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The linear charge density on a dielectric ring of radius $R$ varies with $\theta $ as $\lambda \, = \,{\lambda _0}\,\cos \,\,\theta /2,$ where $\lambda _0$ is constant. Find the potential at the centre $O$ of ring. [in volt]
$\lambda _0\,\,R$
$\frac {\lambda _0\,R}{2}$
$\frac {\lambda _0}{4\pi \epsilon _0R }$
zero
Solution
The charge on the infinitesimal elements of arc which subtend an angle $d\theta $ at the center of the ring.
$\mathrm{dQ}=\lambda \mathrm{Rd} \theta=\lambda_{0} \cos \frac{\theta}{2} \mathrm{Rd} \theta$
Potential at the centre of ring due to charge $\mathrm{dQ}$
$\mathrm{dV}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{d} \mathrm{Q}}{\mathrm{R}}=\frac{\lambda_{0} \cos \frac{\theta}{2} \cdot \mathrm{Rd} \theta}{4 \pi \varepsilon_{0} \mathrm{R}}$
$\mathrm{V}=\int \mathrm{d} \mathrm{V} \Rightarrow \mathrm{V}=\frac{\lambda}{4 \pi \varepsilon_{0}} \int_{0}^{2 \pi} \cos \frac{\theta}{2} \mathrm{d} \theta$
$=\frac{\lambda}{4 \pi \varepsilon_{0}}\left[\frac{\sin \frac{\theta}{2}}{\frac{1}{2}}\right]_{0}^{2 \pi}=0 \mathrm{\,V}$
