1) ![+2.19\mu C](https://tex.z-dn.net/?f=%2B2.19%5Cmu%20C)
The electrostatic force between two charges is given by
(1)
where
k is the Coulomb's constant
q1, q2 are the two charges
r is the separation between the charges
When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of
![\frac{Q}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7B2%7D)
where Q is the total charge between the two spheres.
So we can actually rewrite the force as
![F=k\frac{(\frac{Q}{2})^2}{r^2}](https://tex.z-dn.net/?f=F%3Dk%5Cfrac%7B%28%5Cfrac%7BQ%7D%7B2%7D%29%5E2%7D%7Br%5E2%7D)
And since we know that
r = 1.41 m (distance between the spheres)
![F= 21.63 mN = 0.02163 N](https://tex.z-dn.net/?f=F%3D%2021.63%20mN%20%3D%200.02163%20N)
(the sign is positive since the charges repel each other)
We can solve the equation for Q:
![Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C](https://tex.z-dn.net/?f=Q%3D2%5Csqrt%7B%5Cfrac%7BFr%5E2%7D%7Bk%7D%7D%3D2%5Csqrt%7B%5Cfrac%7B%280.02163%29%281.41%29%5E2%7D%7B8.98755%5Ccdot%2010%5E9%7D%7D%7D%3D4.37%5Ccdot%2010%5E%7B-6%7D%20C)
So, the final charge on the sphere on the right is
![\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7B2%7D%3D%5Cfrac%7B4.37%5Ccdot%2010%5E%7B-6%7D%20C%7D%7B2%7D%3D2.19%5Ccdot%2010%5E%7B-6%7DC%3D%2B2.19%5Cmu%20C)
2) ![q_1 = +6.70 \mu C](https://tex.z-dn.net/?f=q_1%20%3D%20%2B6.70%20%5Cmu%20C)
Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that
(we put a negative sign since the force is attractive, which means that the charges have opposite signs)
r = 1.41 m is the separation between the charges
And also,
![q_2 = Q-q_1](https://tex.z-dn.net/?f=q_2%20%3D%20Q-q_1)
So we can rewrite eq.(1) as
![F=k \frac{q_1 (Q-q_1)}{r^2}](https://tex.z-dn.net/?f=F%3Dk%20%5Cfrac%7Bq_1%20%28Q-q_1%29%7D%7Br%5E2%7D)
Solving for q1,
![Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0](https://tex.z-dn.net/?f=Fr%5E2%3Dk%20%28q_1%20Q-q_1%5E2%7D%29%5C%5Ckq_1%5E2%20-kQ%20q_1%20%2BFr%5E2%20%3D%200)
Since
, we can substituting all numbers into the equation:
![8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0](https://tex.z-dn.net/?f=8.98755%5Ccdot%2010%5E9%20q_1%5E2%20-3.93%5Ccdot%2010%5E4%20q_1%20-0.141%20%3D%200)
which gives two solutions:
![q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C](https://tex.z-dn.net/?f=q_1%20%3D%206.70%5Ccdot%2010%5E%7B-6%7D%20C%5C%5Cq_2%20%3D%20-2.34%5Ccdot%2010%5E%7B-6%7D%20C)
Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is
![q_1 = +6.70 \mu C](https://tex.z-dn.net/?f=q_1%20%3D%20%2B6.70%20%5Cmu%20C)