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Location of at Equilibrium

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kakar133

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Hello every one,

can some one help me about this question.

Two positive charges Q1 and Q2 are located at (0,0) and (1,1) respectively, on the xy plane. where should on put a negative charge Q3 with |Q3|=4|Q1| so that positive charge Q2 would be at at equilibrium ?
 

OK, we are speaking here of instable equilibrium, that means the sum of forces on Q2 will be zero statically in the first time instant then the system will evolve.
First of all the add charge Q3 must stay on the line where Q1 and Q2 are placed, otherwise a non compesated force will be created. We can note that, in this case, the line connecting Q1 and Q2 is y=x.
If we place the charge on the first quadrant, then Q2 will be repulsed by Q1 and attracted by Q3 then the two forces will be directed on the same way so their sum will not be zero. If instead we place the charge on the third quadrant, Q2 will be replused by Q1 and attracted by Q3 with the two forces directed in opposite direction. Then this is our case.



In order to calculate the position we have to know that the force between two charges "Qa" and "Qb" place at a distance "dab" is proportional to:

f\[\propto\]Qa*Qb/dab2

the distance between Q1 and Q2 is d12=sqrt((0-1)2+(0-1)2)=sqrt(2) then f12\[\propto\]Q1*Q2/sqrt(2)

now the distance Q2 and Q3 is d23=sqrt((x-1)2+(x-1)2)=(x-1)*sqrt(2) then f23\[\propto\]Q2*Q3/[(x-1)*sqrt(2)]

Since we want the equilibrium we have to pose |f12|=|f23| we already find the quadrant. Thus:

Q1*Q2/2=Q2*Q3/[(x-1)2*2]

since |Q1|=|Q2|=Q and |Q3|=4*Q

Q2/2=4*Q2/[(-x-1)2*2]

solving with respect to x (we need only the positive solution) we obtain x=1 and since y=x then y=1
 
Last edited:
it means that both Q2 and Q3 are at the same position.
albbg said:
OK, we are speaking here of instable equilibrium, that means the sum of forces on Q2 will be zero statically in the first time instant then the system will evolve.
First of all the add charge Q3 must stay on the line where Q1 and Q2 are placed, otherwise a non compesated force will be created. We can note that, in this case, the line connecting Q1 and Q2 is y=x.
If we place the charge on the first quadrant, then Q2 will be repulsed by Q1 and attracted by Q3 then the two forces will be directed on the same way so their sum will not be zero. If instead we place the charge on the third quadrant, Q2 will be replused by Q1 and attracted by Q3 with the two forces directed in opposite direction. Then this is our case.



In order to calculate the position we have to know that the force between two charges "Qa" and "Qb" place at a distance "dab" is proportional to:

f\[\propto\]Qa*Qb/dab2

the distance between Q1 and Q2 is d12=sqrt((0-1)2+(0-1)2)=sqrt(2) then f12\[\propto\]Q1*Q2/sqrt(2)

now the distance Q2 and Q3 is d23=sqrt((x-1)2+(x-1)2)=(x-1)*sqrt(2) then f23\[\propto\]Q2*Q3/[(x-1)*sqrt(2)]

Since we want the equilibrium we have to pose |f12|=|f23| we already find the quadrant. Thus:

Q1*Q2/2=Q2*Q3/[(x-1)2*2]

since |Q1|=|Q2|=Q and |Q3|=4*Q

Q2/2=4*Q2/[(-x-1)2*2]

solving with respect to x (we need only the positive solution) we obtain x=1 and since y=x then y=1
Click to expand...
 

No, because I used -x, then Q3(-1,-1). Don't forget we know the charge have to stay in the third quadrant.
 
Q2/2=4*Q2/[(-x-1)2*2] ?
how you obtained this step from the previous one , I mean (-x-1)^2. In the previous step it was (x-1)^2.
albbg said:
No, because I used -x, then Q3(-1,-1). Don't forget we know the charge have to stay in the third quadrant.
Click to expand...
 

You are right.
At first I used "x" as position for Q3, and I sent the post. After it seemed to me more understsadable to use "-x" so I corrected the equation. But I corrected only the last one, forgetting to update all the equations I wrote.
 
Thank you very much.
albbg said:
You are right.
At first I used "x" as position for Q3, and I sent the post. After it seemed to me more understsadable to use "-x" so I corrected the equation. But I corrected only the last one, forgetting to update all the equations I wrote.
Click to expand...
 

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