JEE Main 2021 (27 Aug, Shift 1) Physics Q11: Electric Field

Question

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 :

Accepted Answer

Option 2: $\mathrm{E}=\frac{\sigma}{2\varepsilon_0}\left(1-\frac{\mathrm{Z}}{(\mathrm{Z}^2+\mathrm{R}^2)^{1/2}}\right)$

Answer

$\mathrm{E}=\frac{\sigma}{2\varepsilon_0}\left(1+\frac{\mathrm{Z}}{(\mathrm{Z}^2+\mathrm{R}^2)^{1/2}}\right)$

Answer

$\mathrm{E}=\frac{\sigma}{2\varepsilon_0}\left(1-\frac{\mathrm{Z}}{(\mathrm{Z}^2+\mathrm{R}^2)^{1/2}}\right)$

Answer

$\mathrm{E}=\frac{2\varepsilon_0}{\sigma}\left(\frac{1}{(\mathrm{Z}^2+\mathrm{R}^2)^{1/2}}+\mathrm{Z}\right)$

Answer

$\mathrm{E}=\frac{\sigma}{2\varepsilon_0}\left(\frac{1}{(\mathrm{Z}^2+\mathrm{R}^2)}+\frac{1}{\mathrm{Z}^2}\right)$

JEE Main 27 August 2021, Shift 1, Physics Q11

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