A proton is located at the point (x = 1.0 nm,y = 0.0 nm)and an electron is located at the point $( \mathrm { x } = 0.0 \mathrm {~nm}$ $y = 4.0 \mathrm {~nm} )$ 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^-19 C)

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Quizplus · Accepted Answer

The electrostatic force between two charges is given by Coulomb's law: $F = k \frac{|q_1 q_2|}{r^2}$, where $k = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, $q_1$ and $q_2$ are the magnitudes of the charges, and $r$ is the distance between them. The charges of a proton and an electron are equal in magnitude but opposite in sign, $|e| = 1.6 \times 10^{-19} \, \text{C}$. The distance $r$ between the proton and the electron can be found using the Pythagorean theorem, given their coordinates: $r = \sqrt{(1.0 \, \text{nm} - 0.0 \, \text{nm})^2 + (0.0 \, \text{nm} - 4.0 \, \text{nm})^2} = \sqrt{1^2 + 4^2} = \sqrt{17} \, \text{nm} = \sqrt{17} \times 10^{-9} \, \text{m}$.Substituting the values into Coulomb's law:$$F = \frac{9.0 \times 10^9 \times (1.6 \times 10^{-19})^2}{(\sqrt{17} \times 10^{-9})^2} = \frac{9.0 \times 10^9 \times 2.56 \times 10^{-38}}{17 \times 10^{-18}} = \frac{23.04 \times 10^{-29}}{17 \times 10^{-18}} = \frac{23.04}{17} \times 10^{-11} \, \text{N}$$This calculation simplifies to approximately $1.4 \times 10^{-11} \, \text{N}$, which matches choice A.

A Proton Is Located at the Point (X = 1 (x=0.0 nm( \mathrm { x } = 0.0 \mathrm {~nm}(x=0.0 nm

A Proton Is Located at the Point (X = 1 $( \mathrm { x } = 0.0 \mathrm {~nm}$