Question

An infinite plane sheet of charge having uniform surface charge density \( +\sigma_s \, \text{C/m}^2 \) is placed on the \( x \)-\( y \) 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 the \( y \)-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 ______.

Answer 1

Correct Answer: 16

Electric Field Due to the Infinite Plane Sheet of Charge:

The electric field \(E_s\) due to an infinite plane sheet of charge with surface charge density \(\sigma\) is given by:

\[ E_s = \frac{\sigma}{2\epsilon_0} \]

Electric Field Due to the Line Charge:

The electric field \(E_\lambda\) at a perpendicular distance \(r\) from an infinitely long line charge with linear charge density \(\lambda_e\) is:

\[ E_\lambda = \frac{\lambda_e}{2\pi\epsilon_0 r} \]

where \(r = 4 - 2 = 2\, \text{m}\) (the distance from the line charge at \(z = 4\, \text{m}\) to the point \((0, 0, 2)\)).

Substitute Values and Simplify:

Given \(|\sigma| = 2|\lambda_e|\), we substitute this into the expressions for \(E_s\) and \(E_\lambda\):

\[ E_s = \frac{\sigma}{2\epsilon_0} = \frac{2\lambda_e}{2\epsilon_0} = \frac{\lambda_e}{\epsilon_0} \]
\[ E_\lambda = \frac{\lambda_e}{2\pi\epsilon_0 \times 2} = \frac{\lambda_e}{4\pi\epsilon_0} \]

Calculate the Ratio of the Electric Fields:

The ratio of the magnitudes of electric fields \(\frac{E_s}{E_\lambda}\) is:

\[ \frac{E_s}{E_\lambda} = \frac{\frac{\lambda_e}{\epsilon_0}}{\frac{\lambda_e}{4\pi\epsilon_0}} = 4\pi \]

Therefore,

\[ \frac{E_s}{E_\lambda} = \pi \sqrt{16} : 1 \]

Comparing with \(\pi \sqrt{n} : 1\), we find \(n = 16\).

Conclusion:

The value of \(n\) is 16.

Answer 2

Step 1: Given information.
An infinite plane sheet of charge with uniform surface charge density \( +\sigma_s \, \text{C/m}^2 \) is placed on the x–y plane.
An infinitely long line charge with uniform linear charge density \( +\lambda_e \, \text{C/m} \) is placed parallel to the y-axis at \( z = 4 \, \text{m} \).
We are to find the ratio of magnitudes of the electric field due to the sheet and the line charge at point \( (0, 0, 2) \), given that \( |\sigma_s| = 2|\lambda_e| \).

Step 2: Electric field due to the infinite sheet of charge.
The electric field due to an infinite sheet of charge having surface charge density \( \sigma_s \) is constant in magnitude and is given by:
\[ E_{\text{sheet}} = \frac{\sigma_s}{2\varepsilon_0} \] Direction: Perpendicular to the sheet (along the z-axis).

Step 3: Electric field due to the line charge.
For an infinite line charge, the electric field at a perpendicular distance \( r \) from the line is given by:
\[ E_{\text{line}} = \frac{\lambda_e}{2\pi \varepsilon_0 r} \] Here, the line is at \( z = 4 \, \text{m} \) and the point of observation is \( (0, 0, 2) \).
Thus, the perpendicular distance between them is: \[ r = 4 - 2 = 2 \, \text{m}. \] Therefore: \[ E_{\text{line}} = \frac{\lambda_e}{2\pi \varepsilon_0 (2)} = \frac{\lambda_e}{4\pi \varepsilon_0}. \]
Step 4: Ratio of magnitudes of electric fields.
We need: \[ \frac{E_{\text{sheet}}}{E_{\text{line}}} = \frac{\frac{\sigma_s}{2\varepsilon_0}}{\frac{\lambda_e}{4\pi \varepsilon_0}} = \frac{\sigma_s}{2\varepsilon_0} \times \frac{4\pi \varepsilon_0}{\lambda_e} = \frac{2\pi \sigma_s}{\lambda_e}. \] Given \( |\sigma_s| = 2|\lambda_e| \), we get: \[ \frac{E_{\text{sheet}}}{E_{\text{line}}} = 2\pi \times 2 = 4\pi. \] Hence, the ratio is: \[ E_{\text{sheet}} : E_{\text{line}} = 4\pi : 1 = \pi \sqrt{n} : 1. \] \[ \pi \sqrt{n} = 4\pi \Rightarrow \sqrt{n} = 4 \Rightarrow n = 16. \]
Final Answer:
\[ \boxed{n = 16} \]