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## Homework Statement

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

(a) What is [tex]\frac Qq[/tex] 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

[tex]F_E = k\frac{q_1 q_2}{r^2}[/tex]

## The Attempt at a Solution

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

First I find the net force on particle 1:

[tex]\sum F_1 = F_{21} + F_{31} + F_{41}[/tex]

[tex]= k\frac{qQ}{a^2} + k\frac{qQ}{a^2} + k\frac{Q^2}{(a\sqrt{2})^2}[/tex]

[tex]= 2k\frac{qQ}{a^2} + k\frac{Q^2}{2a^2}[/tex]

[tex]= 2k\frac{2qQ + Q^2}{2a^2}[/tex]

[tex] = k\frac{2qQ + Q^2}{a^2}[/tex]

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

[tex]\sum F_Q = 2k\frac{2qQ + Q^2}{a^2} = 0[/tex]

[tex]= \frac{2k}{a^2}(2qQ + Q^2) = 0[/tex]

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

[tex]2qQ + Q^2 = 0[/tex]

[tex]Q^2 = -2qQ[/tex]

[tex]Q = -2q[/tex]

[tex]\frac Qq = -2.0[/tex]

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.