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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 :-
${E}=\frac{\sigma}{2 \varepsilon_{0}}\left(1-\frac{{Z}}{\left({Z}^{2}+{R}^{2}\right)^{1 / 2}}\right)$
${E}=\frac{\sigma}{2 \varepsilon_{0}}\left(1+\frac{{Z}}{\left({Z}^{2}+{R}^{2}\right)^{1 / 2}}\right)$
${E}=\frac{2 \varepsilon_{0}}{\sigma}\left(\frac{1}{\left({Z}^{2}+{R}^{2}\right)^{1 / 2}}+{Z}\right)$
${E}=\frac{\sigma}{2 \varepsilon_{0}}\left(\frac{1}{\left({Z}^{2}+{R}^{2}\right)}+\frac{1}{{Z}^{2}}\right)$
Solution
Consider a small ring of radius ${r}$ and thickness dr on disc.
$[Image]$
area of elemental ring on disc
${d} {A}=2 \pi {rdr}$
charge on this ring ${dq}=\sigma {d} {A}$
${d} {E} z=\frac{{kdqz}}{\left({z}^{2}+{r}^{2}\right)^{3 / 2}}$
$E=\int_{0}^{\mathbb{R}} {dE}_{z}=\frac{\sigma}{2 \epsilon_{0}}\left[1-\frac{{z}}{\sqrt{{R}^{2}+{z}^{2}}}\right]$
