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