Question

Four point charges have the same magnitude of 2.36 10-12 C and are fixed to the corners of a square that is 3.65 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

Answer 1

Let's assume the following configuration: top left is a positive charge, top right is a positive charge, bottom left is a negative charge, and bottom right is a positive charge.

To simply things, try to look to see if the electric field of some charges will cancel with some of the other charges. The electric field of the positive charge on the top left should cancel out with the electric field of the positive charge of the bottom right. This leaves with just the positive charge and the negative charge at opposite corners. If you draw their lines, you'll notice that they don't cancel each other out, but add to themselves. So the strength net electric field will be twice the strength of the electric field caused by one charge.

So here is the electric field for one charge

E=ke*q/(r^2)

where ke=8.99 x 10^9 Nm^2/C^2
q=2.36 x 10^-12 C
r needs to be found

r is the distance from the charge to the center of the square. this can easily be found through Pythagorean theorem r=sqrt( 2 * (0.365m)^2) =0.516m

So plug all these numbers in the above formula and then multiply that number by 2, and that will be your answer

E=ke*q/(r^2) =0.07879Nc

net Electric field=2E=0.15758NC