Answer:
[tex]F_x==101.48*10^{-6} N[/tex]
[tex]F_{y}=-53.8*10^{-6}N[/tex]
Explanation:
We first have to get the distance and the angle between the points and the new charge.
to obtain the distance we have to do this:
[tex]X'=x2-x1 \\Y'=y2-y1\\d=\sqrt{X'^2+Y'^2[/tex]
the distance from charge of -3.35nC
[tex]X'=2.8cm-(0)=2.8*10^{-2}m\\Y'=3.85cm-(0)=3.85*10^{-2}m\\d_{3.35} =\sqrt{(2.8*10^{-2})^2+(3.85*10^{-2})^2} =4.76*10^{-2}m[/tex]
and the angle:
θ[tex]=tg^{-1} (\frac{3.85*10^{-2}}{2.8*10^{-2}} )=53.97^{o}[/tex]
the distance from charge of -2.45nC
[tex]X'=2.8cm-(0)=2.8*10^{-2}\\Y'=3.85cm-(3.85cm)=0 m\\d_{2.45} =\sqrt{(2.8*10^{-2})^2+(0)^2} =2.8*10^{-2}m[/tex]
and the angle:
θ[tex]=tg^{-1} (\frac{0}{2.8*10^{-2}} )=0^{o}[/tex]
Now we have to calculate the force applying coulomb:
[tex]F=K*\frac{qq'}{d^2} \\K=9*10^9 N.M^2/C^2[/tex]
The X component is equal to the magnitud of the force times Cos(θ):
[tex]F_{x1} =9*10^9*\frac{3.35*10^{-9}*5.00*10^{-9}}{(4.76*10^{-2})^2} cos(53.97)\\F_{x1}=39.14*10^{-6}N[/tex]
because the charges are opposites the resultant force is negative
[tex]F_{x1}=-39.14*10^{-6}N[/tex]
[tex]F_{x2} =9*10^9*\frac{2.45*10^{-9}*5.00*10^{-9}}{(2.8*10^{-2})^2} cos(0)\\F_{x2}=140.62*10^{-6}N[/tex]
So the net force in X is:
[tex]F_x=140.62*10^{-6}-39.14*10^{-6}=101.48*10^{-6} N[/tex]
The Y component is equal to the magnitud of the force times Sin(θ):
[tex]F_{y1} =9*10^9*\frac{3.35*10^{-9}*5.00*10^{-9}}{(4.76*10^{-2})^2} sin(53.97)\\F_{y1}=53.8*10^{-6}N[/tex]
because the charges are opposites the resultant force is negative
[tex]F_{y1}=-53.8*10^{-6}N[/tex]
[tex]F_{y2} =9*10^9*\frac{2.45*10^{-9}*5.00*10^{-9}}{(2.8*10^{-2})^2} sin(0)\\F_{y2}=0}N[/tex]
So the net force on Y is:
[tex]F_{y}=-53.8*10^{-6}N[/tex]
