A proton is located at the point (x = 1.0 nm,y = 0.0 nm)and an electron is located at the point (x=0.0nm, y= 4.0nm). Find the magnitude of the electrostatic force that each one exerts on the other.(k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²,e = 1.6 × 10^-19C) A)1.4 × 10^-11 N B)5.3 × 10^-18 N C)5.3 × 10⁸ N D)5.9 × 10^-15 N

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Quizplus · Accepted Answer

The magnitude of the electrostatic force can be calculated using Coulomb's law: $F = k \frac{|q_1q_2|}{r^2}$, where $k = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, $q_1 = q_2 = 1.6 \times 10^{-19} \, \text{C}$, and $r$ is the distance between the proton and the electron. The distance $r$ can be found using the Pythagorean theorem, given the positions of the proton and electron: $r = \sqrt{(1.0 \, \text{nm})^2 + (4.0 \, \text{nm})^2} = \sqrt{1 + 16} = \sqrt{17} \, \text{nm} = \sqrt{17} \times 10^{-9} \, \text{m}$. Substituting these values into Coulomb's law gives $F = 9.0 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(\sqrt{17} \times 10^{-9})^2}$, which calculates to approximately $1.4 \times 10^{-11} \, \text{N}$.

A Proton Is Located at the Point (X = 1