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

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.