A semicircular ring of radius a has a charge | vedclass

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(D) The charge on a small element of the ring of angular width $d	heta$ is given by $dq = \lambda \cdot dl$,where $dl = a \, d	heta$.Given $\lambda

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Zero

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(D) The charge on a small element of the ring of angular width $d	heta$ is given by $dq = \lambda \cdot dl$,where $dl = a \, d	heta$.Given $\lambda = {\lambda _0} \cos 	heta$,we have $dq = (\lambda _0 \cos 	heta) (a \, d	heta)$.The semicircular ring spans from $	heta = 0$ to $	heta = \pi$.Therefore,the total charge $Q$ is:$Q = \int_{0}^{\pi} \lambda _0 a \cos 	heta \, d	heta$$Q = \lambda _0 a \int_{0}^{\pi} \cos 	heta \, d	heta$$Q = \lambda _0 a [\sin 	heta]_{0}^{\pi}$$Q = \lambda _0 a (\sin \pi - \sin 0)$$Q = \lambda _0 a (0 - 0) = 0$.Thus,the total charge on the ring is zero.

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

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