A thin metallic wire having cross sectional area of $10^{- 4} m^{2}$ is used to make a ring of radius $30 cm$ . A positive charge of $2 π C$ 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, $\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9}$ SI units)

Question

A thin metallic wire having cross sectional area of $10^{- 4} m^{2}$ is used to make a ring of radius $30 cm$ . A positive charge of $2 π C$ 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, $\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9}$ SI units)

Aicharya · Accepted Answer

The linear charge density of the ring is given by

$λ = \frac{Q}{2 π R} . . . (1)$

With reference to the above diagram, for the equilibrium condition of the ring, it can be written that

$2 T s i n (\frac{d θ}{2}) = \frac{1}{4 π ε_{0}} \frac{q_{0}}{R^{2}} \cdot [λ (R d θ)]$

$\Rightarrow T = \frac{1}{4 π ε_{0}} \frac{q_{0} Q}{(R^{2}) \times 2 π}$ [As for very small angles $\sin (\frac{d θ}{2}) = \frac{d θ}{2}$ ]

$= \frac{(9 \times 10^{9}) (2 π \times 30 \times 10^{- 12})}{(0.30)^{2} \times 2 π}$

$= \frac{9 \times 10^{- 3} \times 30}{9 \times 10^{- 2}} = 3 N$

A thin metallic wire having cross sectional area of 10 - 4 m 2 is used to make a ring of radius 30 cm . A…

Solution