A uniformly charged disc of radius R having surface charge density σ… | JEE Main 2021 (27 Aug, Shift 1)

Q33JEE Main 2021 · 27 AUG · SHIFT 1Medium

Electric Field & Field Lines

A uniformly charged disc of radius R having surface charge density $\sigma$ is placed in the xy plane with its center at the origin. Find the electric field intensity along the z-axis at a distance Z from origin :-

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Answer: (a)

Split the disc into rings of radius $r$ , width $dr$ : $dE = \dfrac{k(\sigma 2\pi r\,dr)Z}{(Z^2+r^2)^{3/2}}$ .
Integrating from $0$ to $R$ : $E = \dfrac{\sigma Z}{2\varepsilon_0}\left[\dfrac{1}{Z} - \dfrac{1}{\sqrt{Z^2+R^2}}\right]$
$E = \dfrac{\sigma}{2\varepsilon_0}\left(1 - \dfrac{Z}{\sqrt{Z^2+R^2}}\right)$
So the answer is (a).

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Electric Field & Field Lines question from JEE Main 2021 (27 Aug, Shift 1)

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