An infinite plane sheet of charge having uniform surface charge density +\sigma_s C / m^2 is placed on x - y plane. Another infinitely long line charge having uniform linear charge density +\lambda_e C / m is placed at z=4 ~m plane and parallel to y -axis. If the magnitude values |\sigma_s|=2|\lambda_e| then at point (0,0,2) , the ratio of magnitudes of electric field values due to sheet charge to that of line charge is \pi \sqrtn: 1 . The value of n is \underline\hspace2cm .

Electrostatics PYQ — JEE Main 2024 · 4 Apr · Shift 1 · Q82 with Solution

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Correct answer: 16

An infinite plane sheet with surface charge density $\sigma_s$ lies on the $x$ - $y$ plane, i.e. at $z=0$ .

The magnitude of electric field due to an infinite sheet is

E_{\text{sheet}}=\frac{|\sigma_s|}{2\varepsilon_0}.

At the point $(0,0,2)$ , this magnitude remains the same.

The line charge is parallel to the $y$ -axis and lies in the plane $z=4$ . Since no $x$ -coordinate is mentioned, it passes through $x=0, z=4$ .

So the line is along $y$ through $(0,4)$ in the $x$ - $z$ cross-section.

The point is $(0,0,2)$ . Its perpendicular distance from the line is

r = |4-2|=2\,\text{m}.

The electric field magnitude due to an infinite line charge is

E_{\text{line}}=\frac{|\lambda_e|}{2\pi\varepsilon_0 r}.

Thus,

E_{\text{line}}=\frac{|\lambda_e|}{2\pi\varepsilon_0\cdot 2} =\frac{|\lambda_e|}{4\pi\varepsilon_0}.

Given

|\sigma_s|=2|\lambda_e|.

Now,

\frac{E_{\text{sheet}}}{E_{\text{line}}} =\frac{\dfrac{|\sigma_s|}{2\varepsilon_0}}{\dfrac{|\lambda_e|}{4\pi\varepsilon_0}} =\frac{|\sigma_s|}{2\varepsilon_0}\cdot \frac{4\pi\varepsilon_0}{|\lambda_e|} =2\pi\frac{|\sigma_s|}{|\lambda_e|}.

Using $|\sigma_s|=2|\lambda_e|$ ,

\frac{E_{\text{sheet}}}{E_{\text{line}}}=2\pi\cdot 2=4\pi.

This is given in the form

\pi\sqrt{n}:1.

So,

\pi\sqrt{n}=4\pi \Rightarrow \sqrt{n}=4 \Rightarrow n=16.

\boxed{16}

The derived answer matches the stored correct answer.