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