# Homework Help: Electric fields and forces help

1. Mar 22, 2004

### mindhater

There are 4 equal charges of 2 micro Coulumbs in a shape of a square. 2 positives line up on 1 side, which is the left side and 2 negatives line up on the other side, which is the left. Again, the charges are of equal magnitude.

The quesiton asks for the magnitude of the electric field at the center...

I was thinking it's zero, but i'm not quite sure if that's true.

The other question is if you put a -4 micro Coulumb charge at the center, what will be the magnitude force on it and direction?

I know the 2 positive charges are attracted to the negative charge, therefore there are two vector arrows coming out of the postive charges into the -4 micro C charge, but I don't which direction the arrow goes when it comes to the 2 negatives. Is there a arrow that comes out of the -4 charge towards the negative charage, or an arrow from the negative charge to the -4 charge. Since it's repulsive i don't know which one.

Any help is appreciated...thx

2. Mar 23, 2004

### HallsofIvy

Since nothing is said about the size of the square, let's assume that it has side 2 and so distance to the center from each vertex *radic;(2). Set up a coordinate system so that the four corners are (1,1), (1,-1), (-1,1), and (-1,-1) with the two positive charges at the first two points. Assuming a "+1" test charge at the center, the force due to the point at (1,1) has magnitude 2/2= 1 and has components (1/&radic(2))(-i- j) (Since it will "push" a positive test charge. Similarly, the force due to the charge at (1,-1) is (1/&radic(2))(-i+ j) while the forces due to the charges at (-1,1) and (-1,-1) are (1/&radic(2))(-i- j) and (1/&radic(2))(-i+ j) respectively (Since they will "pull" the test charge toward them). The total force on a test charge, and so the field strength, will be the sum: (1/&radic(2))((-1-1-1-1)i+(1-1+1-1)j)= -4i.

Putting a charge of -4 in the center gives a force of (-4)(-4i)= 16i. That is, the two negative charges repel while the two positive charges attract and it moves in the positive x-direction.

By the way, thinking "I know the 2 positive charges are attracted to the negative charge" is misleading. The two positive charges are fixed in place and aren't moving. It is the negative charge in the middle that is being attracted. That could cause you to have your vectors reversed.

3. Mar 23, 2004

### Chen

It's kind of like a double electric dipole... isn't it?