A proton is located at the point (x = 1.0 nm, y = 0.0 nm)and an electron is located at the point $( x = 0.0 \mathrm {~nm} ,$ $y = 4.0 \mathrm {~nm} )$ Find the magnitude of the electrostatic force that each one exerts on the other. (k = 1/4??₀ = 9.0 × 10⁹ N ? m²/C², e = 1.6 × 10^-19 C

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The magnitude of the electrostatic force between two point charges is given by Coulomb's law:$$F = k\frac{q_1q_2}{r^2}$$where k is Coulomb's constant, q_1 and q_2 are the charges of the two particles, and r is the distance between them.In this problem, we have a proton and an electron, which have charges of +e and -e, respectively. The distance between them is:$$r = \sqrt{(1.0 \mathrm{~nm})^2 + (4.0 \mathrm{~nm})^2} = 4.12 \mathrm{~nm}$$Substituting these values into Coulomb's law, we get:$$F = (9.0 imes 10^9 \mathrm{~N \cdot m^2/C^2})\frac{(1.6 imes 10^{-19} \mathrm{~C})^2}{(4.12 imes 10^{-9} \mathrm{~m})^2} = 1.4 imes 10^{-11} \mathrm{~N}$$Therefore, the magnitude of the electrostatic force that each particle exerts on the other is 1.4 × 10^-11 N.

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

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