In Fig. 23-36, eight charged particles form a square array; charge q = +e and di
ID: 1375057 • Letter: I
Question
In Fig. 23-36, eight charged particles form a square array; charge q = +e and distance d = 3.5 cm. What are the magnitude and direction of the net electric field at the center?
I know the answer is 3.5 x 10^-6 and 0 degrees. Can you please explain what equation you use and why to find that?
In Fig. 23-36, eight charged particles form a square array; charge q = +e and distance d = 3.5 cm. What are the magnitude and direction of the net electric field at the center? I know the answer is 3.5 x 10^-6 and 0 degrees. Can you please explain what equation you use and why to find that?Explanation / Answer
The forces due to the opposite corners cancel.
The forces due to the top and bottom charges cancel.
Thus, what is left is only the charges in the middle, +q and -2q.
Thus, by Coulomb's law,
E = kq1/r1^2 + kq2/r2^2
Here, q1 = +q, q2 = -2q, r1 = r2 = 3.5 cm = 0.035 m.
Thus,
E = 3.5E-6 N/C, towards the positive x. [answer]