Question 77 of 311

A thin metallic wire having cross sectional area of ${10}^{-4}{m}^{2}$ is used to make a ring of radius $30cm$. A positive charge of $2\pi C$ is uniformly distributed over the ring, while another positive charge of $30pC$ is kept at the centre of the ring. The tension in the ring is _______N; provided that the ring does not get deformed (neglect the influence of gravity).
(Given, $\dfrac{1}{4\pi {ϵ}_{0}}=9\times {10}^{9}$ SI units)

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Answer: 3

The linear charge density of the ring is given by
$\lambda =\dfrac{Q}{2\pi R}...\left(1\right)$
With reference to the above diagram, for the equilibrium condition of the ring, it can be written that
$2Tsin\left(\dfrac{d\theta }{2}\right)=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{{q}_{0}}{{R}^{2}}\cdot \left[\lambda \left(Rd\theta \right)\right]$
$\Rightarrow T=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{{q}_{0}Q}{\left({R}^{2}\right)\times 2\pi }$ [As for very small angles $\sin \left(\dfrac{d\theta }{2}\right)=\dfrac{d\theta }{2}$]
$=\dfrac{\left(9\times {10}^{9}\right)\left(2\pi \times 30\times {10}^{-12}\right)}{(0.30{)}^{2}\times 2\pi }$
$=\dfrac{9\times {10}^{-3}\times 30}{9\times {10}^{-2}}=3N$