An infinite plane sheet of charge having uniform surface charge density \(+\sigma_5 \mathrm{C} / \mathrm{m}^2\) is placed on \(\mathrm{x}-\mathrm{y}\) plane. Another infinitely long line charge having uniform linear charge density \(+\lambda_e \mathrm{C} / \mathrm{m}\) is placed at \(z=4 \mathrm{~m}\) plane and parallel to \(y\)-axis. If the magnitude values \(\left|\sigma_s\right|=2\left|\lambda_{\mathrm{e}}\right|\) then at point \((0,0,2)\), the ratio of magnitudes of electric field values due to sheet charge to that of line charge is \(\pi \sqrt{\mathrm{n}}: 1\). The value of \(n\) is _______.

Explanation:
Let us calculate the ratio of the electric fields step-by-step based on the Electric Charges and Fields chapter of the NCERT Class 12 Physics textbook:

Step 1: Electric field due to an infinite plane sheet of charge (\(E_s\))
The magnitude of the electric field produced by an infinite plane sheet of charge with uniform surface charge density \(\sigma_s\) at any given point is independent of its distance from the sheet and is expressed as:
\(E_s = \frac{\sigma_s}{2\varepsilon_0}\)

Step 2: Electric field due to an infinitely long straight line charge (\(E_l\))
The magnitude of the electric field produced by an infinitely long line charge with uniform linear charge density \(\lambda_e\) at a perpendicular distance \(r\) from the line is given by:
\(E_l = \frac{\lambda_e}{2\pi\varepsilon_0 r}\)

Step 3: Determine the perpendicular distance (\(r\)) from the line charge to the point
- The infinite sheet lies on the flat \(x-y\) plane where \(z = 0\).
- The line charge is parallel to the \(y\)-axis on the \(z = 4\text{ m}\) plane, meaning it effectively passes through the cross-section coordinates \(x = 0, z = 4\text{ m}\).
- The target test point is located at \((0,0,2)\), where \(z = 2\text{ m}\).
Thus, the absolute perpendicular distance \(r\) from the line charge to the test point is:
\(r = |4 - 2| = 2\text{ m}\)

Step 4: Calculate the ratio of the magnitudes of the electric fields
Taking the ratio of the sheet's field strength (\(E_s\)) to the line's field strength (\(E_l\)):
\(\frac{E_s}{E_l} = \frac{\left(\frac{\sigma_s}{2\varepsilon_0}\right)}{\left(\frac{\lambda_e}{2\pi\varepsilon_0 r}\right)} = \frac{\sigma_s \cdot \pi r}{\lambda_e}\)

Substituting the given relationship \(|\sigma_s| = 2|\lambda_e|\) and the calculated distance \(r = 2\text{ m}\) into the equation:
\(\frac{E_s}{E_l} = \frac{(2\lambda_e) \cdot \pi \cdot 2}{\lambda_e} = 4\pi = \frac{4\pi}{1}\)

Step 5: Compute the final value of \(n\)
We are given that this field magnitude ratio is equal to \(\pi\sqrt{n} : 1\):
\(\pi\sqrt{n} = 4\pi\)
\(\sqrt{n} = 4\)
Squaring both sides of the relation yields:
\(n = 16\)

Hence, Option A is the correct choice.