Net Electrostatic Force on particles

[Aryth]

Homework Statement

Four particles form a square. The charges are q_1 = q_4 = Q and q_2 = q_3 = q.

(a) What is \frac Qq 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 0? Explain.

The particles are arranged as such:

1 2
3 4

All separated by a distance a, the axes are drawn as standard up = +y and right = +x.

Homework Equations

F_E = k\frac{q_1 q_2}{r^2}

The Attempt at a Solution

(The way I solve for Force, I say that the force from 2 to 1 as F_21)

First I find the net force on particle 1:

\sum F_1 = F_{21} + F_{31} + F_{41}

= k\frac{qQ}{a^2} + k\frac{qQ}{a^2} + k\frac{Q^2}{(a\sqrt{2})^2}

= 2k\frac{qQ}{a^2} + k\frac{Q^2}{2a^2}

= 2k\frac{2qQ + Q^2}{2a^2}

= k\frac{2qQ + Q^2}{a^2}

I determined that the net force on 4 must be the same, so the net force on Q is then:

\sum F_Q = 2k\frac{2qQ + Q^2}{a^2} = 0

= \frac{2k}{a^2}(2qQ + Q^2) = 0

I determined that (2qQ + Q^2) must be 0, so:

2qQ + Q^2 = 0

Q^2 = -2qQ

Q = -2q

\frac Qq = -2.0

Is that right? I don't even know how to do part b and I know I forgot something in part a that is important... Which is exactly why I'm here asking for help.

 

[LowlyPion]

Welcome to PF.

That looks like the right result.

By symmetry wouldn't the only way the they could all be 0 net force would be if they were all equal charge magnitudes?

 

[rl.bhat]

In the above calculation, the fourth step is wrong.
It should be k[4qQ +Q^2]/2a^2

 

[Aryth]

Thanks for the help guys. And I can't believe I missed that little error.

So, my \frac Qq is actually -4.0 and not -2.0.

Hey LowlyPion, would you be willing to tell me why (b) is the answer it is. I can kinda see it but I can't make it make sense to me. And thanks for the welcome.

 

[gabbagabbahey]

Other than the trivial case (Q=q=0), I don't imagine there is a way that the force on all 4 particles could be zero.

If you were told that instead of the force on particles 1 and 4 being zero, the force on 2 and 3 was zero you could apply the same method as in part a and obtain (due to symmetry) \frac{q}{Q}=-4 right?

If the force on all four is zero,. then both cases would have to apply simultaneously. Is it possible for \frac{q}{Q}=-4 and \frac{Q}{q}=-4 to both be true?

 

[Aryth]

Ah, that makes perfect sense.

Thank you very much for the help. I appreciate it.

 

[gabbagabbahey]

Welcome