Question

An infinitely large non-conducting plane of uniform surface charge density $\alpha$ has circular of certain radius carved out from it. The elastic field at a point which is at a distance 'a' from the centre of the aperture (perpendicular to the plane) is $\frac{\sigma }{2\sqrt{2}{ϵ}_0}$. Find the radius of aperture:

• A. $a\sqrt{2}$
• B. $a$
• C. $2a$
• D. $3a$

Answer 1

The correct option is B $a$

$\begin{array}{c}\frac{\sigma }{2E_0}\\ =\frac{\sigma }{2E_0}\left(1-\frac{a}{{\left(a^2+R^2\right)}^{\frac{1}{2}}}\right)\\ =\frac{-\sigma }{2E_0}\left(1-\frac{a}{{\left(a^2+R^2\right)}^{\frac{1}{2}}}\right)\\ Effectiveelectricfieldwillbe\\ =\frac{\sigma }{2E_0}+\frac{-\sigma }{2E_0}\left(1-\frac{a}{{\left(a^2+R^2\right)}^{\frac{1}{2}}}\right)\\ =\frac{\sigma }{2E_0}\left(\frac{a}{{\left(a^2+R^2\right)}^{\frac{1}{2}}}\right)\\ =\frac{\sigma }{2E_0}\left(\frac{a}{{\left(a^2+R^2\right)}^{\frac{1}{2}}}\right)=\frac{\sigma }{2\sqrt{3E_0}}\\ \Rightarrow R=a\\ \therefore theradiusofcircularaperturewillbea\\ Hence,\\ optionBiscorrectanswer.\end{array}$