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

To find the potential difference (V) between points M and N, we can use the formula for work done (W) in moving a charge (Q) in an electric field:

$W = Q imes V$

Where:

• W is the work done (in joules)
• Q is the charge (in coulombs)
• V is the potential difference (in volts)

Given:

• W = 1.5 mJ = 1.5 \times 10^{-3} J
• Q = 30 µC = 30 \times 10^{-6} C

Rearranging the formula to solve for V gives:

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

Substituting the values:

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

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