JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 \({E}=\frac{\sigma}{2 \varepsilon_{0}}\left(1-\frac{{Z}}{\left({Z}^{2}+{R}^{2}\right)^{1 / 2}}\right)\)
- B \({E}=\frac{\sigma}{2 \varepsilon_{0}}\left(1+\frac{{Z}}{\left({Z}^{2}+{R}^{2}\right)^{1 / 2}}\right)\)
- C \({E}=\frac{2 \varepsilon_{0}}{\sigma}\left(\frac{1}{\left({Z}^{2}+{R}^{2}\right)^{1 / 2}}+{Z}\right)\)
- D \({E}=\frac{\sigma}{2 \varepsilon_{0}}\left(\frac{1}{\left({Z}^{2}+{R}^{2}\right)}+\frac{1}{{Z}^{2}}\right)\)
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 \({dq}=\sigma {d} {A}\) \({d} {E} z=\frac{{kdqz}}{\left({z}^{2}+{r}^{2}\right)^{3 / 2}}\)…
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