Answer:
a) [tex]F=-6.25\times10^{-6}N[/tex], attractive.
b) [tex]F=6.85\times10^{-7}N[/tex], repulsive.
Explanation:
We use Coulomb's Law to calculate the electrostatic force between 2 charges [tex]q_1[/tex] and [tex]q_2[/tex] separated a distance r:
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Where [tex]k=8.99\times10^9Nm^2/C^2[/tex] is Coulomb's constant.
a) At the beginning we then have:
[tex]F=\frac{kq_1q_2}{r^2}=\frac{(8.99\times10^9Nm^2/C^2)(12\times10^{-9}C)(-23\times10^{-9}C)}{(0.63 m)^2}=-6.25\times10^{-6}N[/tex]
Since their signs are different it will be attractive.
b) The total charge must be conserved, which is:
[tex]q=q_1+q_2=12\times10^{-9}C-23\times10^{-9}C=-11\times10^{-9}C[/tex]
So now each charge will have a charge half this value [tex]q'=-5.5\times10^{-9}C[/tex]
And the force will be:
[tex]F=\frac{kq'q'}{r^2}=\frac{(8.99\times10^9Nm^2/C^2)(-5.5\times10^{-9}C)(-5.5\times10^{-9}C)}{(0.63 m)^2}=6.85\times10^{-7}N[/tex]
Since their signs are the same it will be repulsive.
