# equilibrium between two charges

Two free point charges +q and +4q are a distance L apart. A third charge is placed so that the entire system is in equilibrium. Find the sign, magnitude, and location of the third charge. Below is the picture that my instructor gave me.

What I don't understand is that how is he able to find that r31 = L/3 ?

Second question is that why did he picked so that the net charge on the first force is zero? why not choose another one? Is it just a matter of choice?
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Commented:
Here is the detail:

In the second equation multiply both sides by r32Xr31 and you get r32 = 2 X r31

So r31 = r32 / 2  which is the same as 2 X r31 = r32

We substitute this into other given equation,  r32 + r31 = L, and get the following:

2 X r31 + r31 = L

3 X r31 = L

therefore

r31 = L/3

and r32 = 2 X r31

so that

r32 = 2/3 L

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Commented:
From the second equation, r32 = 2 X r31

Also, it said that  r32 +r31= L   therefore r32 = 2L/3

That's 2/3 L for r32 and 1/3 L for r31
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Commented:
"why did he picked so that the net charge on the first force is zero? why not choose another one? Is it just a matter of choice?"

Remember that when the right magnitude third charge is placed at the right location, the system is in equilibrium.  That means that the net force at each of the three points is zero.

The net force at each of the points must be zero for the equilibrium to work.
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Author Commented:
well it says that the second and third charge exert a force on the first one... why doesn't he say the first and the second charge exert a force on the third charge and this equals to 0
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Commented:

"why doesn't he say the first and the second charge exert a force on the third charge and this equals to 0"

Even though that's true (the new net resultant force on the third one is zero, as it is on all of them), he/she is just talking about the other two because that is what he/she needs to explain the equations that are being presented to you.

They all exert forces on each other, and in an equilibrium they all net-out to zero at each point.
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Commented:
"They all exert forces on each other, and in an equilibrium they all net-out to zero at each point."

I believe that the forces on the two end charges do not sum to zero. That does not change the analysis given. They are just pinned to their positions.
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Commented:
aburr,

so...

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Commented:
"I believe that the forces on the two end charges do not sum to zero. That does not change the analysis given. They are just pinned to their positions."
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Point 1. My comment quoted above is incorrect. WaterStreet is completely correct (I got a sign wrong). The problem is a good example of shielding

yes, by you. My comments should all be erased

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Author Commented:
just to confirm that I understand the concept here.. when they say that the system is equilibrium therefore it means that the sum of all forces on each of the charges adds up to 0 from all of the surrounding charges around it?
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Commented:
Yes
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Author Commented:
Therefore can't we solve this problem as well by saying that the first and the second charges both exert a force on the third charge and the net charge on third charge must be equal to 0.
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Commented:
No.  Why do you say that?
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Author Commented:
well based on my previous comment:

just to confirm that I understand the concept here.. when they say that the system is equilibrium therefore it means that the sum of all forces on each of the charges adds up to 0 from all of the surrounding charges around it?
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Commented:
System equilibrium means that the resultant (net) force on each point, from the others, is 0.
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Author Commented:
that's what I am saying:

the resultant force on first charge due to charge 2 and 3 is 0
the resultant force on second charge due to charge 1 and 3 is 0
the resultant force on third charge due to charge 1 and 2 is 0

right?
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Commented:
Yes,
Yes,
Yes
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Commented:
and right
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Author Commented:
the solution uses this condition to solve the problem:

the resultant force on first charge due to charge 2 and 3 is 0

what my question is that this problem can also be solved using the other two conditions that I mentioned above
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Commented:
Yes, I think so.  Your instructor only had to pick one to solve the problem
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Author Commented:
Ok, I just want to try using the other solutions and see if it gives the same result as well
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Commented:
You are ambitious, and want to make sure you understand!

What is your course of study?
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Author Commented:
Do you mean my major?
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Commented:
Yes

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Author Commented:
I am doing a minor in computer engineering, and will have a test soon about this electric charge, field, and continuous charges.. my calculus skills aren't that good and therefore I will practice that later and will strenghten on the concepts first... I will ask questions about continuous charge in a few minutes in another topic
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