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I am trying to solve this problem where two point charges are infinite distance apart(horizontally) but separated by some finite distance vertically. Both are positively charged and projected horizontally, anti parallely with some velocity. I am required to find the minimum distance of separation. How can I set up this problem? I am not able to figure out how to describe the motion of the particles because the forces will act all throughout the motion and there will be perpendicular and parallel components of this force which keep changing direction. I am just a grade 12 student and this is an exam problem. To me this problem seems extremely complex, so am I missing out on something? Does the problem have a trivial solution?

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  • $\begingroup$ Your setup isn't clear. What on earth do you mean by "infinite distance apart(horizontally) but separated by some finite distance vertically" ? You would think this is relative to some coordinate system, but then you add "Both are positively charged and projected towards towards each other with some velocity", so why on earth isn't the coordinate system just a single axis aligned with the 2 charges ? Maybe you meant to say that they are projected towards each other horizontally, and not directly towards each other. If so this isn't clear. A diagram would help. $\endgroup$ – ticster Jul 22 '14 at 12:44
  • $\begingroup$ Sorry, I have only my ipad, so no diagram. What I mean is they are at infinite distance but not on the same line so as they come closer they won't be moving exactly head on but with a finite distance between them. If they were on the same line, that would be easy, but here there would be a perpendicular component of force $\endgroup$ – user34304 Jul 22 '14 at 12:47
  • $\begingroup$ Hint to the question (v1): Look up Rutherford scattering. $\endgroup$ – Qmechanic Jul 22 '14 at 12:48
  • $\begingroup$ @ticster And sorry not projected towards each other but projected horizontally, antiparallely like u said in the end $\endgroup$ – user34304 Jul 22 '14 at 12:52
  • $\begingroup$ @Qmechanic Rutherford scattering might be more than what he's bargaining for. He just needs to calculate the trajectories where the initial separations is L, find the closest point in those trajectories, and see what it is as $L \rightarrow \infty$. $\endgroup$ – ticster Jul 22 '14 at 13:28
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If you look at the two particles in the centre of mass frame you'll see something like:

Closest approach

Your task is to find $b$. In the COM frame both particles move in an effective central potential, and the two trajectories are symmetric. I'm going to ignore the green one and just show how to calculate $b$ for the red trajectory. The closest approach is then $2b$.

In physics generally, and especially in exam questions, the best starting point is to look helpful conservation laws. In this case we know angular momentum is conserved because the effective potential is spherically symmetric. At the starting point the angular momentum is:

$$ L_0 = \frac{mv_0}{d} $$

and at the moment of closest approach:

$$ L = \frac{mv}{b} $$

and since $L_0 = L$ we have:

$$ \frac{v_0}{d} = \frac{v}{b} \tag{1} $$

where $v$ is the initial velocity and $v$ is the velocity at the moment of closest approach.

We also know energy is conserved. The potential energy is initially zero, and at the moment of closest approach it's:

$$ V = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{2b} $$

where $Q$ is the particle charge. Since energy is conserved we have:

$$ \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + \frac{1}{4\pi\epsilon_0}\frac{Q^2}{2b} \tag{2} $$

Now, you have two unknowns, $v$ and $b$, and two equations, (1) and (2), so you can eliminate $v$ and calculate what $b$ is. Since this is basically a homework problem I'll the rest to you, but feel free to shout if you're still stuck.

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  • $\begingroup$ Thanks a lot!! I am asking just out of curiosity, but suppose I wanted to find out the equation of trajectory, how would you go about doing that? I just want to know the general method $\endgroup$ – user34304 Jul 22 '14 at 16:01
  • $\begingroup$ A doubt, sorry. I didn't really get what you mean by " in the com frame the particles move in an effective central potential" $\endgroup$ – user34304 Jul 22 '14 at 16:10
  • $\begingroup$ @user34304: the equation for the trajectory is surprisingly fiddly to calculate. For an attractive force (unlike charges) it's the same equation as for gravity, and Googling will find lots of articles on calculating orbits. Be prepared for some hard work though! $\endgroup$ – John Rennie Jul 22 '14 at 16:52
  • $\begingroup$ @user34304: in the centre of mass frame the line joining the two particles always passes through the centre of mass - obviously, because the centre of mass is halfway between the two particles. That means for both particles the force they feel always points towards the centre of mass. So it is as if they are moving in a central field centred on the centre of mass. $\endgroup$ – John Rennie Jul 22 '14 at 16:54

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