Charges are Great Need more assistance

In summary, to keep the three charges in place, the ratio q_3/q must be equal to .03 and the x and y components must both be .03.
  • #1
Saladsamurai
3,020
7
[SOLVED] Charges are Great! Need more assistance!

Homework Statement


Picture2-1.png


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.
 
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  • #2
Okay, I looked it up on Cramster (which I absolutely hate doing) and in their solution they assume that [itex]F_{31}=F_{21}[/itex] , but I cannot reason that out.

Any ideas?
 
  • #3
Good Morning! Maybe I'll have some luck, now let's see here. . .:smile:
 
  • #4
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.
 
  • #5
kamerling said:
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?
 
  • #6
Saladsamurai said:
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.
 
  • #7
What are the expressions for the total force on each charge? For each one just add the forces from each of the other two.
 
  • #8
You mean like [itex]F_{1,net}=F_{12}+F_{13}[/itex]?
 
  • #9
Sure, why not? Now, what are [tex]F_{12}[/tex] and [tex]F_{13}[/tex]?
 
  • #10
[tex]F_{12}=\frac{k|q||4q|}{L^2}[/tex]

and

[tex]F_{13}=\frac{k|q||q_3|}{(L-x)^2}[/tex]

Okay, I have values for L and X. What about the [itex]F_{1,net}[/itex] 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?
 
Last edited:
  • #11
I cannot get this to come out right! I am trying to solve for the ratio [itex]\frac{q_3}{q}[/itex]Using
[tex]F_{1,net}=F_{12}+F_{13}[/tex]

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

I thought I could say that [itex]F_{1,net}=0[/itex] And then solve the above for

[tex]\frac{q_3}{q}[/tex]

But this is not working?
 
  • #12
Got IT! r_2 not equal to L-x. . . duh. It's just x!

Thanks everyone!
 

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