Coulomb's Law: Practice Problem & Solution
1. The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \AA$ apart is. ($m_e \simeq 9 \times 10^{-31} \text{ kg}, e = 1.6 \times 10^{-19} \text{ C}$) (Take $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2}$) (2020-Covid)
Solution Explained:
To solve this problem, we apply the core principles of Coulomb's Law. Understanding the underlying formula is key to arriving at the correct answer below:
Using Coulomb's law, $F = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$. Acceleration $a = \frac{F}{m_e}$. Plugging in values gives $a = 10^{22} \text{ m/s}^2$.
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