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  4. Figure shows a annular disc of outer radius R inner radius R, and uniform surface charge density. Total charge is q. The potential at the centre is (k q/r')(k = (1/4 π (ϵ )0)) where r' is <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-mxpucrujdfbjkmrq.jpg />

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Q. Figure shows a annular disc of outer radius $R$ inner radius $R,$ and uniform surface charge density. Total charge is $q.$ The potential at the centre is $\frac{k q}{r^{'}}\left(k = \frac{1}{4 \pi \left(\epsilon \right)_{0}}\right)$ where $r^{'}$ is
Question

NTA AbhyasNTA Abhyas 2020

Solution:

$V=k\displaystyle \int _{x = r}^{R}\frac{\sigma \cdot 2 \pi x d x}{x}=k\cdot \sigma 2\pi \left(\right.R-r\left.\right)$