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. ) 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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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 (9.0 × 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:$$F = (9.0 	imes 10^9 	ext{ N·m}^2/	ext{C}^2)\frac{(1.6 	imes 10^{-19} 	ext{ C})^2}{(4.12 	imes 10^{-9} 	ext{ m})^2} = 1.4 	imes 10^{-11} 	ext{ N}$$Therefore, the attractive Coulombic force between the proton and the electron is 1.4 × 10^-11 N.

A Proton Is Located at X = 1