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

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]

Answer 2

The force of repulsion between charges will be 2.309 N.

Step-by-step explanation:

The number of electrons in a copper penny =N =
`10* 10^(23)`

Charge on an electron =
`e=-1.602* 10^(-19) C`

Total charge of the N electrons = Q

`Q=N* e`

`Q=10* 10^(23)* 1.602* 10^(-19)C=-1.602* 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* 10^(-5)C`

`q=Q`

Let the force of repulsion be F.

Distance between the Q and q charges = r =1 m

`F=k(Qq)/(r^2)` (Coulomb law)

`F=9* 10^9 Nm^2/C^2* ((-1.602* 10^(-5) C)* (-1.602* 10^(-5) C))/((1m)^2)`

`F=2.309 N`

The force of repulsion between charges will be 2.309 N.