5
$\begingroup$

What will be final velocity of three charges q, q and 2q kept along an equilateral triangle of side r at infinite distance. All three masses are equal.

I tried to conserve Total Energy

$$\frac{2kq^2}{r} + \frac{2kq^2}{r} + \frac{kq^2}{r} = \frac{mv^2}{2}+ \frac{mv^2}{2} + \frac{m(v_{2})^2}{2})$$

$$\frac{5kq^2}{r} = m(v^2 + \frac{v_{2}^2}{2}$$

$$\sqrt{\frac{10kq^2}{rm}-2v^2}= v_{2}$$

Conserving momentum gave

$$v_{2} = v\sqrt{2(1+\cos\theta)}\tag{1}$$ where $\theta$ is angle b\w velocity of q & q.

When I tried to make some graph predicting their motion and speed in different direction was like hell.

I could not get any further.

As helped by Fire I used COM along y direction.

$$my + my - my' = 0$$ $$ y' = 2y$$ $$dy'/dt = 2dy/dt $$ $$ v_{2}= 2 v_{y}$$ $$v_{2}= 2 v\cos\left(\frac{\theta}{2}\right)\tag{2}$$

I can't imagine if charges would move along straight line or in curves. Will $\theta$ change or not? Can all three have same velocity at infinity?

Someone told me it uses Taylor series.

$\endgroup$
9
  • 2
    $\begingroup$ I take back the previous comment. The constraint is absolutely critical in this, and if you are going to use any kind of conservation it tells you things not clearly specified in the problem statement. This may be a very cute (i.e. deeply evil) exercise. $\endgroup$ Feb 2, 2016 at 4:16
  • $\begingroup$ I guess some information on the masses is needed. $\endgroup$
    – akhmeteli
    Feb 2, 2016 at 6:09
  • 1
    $\begingroup$ as dmckee said, it is likely to be an evil problem. I just tried writing the equation of motion for the q-particles and solve it numerically. Interestingly, their orbits look optically very much linear (!?). But maybe it is just an error in coding, I did not have time to check the detail carefully, though. $\endgroup$
    – cnguyen
    Feb 9, 2016 at 13:10
  • 1
    $\begingroup$ I find odd that @David Z closed this question as off-topic. It sounds like a homework, but it is rather non-trivially interesting. The questioner did show effort in solving it, too. $\endgroup$
    – cnguyen
    Feb 11, 2016 at 8:35
  • 2
    $\begingroup$ @ophelia Being nontrivially interesting is not one of the criteria we use to determine whether a question is on topic or not. Hopefully you find it less odd after noting that. If you think the question does not meet the criteria for being on hold, you can make that argument for reopening it. But as I see it, the question is homework-like and does not ask a conceptual question - what it's asking is basically "what do I do next?" $\endgroup$
    – David Z
    Feb 11, 2016 at 8:53

4 Answers 4

2
$\begingroup$

It might help you to think about the symmetry of the situation.

enter image description here

First in the application of conservation of momentum and then what the trajectories of the charges must be to keep the centre of mass $C$ at the same position. This will give you a connection between $v_y$ and $v_x$.

$\endgroup$
5
  • $\begingroup$ I already had that figure in mind but couldn't get connection b/w vx and vy $\endgroup$ Feb 2, 2016 at 7:57
  • $\begingroup$ What must happen to the three masses if the centre of mass stays in the same position? Move just one of of the masses and note the effect on the position of the centre of mass. Do the same for two masses. $\endgroup$
    – Farcher
    Feb 2, 2016 at 8:52
  • $\begingroup$ Sorry!! but still got nothing $\endgroup$ Feb 2, 2016 at 9:20
  • $\begingroup$ @AnubhavGoel - Farcher's answer is correct... $\endgroup$ Feb 9, 2016 at 16:17
  • $\begingroup$ @honeste_vivere He is always correct. But hint is not getting me anywhere. $\endgroup$ Feb 9, 2016 at 17:05
2
$\begingroup$

Consider a more general case in which the three charges are $\lambda q, q$, and $q$, with $\lambda\geq 0$. Let $A$, $B$, and $C$ be the respective positions of the charges. By symmetry, we have $|AB|=|AC|$. Due to the conservation of momentum, the center of mass $O$ of the system is stationary.

Let $\theta$ be the angle $OBC$, and $\alpha$ be the angle $ABO$. Simple trigonometric manipulations give the relation

$$\tan\alpha=\frac{2\tan\theta}{3\tan^2\theta+1}.$$

Notice that $\theta=\alpha=\pi/6$ at $t=0$.

