Answer:
The net charge is ![67.89 \mu C](https://tex.z-dn.net/?f=67.89%20%5Cmu%20C)
Solution:
As per the question:
Mass of the plastic bag, m = 12.0 g = ![12.0\times 10^{-3}\ kg](https://tex.z-dn.net/?f=12.0%5Ctimes%2010%5E%7B-3%7D%5C%20kg)
Magnitude of electric field, E = ![10^{3}\ N/C](https://tex.z-dn.net/?f=10%5E%7B3%7D%5C%20N%2FC)
Angle made by the string, ![\theta = 30^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2030%5E%7B%5Ccirc%7D)
Now,
To calculate the net charge, Q on the ball:
Vertical component of the tension in the string, ![T = Tcos\theta](https://tex.z-dn.net/?f=T%20%3D%20Tcos%5Ctheta)
Horizontal component of the tension in the string, ![T = Tsin\theta](https://tex.z-dn.net/?f=T%20%3D%20Tsin%5Ctheta)
Now,
Balancing the forces in the x-direction:
![Tsin\theta = QE](https://tex.z-dn.net/?f=Tsin%5Ctheta%20%3D%20QE)
(1)
Balancing the forces in the y-direction:
![Tcos\theta = mg](https://tex.z-dn.net/?f=Tcos%5Ctheta%20%3D%20mg)
where
g = acceleration due to gravity = ![9.8\ m/s^{2}](https://tex.z-dn.net/?f=9.8%5C%20m%2Fs%5E%7B2%7D)
Thus
![T = \frac{mg}{cos\theta }](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7Bmg%7D%7Bcos%5Ctheta%20%7D)
![T = \frac{12.0\times 10^{-3}\times 9.8}{cos30^{\circ}} = 0.1357\ N](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B12.0%5Ctimes%2010%5E%7B-3%7D%5Ctimes%209.8%7D%7Bcos30%5E%7B%5Ccirc%7D%7D%20%3D%200.1357%5C%20N)
Use T = 0.1357 N in eqn (1):
![Q = {0.1357\times sin30^{\circ}}{10^{3}} = 6.789\times 10^{- 5}\ C](https://tex.z-dn.net/?f=Q%20%3D%20%7B0.1357%5Ctimes%20sin30%5E%7B%5Ccirc%7D%7D%7B10%5E%7B3%7D%7D%20%3D%206.789%5Ctimes%2010%5E%7B-%205%7D%5C%20C)
![Q = 67.89\times 10^{- 5}\ C = 67.89\mu C](https://tex.z-dn.net/?f=Q%20%3D%2067.89%5Ctimes%2010%5E%7B-%205%7D%5C%20C%20%3D%2067.89%5Cmu%20C)