# Homework Help: Electric Forces HELP!

1. Jan 29, 2005

### NotaPhysicsMan

Ok,

the problem states: there are four charges, each with a magnitude of 2.0 uC. Two are positive and two are negative. The charges are fixed to the corners of a 0.30 m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.

Anyway. I think that it is like this + -

+ -

So since the only attraction I see is : one positive to two negs, one across and one diagonal, no vertical but repulsive force.

So I do my components... Fx=kQq/r^2 = 8.99x10^9 (2.0x10^-6C)(2.0x10^-6C)/(0.30^2) = 0.3995 N. Since I got my x component and my y, essentially same, I do pythagoras and get the answer 0.56497 N for the net force.

Ideas?

Thanks!

2. Jan 29, 2005

### genxhis

There are three forces acting on each charge. Your answer gives the magnitude after accounting only for the two nearby attractive charges. But you need to also include the repulsive charge which is across the diagonal.

3. Jan 29, 2005

### NotaPhysicsMan

diagonal? wouldnt' that be a attractive force?

4. Jan 29, 2005

### NotaPhysicsMan

So there are three forces one across, one diagonal and one vertical (repulsion). The repulsion force is the same as the horizontal attraction, which is F= 0.3995 N. The diagonal force, is 0.200N found by finding the length of the diagonal line. So to find the net charge, I have to break down the 0.200 N into it's components right? I get approx. F=0.141 N for both x and y components. I add up the forces. For Fx= 0.3995+0.141 N= .5405N. And the Fy=0.3995-0.141 N = 0.2585 N. Use pythagoras and get Square Root of (0.3589)= 0.5990 N?

5. Jan 29, 2005

### Staff: Mentor

Note that this implies that the charges are arranged so that like charges face each other across a diagonal, like this:
+ -
- +