Electrostatic Force

1. Jan 16, 2009

Aryth

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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.

2. Jan 16, 2009

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?

3. Jan 16, 2009

rl.bhat

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

4. Jan 16, 2009

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.

5. Jan 16, 2009

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?

6. Jan 16, 2009

Aryth

Ah, that makes perfect sense.

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

7. Jan 16, 2009

Welcome