Four particles form a square with different charges

[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

 

[saunderson]

Are you familiar with vector notation?

 

[jr662]

kind of not really

 

[jr662]

i think i got it but i don't 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°

 

[saunderson]

i think a sketch would help you to understand:

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

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!
Hope i could help you!?

 

[jr662]

thanks