Charges are Great Need more assistance

[Saladsamurai]

[SOLVED] Charges are Great! Need more assistance!

Homework Statement

In the figure particle 1 of +q and particle 2 of charge +4q are held at a separation of L=.09 m If particle 3 of charge q_3 is to be placed such that the three particles remain in place when released, what must be (a) the ratio q_3/q and the (b) x and (c) y components?



Okay, I solved b and c easy enough. There is no y component and x=.03.

I am having trouble with a. Any hints? I know it must be a similar procedure, that is, I need to use the fact that there is NO NET force...but I can't seem to spot the correct relationship such that I am left with the desired variables (they don't cancel).

Word.

 

[Saladsamurai]

Okay, I looked it up on Cramster (which I absolutely hate doing) and in their solution they assume that $F_{31}=F_{21}$ , but I cannot reason that out.

Any ideas?

 

[Saladsamurai]

Good Morning! Maybe I'll have some luck, now let's see here. . .

 

[kamerling]

For your answer of b you used the fact that the net force on q3 must be 0. now use the fact that the net force on q1 and q2 must be 0.

 

[Saladsamurai]

For your answer of b you used the fact that the net force on q3 must be 0. now use the fact that the net force on q1 and q2 must be 0.

Okay, I am kind of with you. So since the net force on all three particles is 0, then. . . crap I lost it.

Can someone give ne a little more of a nudge?

 

[Saladsamurai]

Okay, I am kind of with you. So since the net force on all three particles is 0, then. . . crap I lost it.

Can someone give ne a little more of a nudge?

Anyone? I know it may be obvious to some, but not to me.

 

[belliott4488]

What are the expressions for the total force on each charge? For each one just add the forces from each of the other two.

 

[Saladsamurai]

You mean like $F_{1,net}=F_{12}+F_{13}$?

 

[belliott4488]

Sure, why not? Now, what are $$F_{12}$$ and $$F_{13}$$?

 

[Saladsamurai]

$$F_{12}=\frac{k|q||4q|}{L^2}$$

and

$$F_{13}=\frac{k|q||q_3|}{(L-x)^2}$$

Okay, I have values for L and X. What about the $F_{1,net}$ would that just be 0? It must be since the condition is that they stay in place, thus F_net-0.

Does this sound reasonable?

 

[Saladsamurai]

I cannot get this to come out right! I am trying to solve for the ratio $\frac{q_3}{q}$Using
$$F_{1,net}=F_{12}+F_{13}$$

$$\Rightarrow F_{1,net}=\frac{k|q||4q|}{L^2}+\frac{k|q||q_3|}{(L-x)^2}$$

I thought I could say that $F_{1,net}=0$ And then solve the above for

$$\frac{q_3}{q}$$

But this is not working?

 

[Saladsamurai]

Got IT! r_2 not equal to L-x. . . duh. It's just x!

Thanks everyone!