An infinite plane sheet of charge having unif | vedclass

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(A) The electric field due to an infinite plane sheet of charge is given by $E_S = \frac{|\sigma_s|}{2\epsilon_0}$.At the point $(0, 0, 2)$,the distan

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) The electric field due to an infinite plane sheet of charge is given by $E_S = \frac{|\sigma_s|}{2\epsilon_0}$.At the point $(0, 0, 2)$,the distance from the sheet (in the $x-y$ plane,i.e.,$z=0$) is $r_S = 2 	ext{ m}$.The electric field due to an infinitely long line charge is given by $E_{\ell} = \frac{|\lambda_e|}{2\pi\epsilon_0 r_{\ell}}$.The line charge is at $z=4 	ext{ m}$ and is parallel to the $y$-axis. The point is $(0, 0, 2)$. The perpendicular distance $r_{\ell}$ from the line charge to the point $(0, 0, 2)$ is $|4 - 2| = 2 	ext{ m}$.Given $|\sigma_s| = 2|\lambda_e|$.The ratio of the electric fields is:$\frac{E_S}{E_{\ell}} = \frac{|\sigma_s| / 2\epsilon_0}{|\lambda_e| / 2\pi\epsilon_0 r_{\ell}} = \frac{|\sigma_s|}{2\epsilon_0} 	imes \frac{2\pi\epsilon_0 r_{\ell}}{|\lambda_e|} = \frac{|\sigma_s| \pi r_{\ell}}{|\lambda_e|}$.Substituting the values $|\sigma_s| = 2|\lambda_e|$ and $r_{\ell} = 2 	ext{ m}$:$\frac{E_S}{E_{\ell}} = \frac{2|\lambda_e| 	imes \pi 	imes 2}{|\lambda_e|} = 4\pi$.We are given the ratio as $\pi \sqrt{n} : 1$,so $\pi \sqrt{n} = 4\pi$,which implies $\sqrt{n} = 4$.Therefore,$n = 16$.

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