Now, consider the forces $\mathbf F_{AB}$ and $\mathbf F_{CB}$ on charge $B$ by charges $A$ and $C$, respectively. The angle $\theta$ changes when $\mathbf F_{AB} + \mathbf F_{CB}$ is not solely along $OB$. So, as $t\rightarrow\infty$, we expect $\theta$ to approach a constant value $\theta_\infty$, such that $\mathbf F_{AB} + \mathbf F_{CB}$ is solely along $OB$. That is, $\theta_\infty$ satisfies

$$F_{AB} \sin\alpha_\infty = F_{CB}\sin\theta_\infty.$$

Using the form of the Coulomb's law for $F_{AB}$ and $F_{CB}$, we find that

$$4\lambda\sin\alpha_\infty=\sin\theta_\infty(1+9\tan^2\theta_\infty),$$

which then allows us to solve for $\theta_\infty$.

Finally, with the relative positions of the charges known, the conservation of energy easily yields the final velocities of the three charges.

Lastly, for $\lambda\ll1$ and $\lambda\gg1$, we see that $\theta_\infty=0$ and $\theta_\infty=\pi/2$, respectively. For the specific case of $\lambda=2$, we can numerically obtain $\theta_\infty\approx0.657$.

$\endgroup$
17
  • $\begingroup$ As t→∞, the angle theta approaches a constant value theta∞. This means that the resultant force acting on the charge at B is solely along OB . I did not understand how resultant force is solely along OB. $\endgroup$ Feb 3, 2016 at 16:42
  • $\begingroup$ @AnubhavGoel, if the force is not along $OB$, the angle $\theta$ would change. $\endgroup$
    – leongz
    Feb 3, 2016 at 20:42
  • $\begingroup$ Why force is along OB at infinity. $\endgroup$ Feb 4, 2016 at 7:47
  • $\begingroup$ @AnubhavGoel, if the force is not along $OB$ at $t\rightarrow\infty$, the angle $\theta$ would not be a constant. $\endgroup$
    – leongz
    Feb 9, 2016 at 0:19
  • $\begingroup$ By saying OB here do you mean B at infinity or B when it is in equilateral triangle. $\endgroup$ Feb 9, 2016 at 2:34
1
$\begingroup$

You're only a half-step away. You listed conservation of energy and linear momentum, both of which are due to there being no external forces on the three-charge system. But with no external forces, you know that the center of mass of the system won't accelerate. Since the COM starts at rest, this means that the COM will remain stationary. Think about what this means in terms of the geometry of your problem.

(I'll help you out with this if you don't see it, but I think you will.)

$\endgroup$
16
  • $\begingroup$ Still struck. Please help! $\endgroup$ Feb 2, 2016 at 5:56
  • $\begingroup$ The system of three particles has a center of mass at the center of the triangle, right? Well, if one particle moves radially outward by 1 cm, then the other two must do the same, or else the center of mass of the system moves, and we know that it can't. The big point is that you already have the final angles between the particles because you have the initial angles. Do you see it? $\endgroup$
    – Fire
    Feb 2, 2016 at 16:39
  • $\begingroup$ Why the rest two charges would also move 1 cm? $\endgroup$ Feb 3, 2016 at 9:45
  • $\begingroup$ Consider in above diagram 2q moves upward by 2 cm. Then , q moves down by 1 cm. qmay move let by 1 or 2 or 3 cm. Center of mass remain at same position. $\endgroup$ Feb 3, 2016 at 10:02
  • $\begingroup$ sorry! not let , left by 1, 2 or 3cm. $\endgroup$ Feb 3, 2016 at 13:20
1
$\begingroup$

This is actually a particular case of the three-body problem (https://en.wikipedia.org/wiki/Three-body_problem), but with repulsion, rather than attraction. There are very few exactly solvable cases of the three-body problem, among them - the Lagrange's case, where the three bodies are at the vertexes of an equilateral triangle at each moment. Therefore, I suspect you should try to prove for the initial conditions of your problem that the distances between the charges remain equal (although I cannot be sure that this is so). Some homework...

$\endgroup$
2
  • $\begingroup$ how would I know if distances b/w them remain same? $\endgroup$ Feb 3, 2016 at 9:41
  • $\begingroup$ It does not look like they do, so my suggestion looks wrong. $\endgroup$
    – akhmeteli
    Feb 3, 2016 at 12:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.