# Increase in tension in the wire

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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?

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 $\frac{KqQ}{2\pi R^2}$
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

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