Answer:
The electric field magnitude= 2.27kN/C direction= [tex]230^o[/tex]
The force is [tex]3.63*10^{-16} C[/tex] direction= [tex]230^o[/tex]
Explanation:
Here we have an array of charged points, we have to calculate the net force in the point x=-2.9m and y=1.2m; so we need have to find the distance of all the charged point from this position.
to obtain the distance we have to do this:
[tex]X'=x2-x1 \\Y'=y2-y1\\d=\sqrt{X'^2+Y'^2[/tex]
fot the charge of 5.5uC
[tex]X'=0.9-(-2.9)=3.8m\\Y'=3.4-1.2=2.2\\d_{5.5} =\sqrt{(3.8)^2+(2.2)^2} =4.39m[/tex]
θ[tex]=tg^{-1} (\frac{2.2}{3.8} )=30^{o}[/tex]
and for the charge of -3.6uC
[tex]X'=2.3-(-2.9)=5.2m\\Y'=-1.9-1.2=-3.1\\d_{3.6} =\sqrt{(5.2)^2+(-3.1)^2} =6.05m[/tex]
θ[tex]=tg^{-1} (\frac{3.1}{5.2} )=30.8^{o}[/tex]
Now that we have the distance and the angle, we can calculate the electric field; ausuming a positive charge:
[tex]E=k*\frac{q}{r^2}[/tex]
The net force in X direction:
[tex]E_{x} =k*\frac{q}{r^2}*cos(angle)[/tex]
[tex]E_{x} =9*10^{9} (\frac{3.6*10^{-6}}{6.05^2}cos(30.8) -\frac{5.5*10^{-6}}{4.39^2}cos(30))\\E_{x} =-1.46*10^{3}\frac{N}{C}[/tex]
the net force in Y direction:
[tex]E_{y} =9*10^{9} (-\frac{3.6*10^{-6}}{6.05^2}sin(30.8)-\frac{5.5*10^{-6}}{4.39^2}sin(30) )\\E_{y} =-1.74*10^{3}\frac{N}{C}[/tex]
So the magnitud of the electric field is:
[tex]\sqrt{E_{x} ^2+E_{y} ^2} =2.27kN/C[/tex]
with a direction of:
θ[tex]=tg^{-1}(\frac{1.74*10^3}{1.46*10^3} )=50^o[/tex], because is going down and to the left, we have to add 180, so the direction is=[tex]230^o[/tex]
to obtain the force of a proton we only have to multiply the Electric field times the charge of the proton, that is:
[tex]F=E*q\\F=2.27*10^{3}*1.6*10^{-19}\\F=3.63*10^{-16} C[/tex]
As we calculated the electric field assuming that the charge was a positive, the direction of the force will be the same. [tex]230^o[/tex]
