An infinite plane sheet of charge having uniform surface charge density +σs C/m² is placed on x-y plane. Another infinitely long line charge having uniform linear charge density +λe C/m is placed at z = 4 m plane and parallel to y -axis. If the magnitude values |σs| = 2|λe| then at point (0, 0, 2), the ratio of magnitudes of electric field values due to sheet charge to that of line charge is π√(n): 1. The value of n is _________. Numerical answer type. Work it out, then open the answer.

Answer: 16. Sheet: Es = (σs)/(2ε₀) (independent of distance). Line at z = 4 m is 2 m from (0,0,2): El = (λe)/(2πε₀ (2)) = (λe)/(4πε₀). (Es)/(El) = (σs)/(2ε₀) · (4πε₀)/(λe) = 2π (σs)/(λe) = 4π = π√(16) So n = 16.

An infinite plane sheet of charge having uniform surface charge… | JEE Main 2024 (4 Apr, Shift 1)

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Electric Flux & Gauss's Law

An infinite plane sheet of charge having uniform surface charge density $+\sigma_s\text{ C/m}^{2}$ is placed on $x\text{-}y$ plane. Another infinitely long line charge having uniform linear charge density $+\lambda_e\text{ C/m}$ is placed at $z = 4\text{ m}$ 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 $\pi\sqrt{n} : 1$ . The value of $n$ is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 16

Sheet: $E_s = \frac{\sigma_s}{2\varepsilon_0}$ (independent of distance).
Line at $z = 4$ m is $2$ m from $(0,0,2)$ : $E_l = \frac{\lambda_e}{2\pi\varepsilon_0 (2)} = \frac{\lambda_e}{4\pi\varepsilon_0}$ .
$\frac{E_s}{E_l} = \frac{\sigma_s}{2\varepsilon_0} \cdot \frac{4\pi\varepsilon_0}{\lambda_e} = 2\pi \frac{\sigma_s}{\lambda_e} = 4\pi = \pi\sqrt{16}$
So $n = 16$ .

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