JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2021
- Hard
- MCQ
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 & Solution
Correct Answer
(A) \({E}=\frac{\sigma}{2 \varepsilon_{0}}\left(1-\frac{{Z}}{\left({Z}^{2}+{R}^{2}\right)^{1 / 2}}\right)\)
Step-by-step Solution
Detailed explanation
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 \({dq}=\sigma {d} {A}\)
\({d} {E} z=\frac{{kdqz}}{\left({z}^{2}+{r}^{2}\right)^{3 / 2}}\)
\(E=\int_{0}^{\mathbb{R}} {dE}_{z}=\frac{\sigma}{2 \epsilon_{0}}\left[1-\frac{{z}}{\sqrt{{R}^{2}+{z}^{2}}}\right]\)
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