A semicircular ring of radius 'a' has | vedclass

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Question

$\mathrm{q}=\int_{0}^{\pi} \lambda \mathrm{Rd} 	heta \quad \lambda=\lambda_{0} \cos 	heta$

$\mathrm{q}=\lambda_{0} \mathrm{R} \int_{0}^{\pi} \cos

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Vedclass · Answer

$\mathrm{q}=\int_{0}^{\pi} \lambda \mathrm{Rd} 	heta \quad \lambda=\lambda_{0} \cos 	heta$

$\mathrm{q}=\lambda_{0} \mathrm{R} \int_{0}^{\pi} \cos 	heta \mathrm{d} 	heta$

$\mathrm{q}=\lambda_{0} \mathrm{R}(\sin 	heta)_{0}^{\pi}=0$

A semicircular ring of radius $'a'$ has charge density $\lambda = {\lambda _0}\,\cos \,\theta $ where ${\lambda _0}$ is constant and $'\theta'$ is shown in figure. Then total charge on the ring is

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