Question

Three identical particles, q1, q2, and q3, each with charge q = 5.00 μC, are placed along a circle of radius r = 2.00 m at angles 90, 180, and 270 degrees from the positive x-axis, respectively.
What is the resultant electric field at the center of the circle, point O? All angles are measured counterclockwise from the positive x-axis.

Answer 1

Answer:

11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

Explanation:

$q$ = magnitude of charge on each particle = 5 μC =  5 x 10⁻⁶ C

$r$ = distance of each particle from center of circle = 2 m

$E$ = Magnitude of electric field at the center by each particle

Magnitude of electric field at the center by each particle is given as

$E = \frac{kq}{r^{2} }$

inserting the values

$E = \frac{(9\times10^{9} )(5\times10^{-6})}{2^{2} }\\E = 11.25\times10^{3} NC^{-1}$

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

$E_{res}$ = Resultant electric field = 11250 N/C

Direction: 0 deg counterclockwise from positive x-axis