Answer:
![8.64\cdot 10^{-18} C](https://tex.z-dn.net/?f=8.64%5Ccdot%2010%5E%7B-18%7D%20C)
Explanation:
There are two forces acting on the oil drop:
- The force of gravity, downward, given by
![F_G = mg](https://tex.z-dn.net/?f=F_G%20%3D%20mg)
where m is the mass of the drop and g is the acceleration due to gravity
- The electric force, upward, given by
![F_E = qE](https://tex.z-dn.net/?f=F_E%20%3D%20qE)
where q is the charge of the oil drop and E is the magnitude of the electric field
The oil drop remains stationary, so the two forces are balanced:
![F_G = F_E\\mg = qE](https://tex.z-dn.net/?f=F_G%20%3D%20F_E%5C%5Cmg%20%3D%20qE)
where
![m=5.2898\cdot 10^{-13}kg\\E=6\cdot 10^5 N/C\\g = 9.8 m/s^2](https://tex.z-dn.net/?f=m%3D5.2898%5Ccdot%2010%5E%7B-13%7Dkg%5C%5CE%3D6%5Ccdot%2010%5E5%20N%2FC%5C%5Cg%20%3D%209.8%20m%2Fs%5E2)
Substituting into the previous equation and solving for q, we find the charge of the oil drop:
![q=\frac{mg}{E}=\frac{(5.2898\cdot 10^{-13} kg)(9.8 m/s^2)}{6\cdot 10^5 N/C}=8.64\cdot 10^{-18} C](https://tex.z-dn.net/?f=q%3D%5Cfrac%7Bmg%7D%7BE%7D%3D%5Cfrac%7B%285.2898%5Ccdot%2010%5E%7B-13%7D%20kg%29%289.8%20m%2Fs%5E2%29%7D%7B6%5Ccdot%2010%5E5%20N%2FC%7D%3D8.64%5Ccdot%2010%5E%7B-18%7D%20C)