Answer:
0.208 N
Explanation:
We are given that
[tex]q_1=q_2=2.06\mu C=2.06\times 10^{-6} C[/tex]
[tex]q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C[/tex]
Distance,d=0.41 m
The magnitude of the net electrostatic force experienced by any charge at point 4
Net force,[tex]F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2[/tex]
[tex]F_1=F_3=F[/tex]
[tex]F_{net}=\sqrt{F^2+F^2+0}-F_2[/tex]
[tex]F_{net}=\sqrt 2F-F_2[/tex]
[tex]F=\frac{kq^2}{d^2}[/tex]
[tex]F_2=\frac{Kq^2}{2d^2}[/tex]
[tex]F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})[/tex]
Where [tex]k=9\times 10^9[/tex]
[tex]F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})[/tex]
[tex]F_{net}=0.208 N[/tex]
