Question

The number of electrons in a copper penny is approximately 10*10^23. How large would the force be on an object if it carried this charge and were repelled by an equal charge one meter away?
I am using don't panic volume 2. How do you do this problem step by step?

Answer 1

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.

Answer 2

The charge on the electron is 1.6x10^-19C. So, 10^24 of them will be a  charge of 1.6x10^5C, F = q1xq2/[(4pi epsilon nought)r^2]