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As the question says , i thought about using coulumbs law , but for any small dq charge , all other part will be attracting it , so.how to get the total tension force ( difficulty is in as distance is different for different small charges )

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Actually this look like a homework type, but I think it is jee type question, so most of the student want this answer I think ishould answer it , actually when you distribute a charge $Q$ unifromly each $dl$ length o the wire will experience the repulsive force from rest of the other, so you can do this question by just simply applying work energy theory. Suppose the tension in the ring is $T$, and we stretch the ring by some distance $d\ell$, then the work done by the tension is $dW = T dr$. d finally, because the work done by the tension has to be equal to the change in the self energy we just equate $\delta E$ and $dW$ to get: $$ T 2\pi dr = \frac{dE}{dr} dr $$ So: $$ T = \frac{1}{2\pi} \frac{dE}{dr} $$ So if you know $dE/dr$ this gives you the tension in the ring.

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  • $\begingroup$ Can we say its definitely proportional to (Q/R)^2 ? , If yes can anyone prove this fact ? , Actually T = kQ^2/R^2 , was given as answer ? Can anyone explain pls $\endgroup$ – Euler142857 Oct 4 at 8:03
  • $\begingroup$ It is a derivation which is already on site, you will search it and you get it $\endgroup$ – yuvraj singh Oct 4 at 11:29
  • $\begingroup$ Pls mention the link Sir $\endgroup$ – Euler142857 Oct 4 at 12:21
  • $\begingroup$ $dE/dr$ is self energy of the ring, here is the link, physics.stackexchange.com/questions/16520/… $\endgroup$ – yuvraj singh Oct 4 at 13:13
  • $\begingroup$ Answer comes to be infinite $\endgroup$ – yuvraj singh Oct 4 at 13:14

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