Answer:
122.36
Explanation:
The distance (d) between Charge 1 and 2 can be calculated as:
[tex]d=\sqrt{0.32^2+0.59^2}=0.67m[/tex]
The force between them is given as
[tex]F_1=\frac{1}{4\pi \epsilon_0}\frac{4*18*10^-12}{0.67^2}= 1.44N[/tex]
The angle of this force with positive x-axis is given as
[tex]\theta_1=90^{\circ}+\tan^{-1}\frac{0.32}{0.59}=118.47^{\circ}[/tex]
Now,
The force between 1 and 3 is
[tex]F_2=\frac{1}{4\pi \epsilon_0}\frac{4*2*10^{-12}}{0.79^2}= 0.115N[/tex]
As the force is attractive it is along negative x direction so the angle will be given as = [tex]\theta_2 = 180^{\circ}[/tex]
So the negative x component of the resultant force will be calculated as
= [tex]1.44\cos(180-118.47)^{\circ}+0.115=0.801[/tex]
And the positive y component = [tex]1.44\sin(180-118.47)^{\circ}=1.26[/tex]
So the angle of the resultant with positive x axis will be
[tex]90^{\circ}+\tan^{-1}\frac{0.801}{1.26}=122.36^{\circ}[/tex]
