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I want to compute the maximum separation when 2 unlike charges are shot apart from each other

         <--- (+)     (-)--->
(+)                                 (-)
 |----------------- d ---------------|

How do I do it? Its actually in my lecture notes:

$$E_i = KE_i + U(r_i)$$

$$E_f = U(r_{max})$$

And to compute them, I equate them $E_i = E_f$. And solve for $r_{max}$, but there are 2 charges, so my $KE_i = 2 \times \frac{1}{2} m v^2$? $U(r_i)$ will be just 1 value or also 2? I am thinking 2 since there are 2 separate charges, each should have their own PE/KE?

Hmm, ... or maybe I can get away with using 1 value if they have the same KE/PE at any point in time?


UPDATE: Another similar question

An electron and positron are $100 fm$. Whats the min speed each particle must have to escape from each other?

    <---(-)        (+)--->
         |- 100 fm -|
(-)                         (+)

The answer looks like:

$$KE_i + KE_i - U = 0$$

So my question is:

  • Why 2 KE but only 1 PE (U)?
  • I suppose reason why its $-U$ because they are unlike charges?
  • Is the reason why $KE_f = 0$? because its asking for min escape velocity? So I assume all KE are used up?
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1 Answer 1

You may recall that one way to solve the problem

A mass $m$ passes the table top moving upwards with initial velocity $v$ (in standard gravity). What is the maximum height above the table reached by mass?

is to use energy conservation: you equate the maximum gravitational potential energy relative the table top with the initial kinetic energy of mass. You can do a similar thing here, replacing the constant gravitational field in my problem with the potential appropriate for your own problem.


One could, of course, write down the expression for the force in each problem and then integrate the equations of motion to get the same answer. This is fairly easy in the case of the mass rising in constant field and slightly more involved in the case of the charges---you have noticed you have to make a bunch of choices about the frame of reference and the treatment of the two particles and so on and so forth.

If you insist on doing it that way you should probably make the canonical transformation to the reduced mass ($\mu$) and separation ($R$) variables as in two body orbits.

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Thanks for you explaination. There is 1 part I don't really understand. What do you mean by your last sentence "canonical transformation to the reduced mass (μ) and separation (R) variables as in two body orbits" –  Jiew Meng Apr 27 '13 at 2:54

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