In Fig. 23-36, eight charged particles form a square array; charge q = +e and di
ID: 1375163 • 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? Please plug in the numbers to the equation. I will give 5 stars for a good answer.
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? Please plug in the numbers to the equation. I will give 5 stars for a good answer.Explanation / Answer
Net electric filed due to upper +3q and lower +3q is zero
Net electric filed due to upper -5q and lower -5q is zero
Net electric filed due to upper +q and lower +q is zero
there is a net electric filed due to +q(left center) and -2*q(right center)
Enet = k*q/d^2 + k*2*q/d^2
= 3*k*q/d^2
= 3*9*10^-9*1.6*10^-19/0.035^2
= 3.52*10^-6 N/c (towards right or +x axis)