Question 101 of 311

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:

A$E=\dfrac{2{\epsilon }_{0}}{\sigma }\left(\dfrac{1}{{\left({Z}^{2}+{R}^{2}\right)}^{1/2}}+Z\right)$B$E=\dfrac{\sigma }{2{\epsilon }_{0}}\left(1+\dfrac{Z}{{\left({Z}^{2}+{R}^{2}\right)}^{1/2}}\right)$C$E=\dfrac{\sigma }{2{\epsilon }_{0}}\left(1-\dfrac{Z}{{\left({Z}^{2}+{R}^{2}\right)}^{1/2}}\right)$D$E=\dfrac{\sigma }{2{\epsilon }_{0}}\left(\dfrac{1}{\left({Z}^{2}+{R}^{2}\right)}+\dfrac{1}{{Z}^{2}}\right)$

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Answer: (C) $E=\dfrac{\sigma }{2{\epsilon }_{0}}\left(1-\dfrac{Z}{{\left({Z}^{2}+{R}^{2}\right)}^{1/2}}\right)$

$E={\int }_{0}^{R}\dfrac{1}{4\pi {\epsilon }_{0}}\times \dfrac{\sigma \times 2\pi x\left(dx\right)\times x}{{\left({x}^{2}+{Z}^{2}\right)}^{\dfrac{3}{2}}}=\dfrac{\sigma }{2{\epsilon }_{0}}\left(1-\dfrac{Z}{{\left({Z}^{2}+{R}^{2}\right)}^{1/2}}\right)$