# Four particles form a square with different charges

1. Sep 18, 2009

### jr662

http://www.webassign.net/hrw/hrw7_21-22.gif

Four particles form a square. The charges are q1= q4=Q and q2=q3=q. a) what is Q/q if the net electrostatic force on particles 1 and 4 is zero? B) is there any value of q that makes the net electrostatic force on each of the four particles zero? Explain.

Spending way too much time on this problem!

I know that Fnet = F1 + F2 +...

also F = K (absq1)(absq2)/ r^2

absolutely clueless

2. Sep 18, 2009

### saunderson

Are you familiar with vector notation?

3. Sep 18, 2009

### jr662

kind of not really

4. Sep 18, 2009

### jr662

i think i got it but i dont understand why you are adding here to get the force of q4 on q1 here using cos and sin.....F 41 = − Q⋅Q4πε 0 ⋅ 2a2 cos 45°iˆ + Q⋅Q/4π ε0 ⋅ 2a2 sin 45°

5. Sep 18, 2009

### saunderson

i think a sketch would help you to understand:

http://www.go-krang.de/physicsforums.com[/URL] [Broken] - thread - 338289.png[/PLAIN] [Broken]​

I assumed that
$$q < 0$$​
so the forces
$$\vec F_{12}, \vec F_{13}$$​
are attractive! Of course
$$\vec F_{14}$$​
is repulsive!

So if you are not familiar with vectors, we have to calculate the magnitude of
$$F_{12+13} = \sqrt{F_{12}^2 + F_{13}^2} = \sqrt{2} \cdot F_{12} ~~~~~~~~~~~~~~ \rm{with} ~~~~~~ F_{12} = F_{13}$$​

Cause the forces F_{12+13} and F_{14} are antiparallel the requirement for your question is
$$F_{12+13} = F_{14}$$​

Because of the symmetrie of this problem, that's the only thing you have to show!