Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Increase in tension in the wire

2 votes

A thin circular wire of radius r has a charge Q. If a point charge q is placed at the centre of the ring, then find the increase in tension in the wire.

According to me initial tension should be zero , as no charge is present in the circular wire .
And a charge q is introduced tension should be KQq/r$^2$ which is change in tension

But this wrong ,why?

asked Jan 20, 2017 in Physics Problems by koolman (4,286 points)
Initial tension will **not** be zero, because the charges already on the wire will repel each other. An initial tension is necessary to keep the wire from falling apart ("exploding").

The **radial force** on the wire is kQq/r^2, but the additional tension is the **tangential** force in the wire which is required to counteract this radial force. See the similar question http://physicsproblems.nfshost.com//?qa=946/tangential-tension-on-a-charged-ring-due-to-q-at-center.
So how can we find initial tension
The question does not ask for the initial tension, only the increase in tension.
From that question I am getting tension T=$\frac{KqQ}{4\pi R^2}$
But the answer given as $\frac{KqQ}{2\pi R^2}$
Check your calculation carefully. If you are are still getting the wrong answer please show the steps in your answer.
Please see my answer
Sorry, my mistake. (I made the same mistake in the answer to the earlier question.) If the element of wire subtends angle $\delta\theta$ then the net tension force on it is $2T\sin(\frac12\delta\theta) \approx T\delta\theta$. This takes care of the missing factor of 2 in your answer.
Thanks a lot

1 Answer

2 votes
Best answer

For an element of ring which subtends an angle $ \delta\theta$ ,the charge is $\delta q=q\frac{\delta\theta}{2\pi}$

And radial component of total tension on the element would be $2T\sin(\frac{1}{2}\delta\theta) \approx T\delta\theta$
That must be equal to $\frac{Q\delta q}{4\pi\epsilon_0 R^2}$

And we know $\frac{\delta q}{\delta\theta}=\frac{q}{2\pi}$

Hence $\frac{KQq}{2\pi R^2}=T$
$\frac{KqQ}{2\pi R^2}$=T

answered Jan 23, 2017 by koolman (4,286 points)
selected Jan 24, 2017 by koolman