Answer:
11250 N/C
Direction: 0 deg counterclockwise from positive x-axis
Explanation:
[tex]q[/tex] = magnitude of charge on each particle = 5 μC = 5 x 10⁻⁶ C
[tex]r[/tex] = distance of each particle from center of circle = 2 m
[tex]E[/tex] = Magnitude of electric field at the center by each particle
Magnitude of electric field at the center by each particle is given as
[tex]E = \frac{kq}{r^{2} }[/tex]
inserting the values
[tex]E = \frac{(9\times10^{9} )(5\times10^{-6})}{2^{2} }\\E = 11.25\times10^{3} NC^{-1}[/tex]
From the diagram , we see that being equal and opposite, the electric fields due to charge q₁ and q₃ cancel out.
So net electric field at center is only due to charge q₂ direction towards positive x-direction
So
[tex]E_{res}[/tex] = Resultant electric field = 11250 N/C
Direction: 0 deg counterclockwise from positive x-axis
