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(B) The total charge $q_{	ext{total}}$ on the ring is given by the integral of the charge density $\lambda$ over the circumference:$q_{	ext{total}}

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equal to $qQ / 4\pi \varepsilon_0 r$

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(B) The total charge $q_{	ext{total}}$ on the ring is given by the integral of the charge density $\lambda$ over the circumference:$q_{	ext{total}} = \int_0^{2\pi} \lambda (r d	heta) = \int_0^{2\pi} \frac{q \sin^2 	heta}{\pi r} (r d	heta) = \frac{q}{\pi} \int_0^{2\pi} \sin^2 	heta d	heta$Using the identity $\sin^2 	heta = (1 - \cos 2	heta) / 2$:$q_{	ext{total}} = \frac{q}{\pi} \int_0^{2\pi} \frac{1 - \cos 2	heta}{2} d	heta = \frac{q}{2\pi} [	heta - \frac{\sin 2	heta}{2}]_0^{2\pi} = \frac{q}{2\pi} (2\pi) = q$Since all points on the ring are at a distance $r$ from the centre,the electric potential $V$ at the centre is:$V = \frac{1}{4\pi \varepsilon_0} \int \frac{dq}{r} = \frac{1}{4\pi \varepsilon_0 r} \int dq = \frac{q_{	ext{total}}}{4\pi \varepsilon_0 r} = \frac{q}{4\pi \varepsilon_0 r}$The work done by the electric force in moving a charge $Q$ from the centre to infinity is $W = Q(V_{	ext{centre}} - V_{\infty})$. Since $V_{\infty} = 0$:$W = Q \cdot V = \frac{qQ}{4\pi \varepsilon_0 r}$

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