A thin metallic wire having cross sectional a | vedclass

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Question

$ 2 \mathrm{~T} \sin \frac{\mathrm{d} 	heta}{2}=\frac{\mathrm{kq}_0}{\mathrm{R}^2} \cdot \lambda \mathrm{Rd} 	heta $

$ {\left[\lambda=\frac{\math

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$ 2 \mathrm{~T} \sin \frac{\mathrm{d} 	heta}{2}=\frac{\mathrm{kq}_0}{\mathrm{R}^2} \cdot \lambda \mathrm{Rd} 	heta $

$ {\left[\lambda=\frac{\mathrm{Q}}{2 \pi \mathrm{R}}ight]}$

$ \Rightarrow \mathrm{T}=\frac{\mathrm{Kq} \mathrm{q}_0 \mathrm{Q}}{\left(\mathrm{R}^2ight) 	imes 2 \pi} $

$ =\frac{\left(9 	imes 10^9ight)\left(2 \pi 	imes 30 	imes 10^{-12}ight)}{(0.30)^2 	imes 2 \pi} $

$ =\frac{9 	imes 10^{-3} 	imes 30}{9 	imes 10^{-2}}=3 \mathrm{~N}$

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