0
$\begingroup$

Three equal charges +Q each are placed on the vertices of an equilateral triangle. A charge +q is initially placed at the centre of the triangle. If this charge (+q) is slightly displaced towards a vertex and left free.

What will be its motion?

My working:

Initially its in equilibrium and when its displaced to one end, charges far from it will exert lesser repulsive force and the vertex charge will exert more force since it went closer to it. Hence it goes back to its original position and does SHM.

Am I correct?

$\endgroup$
1
$\begingroup$

I suspect that the Earnshaw's theorem might make such motion unstable. You may consider a similar problem: a positive charge between two positive charges. All your reasoning seems applicable there, but it is intuitively obvious that the motion will be unstable.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What do you mean by unstable? It won't be SHM then? $\endgroup$ – Mixcels Apr 7 '15 at 13:04
  • $\begingroup$ Is it oscillatory motion then? $\endgroup$ – Mixcels Apr 7 '15 at 19:18
  • $\begingroup$ @Mixcels: Imagine a ball "oscillating" on a saddle: its motion will be unstable with respect to ball motion orthogonal to its original trajectory, so the ball will eventually fall from the saddle. So the motion of your charge will be unstable with respect to its motion in the direction orthogonal to its displacement from the center of the triangle. So the charge will probably be driven outside of the triangle, and its motion will not be periodical. $\endgroup$ – akhmeteli Apr 8 '15 at 0:08
  • $\begingroup$ I dont understand why would it go off when repulsive force is pushing it back to the centre. $\endgroup$ – Mixcels Apr 8 '15 at 10:16
  • $\begingroup$ @Mixcels: because the charge's motion will be unstable with respect to displacements in the transverse direction, however small. There is a reason it is difficult, to say the least, to balance an egg on its narrow end. $\endgroup$ – akhmeteli Apr 8 '15 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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