Need Help With Point Charges Problem?

Question

A) Calculate the magnitude of the electric field at the origin due to the following distribution of charges: +q at (x,y) = (a,a), +q at (a,-a), -q at (-a,-a) and -q at (-a,a). Where q = 7.30 × 10-7 C and a = 8.90 cm.

Answer 1

I am going to number the charges from 1-4 in the order you listed them.

If you consider the direction of the electric field at the origin due to each charge you will see that overall, the electric field will point towards the negative x-axis.

The distance between the origin and any charge is a√2 by the special triangle rule.

Charges 1 and 3 and charges 2 and 4 produce the same electric field with equal magnitude and direction. So

E(13) = E(24) = 2(kq/(a√2)^2)

The vertical components of E(13) and E(24) cancel each other out., so the overall electric field is

E = 2E(13)cos 45° = 4(kq/(a√2)^2)(√2/2) = kq√2/a^2

= (8.99 N*m^2/C^2)(7.30 × 10-7 C)√2/(8.90e-2 m)^2

= 1.17e6 N/C