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If I place two positive charges of different magnitudes on the x-axis (one at the origin, one at the some position x), as the two charges repel each other are the forces that they impart on each other the same? Why?

Edit: I'm working on a problem that has that situation, and the problem asks for me to find the position of one of the charges if the other one is at a given position. Why can't I just set the two forces equal to each other and use that to find the position of the other charge?

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Give me one reason...why should they be different? –  Cheeku Mar 9 '13 at 5:52
    
Newton's Third Law. –  jld Mar 9 '13 at 6:39

1 Answer 1

Oh, you need to use co-ordinate transformation if you want to solve a problem like that.

I am assuming that the problem wants you to find the position as a function of time.

Here's a hint:
Let first particle be at $x_1$, and the second particle be at $x_2$. If they have masses $m_1$ and $m_2$. Then, the following equations are valid, for any mutual forces of attraction or repulsion: $$ F = m_1 \ddot{x_1} \\ -F = m_2 \ddot{x_2} $$

If you wanna find what'll happen if particle 1 is at a fixed position, you wanna find what'll happen with respect to particle 1

Use above two equations to get $$ \ddot{x} = -F (\frac{1}{m_1} + \frac{1}{m_2}) $$

where $x$ is the position of particle 2 when you consider particle 1 at a fixed position.

Try this out, put correct values for variables, and you 'should' get the right answer.

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with those equations, do I still need to take into account the charges of each particle? –  Andrew Mar 9 '13 at 17:40

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