A circular lamina of radius 'R' is having surface charge density \( \sigma\,\mathrm{cm^{-2}} \). Electric field at axial distance '2R' is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( E = \displaystyle\frac{\sigma}{2\varepsilon_0}\left(1 - \frac{2}{\sqrt{5}}\right) \)
The disc can be divided into thin rings. On the axis, the field of each ring points along the axis (the perpendicular components cancel by symmetry), so the ring fields simply add.
A ring of radius \( s \) and width \( ds \) carries charge \( dq = \sigma\,2\pi s\,ds \). Its axial field at distance \( x \) from the centre is
\( \displaystyle dE = \frac{1}{4\pi\varepsilon_0}\,\frac{x\,dq}{(s^2 + x^2)^{3/2}} = \frac{\sigma x}{2\varepsilon_0}\,\frac{s\,ds}{(s^2 + x^2)^{3/2}} \)
Integrating from \( s = 0 \) to \( s = R \):
\( \displaystyle E = \frac{\sigma x}{2\varepsilon_0}\left[\frac{1}{x} - \frac{1}{\sqrt{R^2 + x^2}}\right] = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) \)
Putting \( x = 2R \):
\( \displaystyle \frac{x}{\sqrt{x^2 + R^2}} = \frac{2R}{\sqrt{4R^2 + R^2}} = \frac{2}{\sqrt{5}} \)
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{2}{\sqrt{5}}\right) \)