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

11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

Step-by-step 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 = (kq)/(r^(2) )`

inserting the values

`E = ((9*10^(9) )(5*10^(-6)))/(2^(2) )\\E = 11.25*10^(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