A thin metallic wire having cross sectional area of $$10^{-4} \text{ m}^2$$ is used to make a ring of radius $$30$$ cm. A positive charge of $$2\pi$$ C is uniformly distributed over the ring, while another positive charge of $$30$$ pC 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, $$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$$ SI units)

The radius of the ring is
$$R = 30\text{ cm}=0.30\text{ m}.$$

The charge uniformly spread over the ring is
$$Q = 2\pi\text{ C},$$
and the point charge kept at the centre is
$$q = 30\text{ pC}=30\times10^{-12}\text{ C}.$$

Choose a small element of the ring that subtends an angle $$d\theta$$ at the centre.
Length of the element
$$dl = R\,d\theta,$$
charge on the element
$$dq = \frac{Q}{2\pi R}\,dl \;=\; \frac{Q}{2\pi}\,d\theta.$$

The electrostatic force on this element due to the central charge $$q$$ is radially outward and equals
$$dF = \frac{1}{4\pi\varepsilon_0}\,\frac{q\,dq}{R^2} = k\,\frac{q}{R^2}\,\frac{Q}{2\pi}\,d\theta,$$
where $$k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\text{N\,m}^2\text{/C}^2.$$

The element is held in equilibrium by the tension $$T$$ in the ring. The two tension forces at its ends are tangential; their radial components add up to an inward force $$2T\sin\!\left(\frac{d\theta}{2}\right)\;\approx\;T\,d\theta$$ (because $$d\theta$$ is infinitesimal).

Equating the outward electrostatic force and the inward force due to tension:
$$T\,d\theta = k\,\frac{qQ}{2\pi R^{2}}\,d\theta \;\;\Longrightarrow\;\; T = k\,\frac{qQ}{2\pi R^{2}}.$$

In the given data $$Q = 2\pi\,$C, so the factors $$2$$\pi$$$$ cancel, giving a very simple result:
$$T = k\,$$\frac{q}{R^{2}$$}.$$

Substituting the numerical values:
$$T = 9$$\times$$10^{9}\,$$\frac{30\times10^{-12}$$}{(0.30)^{2}} = 9$$\times$$10^{9}$$\times$$30$$\times$$10^{-12}$$\times$$$$\frac{1}{0.09}$$$$ $$\phantom{T} = 9$$\times$$30$$\times$$10^{-3}$$\times$$$$\frac{1}{0.09}$$ = 270$$\times$$10^{-3}$$\times$$$$\frac{1}{0.09}$$ = 0.27$$\times$$$$\frac{1}{0.09}$$ = 3\;$$\text{N}$$.$$

Hence the tension in the metallic ring is $$\mathbf{3\;N}.$$