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Where must a third charge be placed to get a net force of 0

1. Homework Statement
A charge of +2 Q (A)is placed at the origin of a coordinate system and a charge of - Q (B) is placed at a distance d from the origin. How far from the origin must have 3rd charge +Q (C) be placed so that the net force on it is zero newtons.

2. Homework Equations
Fe=kQq/r^2

3. The Attempt at a Solution
The solution to this question says that the Third charge must be placed greater than 2d distance away from the origin.

I tried to solve this problem by finding a distance that would make the absolute value of the force of A on C equal the force of B on C. Making the distance that the Third charge is place morevthan 2d just does not make these forces on C equal. For example placing the Third charge at 3D makes the charge A on C : B on C have the ratio of 2/9d^2 : 1/4d^2. That doesn't seem to make the net force on charge C = zero to me
 
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BvU

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The solution to this question
There is a possibility that the book solution is wrong (typo, relates to a previous version of the exercise, etcetera). Why don't you work out your own solution and verify that it is correct.
Trial and error is a start, but for an answer you should write an equation that you can solve: place C at position x on the AB axis and set up an expression for the force. Does the charge of C matter ?
 
I did try several different distances. In the example above that I listed I tried 3D which should have worked as an answer but it did not. The answer in the book says that a distance greater than 2D should make the net force equal to zero. The charge on C does matter. It is listed as q+. The amount and sign on the charge explains whether or not the other two charges will be attracted or repulsed by charge C.

So are you telling me that the answer in the book is wrong? And that I should disregard it?
 

BvU

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I am asking you to write an equation and see if you can solve it...:rolleyes:
 

haruspex

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The solution to this question says that the Third charge must be placed greater than 2d distance away from the origin.
Please clarify... Does it state a specific distance, and you are telling us that it is greater than 2d, or does it say that any distance > 2d will do (surely not), or that it tells you the distance is greater than 2d without specifying exactly what it is?
If the third, I do not see how you can imply it is wrong merely by checking some other distance > 2d.
 
I am asking you to write an equation and see if you can solve it...:rolleyes:
I actually work out an equation with 3d. You can see it in my original post
 

BvU

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Yes. Now the correct equation, please:
Force at postion x from +2q + Force at position x from -q = 0
 
How do I attach a picture of the question? I'm only seeing options to add a link to a url
 
The answer is literally " greater than 2 D" I understand that. Doesn't necessarily mean that any value greater than 2D would be the answer, but I still have no idea how they came to that conclusion mathematically.
 
Ok. I was able to track down the answer as 3.41d (p286, #22) which apparently can be derived quadratically, but I have no idea how they did that. Any insight?
 

BvU

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Force at postion x from +2q + Force at position x from -q = 0
Apply your own relevant equation

"Greater than 2d" is a hint: there are three regions (left, between, right, if you place the -q at ##x=d##). The first two don't qualify (why not?) so you have to find the solution on the right.
 
I still don't understand, because the equation that would give the answer seems to have two variables.( A/(xd) ^2) -( B/(yd )^2) = x and y are not the same number. I've obviously thought about this a lot and I'm extremely frustrated. Is there any chance you could just tell me how they got to the answer 3.41 d quadratically?


Apply your own relevant equation

"Greater than 2d" is a hint: there are three regions (left, between, right, if you place the -q at ##x=d##). The first two don't qualify (why not?) so you have to find the solution on the right.
Apply your own relevant equation

"Greater than 2d" is a hint: there are three regions (left, between, right, if you place the -q at ##x=d##). The first two don't qualify (why not?) so you have to find the solution on the right.
 
Sorry. My above post had an error instead of the last X the equation should equal 0
 

BvU

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+2q is positioned at the origin.
-q is is positioned at ##x=d##.

Your test charge is positioned on the x axis. What is the force (a vector with magnitude, direction) the +2q exerts on the test charge as a function of x?
Same question for the force from the -q . Add them up, set to 0 and voila...
 
How can I calculate the force charge A or charge B will have on the Third charge without knowing what the distance of the 3rd charge is from charge A and charge B? It's required in the formula to know what r squared is to calculate Force. Without knowing the distance that they are from the Third charge how do I know what force they have on it?

+2q is positioned at the origin.
-q is is positioned at ##x=d##.

Your test charge is positioned on the x axis. What is the force (a vector with magnitude, direction) the +2q exerts on the test charge as a function of x?
Same question for the force from the -q . Add them up, set to 0 and voila...
 
You keep saying to calculate the force that each charge has on the test charge, but I can't calculate force on the test charge without knowing its distance from the test charge.
 

BvU

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Come on !
What is the distance between two charges if one is located at the origin and the other is at position ##x## on the x-axis ?
 

BvU

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Ah! Did you already conclude from symmetry that the solution is to be found on the x-axis ?
 
The distance between the first and second charge is x. It still doesn't tell me where the Third charge will be located.

Come on !
What is the distance between two charges if one is located at the origin and the other is at position ##x## on the x-axis ?
="BvU, post: 5967703, member: 499340"]+2q is positioned at the origin.
-q is is positioned at ##x=d##.

Your test charge is positioned on the x axis. What is the force (a vector with magnitude, direction) the +2q exerts on the test charge as a function of x?
Same question for the force from the -q . Add them up, set to 0 and voila...[/QUOTE]
Come on !
What is the distance between two charges if one is located at the origin and the other is at position ##x## on the x-axis ?
 
If the test charge is placed at 2d then it would feel (1/2d^2)-(1/d^2) = -1/2d^2. So I guess they are technically saying that it would have to be further away than 2d to overcome feeling this negative force (.-1/2d^2.) But it still doesn't explain how they got to 3.41d quadratically
 

BvU

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Please please please answer the question
What is the distance between two charges if one is located at the origin and the other is at position ##x## on the x-axis ?
So we can deal with your inhibition in post #16 and continue from there.
 

BvU

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The distance between the first and second charges is x.
Good.
Now: what is the distance between two charges when the first is placed at ##x=d## and the second at ' ##\ x=x\ ## ' (where ##x>d##)
 
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