.5}{30} \times 10^3 = 0.05 \times 10^3 = 50 \, \text{V}$

Thus, the potential difference between points M and N is 50 V.