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 = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C², e = 1.6 × 10^-19 C)

Question

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 = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C², e = 1.6 × 10^-19 C)

Quizplus · Accepted Answer

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 case, the proton has a charge of +e and the electron has a charge of -e, so the magnitude of the electrostatic force between them is:$$F = k\frac{e^2}{r^2}$$where r is the distance between the proton and the electron.The distance between the proton and the electron is:$$r = \sqrt{(1.0 \mathrm {~nm})^2 + (4.0 \mathrm {~nm})^2} = 4.12 \mathrm {~nm}$$Substituting this value into the equation for the electrostatic force, we get:$$F = (9.0 imes 10^9 \mathrm {~N} \cdot \mathrm {~m}^2/\mathrm {~C}^2)\frac{(1.6 imes 10^{-19} \mathrm {~C})^2}{(4.12 imes 10^{-9} \mathrm {~m})^2} = 1.4 imes 10^{-11} \mathrm {~N}$$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 $( x = 0.0 \mathrm {~nm}$