Electrostatics
Beginner
Coulomb Law
Source: High school physics (Chinese)
Problem Sets:
Electrostatics
Problem
A gold nucleus contains 79 protons, and an alpha particle contains 2 protons. The elementary charge is $e = 1.6 \times 10^{-19}$ C. The mass of an alpha particle is $m_{\alpha} = 6.68 \times 10^{-27}$ kg. An alpha particle is at a distance $r = 6.9 \times 10^{-15}$ m from a gold nucleus.
- What is the magnitude of the electrostatic force on the alpha particle?
- What is the magnitude of its acceleration?
[Q1] $F = 76.5$ N [Q2] $a = 1.1 \times 10^{28}$ m/s$^2$
First, determine the charges of the gold nucleus ($q_{Au}$) and the alpha particle ($q_{\alpha}$).
$$q_{Au} = 79e = 79(1.6 \times 10^{-19} C) = 1.264 \times 10^{-17} C$$ $$q_{\alpha} = 2e = 2(1.6 \times 10^{-19} C) = 3.2 \times 10^{-19} C$$[Q1] Calculate the electrostatic force using Coulomb's Law.
$$F = k \frac{q_{Au} q_{\alpha}}{r^2}$$ $$F = (9.0 \times 10^9 N \cdot m^2/C^2) \frac{(1.264 \times 10^{-17} C)(3.2 \times 10^{-19} C)}{(6.9 \times 10^{-15} m)^2} \approx 76.5 N$$[Q2] Use Newton's Second Law to find the acceleration.
$$a = \frac{F}{m_{\alpha}}$$ $$a = \frac{76.5 N}{6.68 \times 10^{-27} kg} \approx 1.1 \times 10^{28} m/s^2$$