Answer:
answer is 2 option because more force is applied
Answer:
![50.91 \mu C](https://tex.z-dn.net/?f=50.91%20%5Cmu%20C)
Explanation:
The magnitude of the net force exerted on q is known, we have the values and positions for
and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by
on q. Then we can know the magnitude of the force exerted by
about q, finally this will allow us to know the magnitude of ![q_{2}](https://tex.z-dn.net/?f=q_%7B2%7D)
exerts a force on q in +y direction, and
exerts a force on q in -y direction.
![F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\](https://tex.z-dn.net/?f=F_%7B1%7D%3D%5Cfrac%7Bkq_%7B1%7D%20q%20%7D%7Bd%5E2%7D%5C%5CF_%7B1%7D%3D%5Cfrac%7B%288.99%2A10%5E9%29%2825%2A10%5E%7B-6%7DC%29%288.4%2A10%5E%7B-6%7DC%29%7D%7B%280.18m%29%5E2%7D%3D58.26%20N%5C%5C)
The net force on q is:
![F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}](https://tex.z-dn.net/?f=F_%7BT%7D%3DF_%7B1%7D%20-%20F_%7B2%7D%5C%5C25N%3D58.26N-F_%7B2%7D%5C%5CF_%7B2%7D%3D58.26N-25N%3D33.26N%5C%5C%5Cmid%20F_%7B2%7D%20%5Cmid%3D%5Cfrac%7Bkq_%7B2%7Dq%7D%7Bd%5E2%7D)
Rewriting for
:
![q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C](https://tex.z-dn.net/?f=q_%7B2%7D%3D%5Cfrac%7BF_%7B2%7Dd%5E2%7D%7Bkq%7D%5C%5Cq_%7B2%7D%3D%5Cfrac%7B33.26N%280.34m%29%5E2%7D%7B8.99%2A10%5E9%5Cfrac%7BNm%5E2%7D%7BC%5E2%7D%288.4%2A10%5E%7B-6%7DC%29%7D%3D50.91%2A10%5E%7B-6%7DC%3D50.91%20%5Cmu%20C)
Explanation:
Below is an attachment containing the solution.
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