Answer:
Q = 6.33μC
Step-by-step explanation:
To find the value of the charge Q you take into account both gravitational force and electric force over each ball. By symmetry you can use the fact that both balls experiences the same forces. Hence you only take into account the forces for one ball for the x component and y component:
![-Mg+Tcos\theta=0\\\\F_e-Tsin\theta=0](https://img.qammunity.org/2021/formulas/physics/high-school/dadxdjt7ynw30s6pz8726lv8dkrpgpk0vu.png)
M: mass of the ball = 0.09kg
T: tension of the string
F_e: electric force between charges
angle = 45°
The electric force is given by:
![F_e=k(Q^2)/(r^2)](https://img.qammunity.org/2021/formulas/physics/high-school/phbjnq7muhw1f1ndz38l5ep0et0p7xk59n.png)
Q: charge of the balls
r: distance between balls = 2m
You divide both equation in order to eliminate the tension T:
![tan\theta=(F_e)/(Mg)=k(Q^2)/(Mgr^2)](https://img.qammunity.org/2021/formulas/physics/high-school/4ebeqrxa0dxtxtv2m9a2rfz3607xmvfl26.png)
By doing Q the subject of the formula and replacing you obtain:
![Q=\sqrt{(tan\theta Mgr^2)/(k)}=\sqrt{(tan45\°(0.09kg)(2m)^2)/((8.89*10^(9)Nm^2/C^2))}=6.33*10^(-6)C=6.33\mu C](https://img.qammunity.org/2021/formulas/physics/high-school/4iltas9izrgigtw6piouh66bppnxiaeguz.png)
hence, the charge of the balls is 6.33μC