Answer:
E=[8.1X-9.63Y]*10^{3}N/m
Explanation:
Field in the point is the sum of the point charge electric field and the field of the infinite line.
First, we calculate the point charge field:
[tex]E_{Charge}=\frac{1}{4\pi \epsilon_0} *\frac{Q}{||r_p -r||^2} *Unitary vector\\||r_p -r||^2=(x_p-x)^2+(y_p -y)^2=1.25 m^2\\Unitary Vector=\frac{(r_p -r)}{||r_p -r||}=\frac{(x_p-x)X+(y_p -y)Y}{||r_p -r||}\\=\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y \\E_{Charge}=K*\frac{3\mu C}{1.25m^2}*(\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y)[/tex]
It is vectorial, where X and Y represent unitary vectors in X and Y. we recall the Coulomb constant k=[tex]\frac{1}{4\pi \epsilon_0}[/tex] and not replace it yet. Now we compute the line field as follows:
[tex]E_{Line}=\frac{\lambda}{2\pi \epsilon_0 distance} *Unitary Vector\\Unitary Vector=X[/tex] (The field is only in the perpendicular direction to the wire, which is X)
[tex]E_{Line}=\frac{-3\mu C/m*2}{2*2\pi \epsilon_0 5*m}X=K*\frac{6\mu C/m}{ 5*m}(-X)[/tex]
We multiplied by 2/2 in order to obtain Coulomb constant and express it that way. Finally, we proceed to sum the fields.
[tex]E=K*\frac{3\mu C}{1.25m^2}*(\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y)+K*\frac{6\mu C}{ 5*m^2}(-X)\\E=K*[2.15-1.2]X-K*[1.07]Y \mu N/m\\E=K*[0.9X-1.07Y] \mu C/m^2\\E=[8.1X-9.63Y]*10^{3}N/m[/tex]
