Electrostatics Question 5
Question 5 - 2024 (27 Jan Shift 1)
A thin metallic wire having cross sectional area of $10^{-4} \mathrm{~m}^{2}$ is used to make a ring of radius $30 \mathrm{~cm}$. A positive charge of $2 \pi \mathrm{C}$ is uniformly distributed over the ring, while another positive charge of $30 \mathrm{pC}$ is kept at the centre of the ring. The tension in the ring is $\mathrm{N}$; provided that the ring does not get deformed (neglect the influence of gravity). (given, $\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{SI}$ units)
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Answer: (3)
$2 \mathrm{~T} \sin \frac{\mathrm{d} \theta}{2}=\frac{\mathrm{kq}_{0}}{\mathrm{R}^{2}} \cdot \lambda \mathrm{Rd} \theta$
$\left[\lambda=\frac{\mathrm{Q}}{2 \pi \mathrm{R}}\right]$
$\Rightarrow \mathrm{T}=\frac{\mathrm{Kq}_{0} \mathrm{Q}}{\left(\mathrm{R}^{2}\right) \times 2 \pi}$
$=\frac{\left(9 \times 10^{9}\right)\left(2 \pi \times 30 \times 10^{-12}\right)}{(0.30)^{2} \times 2 \pi}$
$=\frac{9 \times 10^{-3} \times 30}{9 \times 10^{-2}}=3 \mathrm{~N}$
