A proton is located at x = 1.0 nm, y = 0.0 nm and an electron is located at x = 0.0 nm,
Y = 4.0 nm. Find the attractive Coulombic force between them. (The value of k is
9) 0 × 10⁹ N · m²/C².)

Question

A proton is located at x = 1.0 nm, y = 0.0 nm and an electron is located at x = 0.0 nm, Y = 4.0 nm. Find the attractive Coulombic force between them. (The value of k is 9)0 × 10⁹ N · m²/C².)

Quizplus · Accepted Answer

The attractive Coulombic force between a proton and an electron is given by:$$F = k\frac{e^2}{r^2}$$where:* k is the Coulomb constant (9 × 10^9 N · m^2/C^2)* e is the elementary charge (1.6 × 10^-19 C)* r is the distance between the proton and the electronIn this problem, the distance between the proton and the electron is:$$r = \sqrt{(1.0 	ext{ nm})^2 + (4.0 	ext{ nm})^2} = 4.12 	ext{ nm} = 4.12 	imes 10^{-9} 	ext{ m}$$Substituting these values into the equation for the Coulombic force, we get:$$F = (9 	imes 10^9 	ext{ N} \cdot 	ext{m}^2/	ext{C}^2)\frac{(1.6 	imes 10^{-19} 	ext{ C})^2}{(4.12 	imes 10^{-9} 	ext{ m})^2} = 1.4 	imes 10^{-11} 	ext{ N}$$Therefore, the attractive Coulombic force between the proton and the electron is 1.4 × 10^-11 N.

A Proton Is Located at X = 1