Answer:
The force of repulsion between charges will be 2.309 N.
Explanation:
The number of electrons in a copper penny =N = [tex]10\times 10^{23}[/tex]
Charge on an electron = [tex]e=-1.602\times 10^{-19} C[/tex]
Total charge of the N electrons = Q
[tex]Q=N\times e[/tex]
[tex]Q=10\times 10^{23}\times 1.602\times 10^{-19}C=-1.602\times 10^{-5}C[/tex]
The force be on an object if it carried this charge and were repelled by an equal charge one meter away.
[tex]Q=-1.602\times 10^{-5}C[/tex]
[tex]q=Q[/tex]
Let the force of repulsion be F.
Distance between the Q and q charges = r =1 m
[tex]F=k\frac{Qq}{r^2}[/tex] (Coulomb law)
[tex]F=9\times 10^9 Nm^2/C^2\times \frac{(-1.602\times 10^{-5} C)\times (-1.602\times 10^{-5} C)}{(1m)^2}[/tex]
[tex]F=2.309 N[/tex]
The force of repulsion between charges will be 2.309 N.
