3.7.3.1
Worked example: Coulomb's law with an alpha particle and gold nucleus
Electric Fields & Potential — AQA A-Level Physics
Worked Example
An alpha particle is placed 2.0 mm from a gold nucleus in a vacuum. Calculate the magnitude of the electric force. Proton number of helium = 2, proton number of gold = 79.
Show Solution
1
List known values
- $r = 2.0 \times 10^{-3}$ m
- $e = 1.60 \times 10^{-19}$ C
- $\varepsilon_0 = 8.85 \times 10^{-12}$ F m$^{-1}$
2
Calculate the charges
Alpha particle: $Q_1 = 2e = 2 \times 1.60 \times 10^{-19}$ C
Gold nucleus: $Q_2 = 79e = 79 \times 1.60 \times 10^{-19}$ C
3
Apply CoulombThe SI unit of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. One coulomb is the chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). transferred by a current of 1 A in 1 second.'s law
$$F = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r^2} = \frac{2 \times 79 \times (1.60 \times 10^{-19})^2}{4\pi \times (8.85 \times 10^{-12}) \times (2.0 \times 10^{-3})^2}$$
4
Evaluate
$$F = 9.1 \times 10^{-21} \text{ N (2 s.f.)}$$
Answer
$F = 9.1 \times 10^{-21}$ N