Answer:
[tex]3.13\times 10^{-5} C[/tex]
Explanation:
We are given that
[tex]q_1=26\mu C=26\times 10^{-6} C[/tex]
[tex]1\mu C=10^{-6} C[/tex]
[tex]y_1=0.21 m[/tex]
[tex]q=8.1\mu C=8.1\times 10^{-6} C[/tex]
[tex]y_2=0.39 m[/tex]
F=28 N
We have to find the magnitude of q2.
We know that
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Where [tex]k=9\times 10^9[/tex]
Using the formula
Force on charge q due to charge q1
[tex]F_1=\frac{9\times 10^9\times 26\times 10^{-6}\times 8.1\times 10^{-6}}{(0.21)^2}[/tex]
[tex]F_1=42.98 N[/tex]
Force on charge q due to point charge q2
[tex]F_2=\frac{9\times 10^9\times q_2\times 8.1\times 10^{-6}}{(0.39)^2}[/tex]
[tex]F_2=4.79\times 10^5 q_2[/tex]
[tex]F=F_1-F_2[/tex]
[tex]28=42.98-4.79\times 10^5q_2[/tex]
[tex]4.79\times 10^5q_2=42.98-28=14.98[/tex]
[tex]q_2=\frac{14.98}{4.79\times 10^5}=3.13\times 10^{-5} C[/tex]