Question

one object has an excess of 74 electrons and another object has 50 less electrons than protons. If the two objects are separated by a distance of .35m, what is the electric force between the two? Is it attractive or repulsive?

Answer 1

Answer:

$-6.94\cdot 10^{-24} N$, attractive

Explanation:

The charge of one electron is

$q_e=-1.6\cdot 10^{-19}C$

And the first object has an excess of 74 electrons, $N_e = 74$, so the net charge of the first object is

$q_1 = N_e q_e = (74)(-1.6\cdot 10^{-19})=-1.18\cdot 10^{-17}C$

The charge of one proton is

$q_p = +1.6\cdot 10^{-19}C$

And the 2nd object has 50 less electrons than protons, $N_p = 50$, so the net charge of the 2nd object is

$q_2=N_p q_p =(50)(+1.6\cdot 10^{-19})=8\cdot 10^{-18}C$

The electric force between two charged objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

$q_1=-1.18\cdot 10^{-17}C$ is the charge on the 1st object

$q_2=+8\cdot 10^{-18}C$ is the charge on the 2nd object

r = 0.35 m is the distance between the two objects

Substituting, we find:

$F=(9\cdot 10^9)\frac{(-1.18\cdot 10^{-17})(8\cdot 10^{-18})}{(0.35)^2}=-6.94\cdot 10^{-24} N$

And the negative sign indicates that the force is attractive (in fact, the electric force between two charges of opposite sign is attractive, while the force between two charges of same sign is repulsive)