# Net Electrostatic Force on particles

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

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
Homework Helper
Gold Member
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?

Ah, that makes perfect sense.

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

gabbagabbahey
Homework Helper
Gold Member
Welcome