Electric Flux Through Rectangular Surface from Line Charge

To find the magnitude of the electric flux through the rectangular surface $ABCD$, we analyze the geometry of the system in the $x$-$z$ plane perpendicular to the line charge.

1. Geometry and Coordinate Setup

• Line Charge: Lies parallel to the $y$-axis at $x = 0$ and $z = \frac{\sqrt{3}}{2}a$. Thus, its cross-sectional position in the $x$-$z$ plane is $P\left(0, \frac{\sqrt{3}}{2}a\right)$.
• Rectangular Surface $ABCD$: Lying in the $x$-$y$ plane ($z = 0$) centered at the origin $(0,0,0)$, with length $L$ along the $y$-axis (from $y = -L/2$ to $y = L/2$) and width $a$ along the $x$-axis (from $x = -a/2$ to $x = a/2$).

2. Angle Subtended by the Surface

Consider the cross-section of the surface in the $x$-$z$ plane, which is represented by the line segment from $E_1\left(-\frac{a}{2}, 0\right)$ to $E_2\left(\frac{a}{2}, 0\right)$.

Let $\theta$ be the angle subtended by this segment at the location of the line charge $P\left(0, \frac{\sqrt{3}}{2}a\right)$. Using trigonometry for the right triangle formed by the origin, $P$, and $E_2$:

$\tan\left(\frac{\theta}{2}\right) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{a/2}{\frac{\sqrt{3}}{2}a} = \frac{1}{\sqrt{3}}$

This gives: $\frac{\theta}{2} = 30^\circ = \frac{\pi}{6} \implies \theta = 60^\circ = \frac{\pi}{3} \text{ radians}$

3. Calculating the Electric Flux using Gauss's Law

For an infinitely long line charge with charge density $\lambda$, the electric field is directed radially outward from the line.

If we construct a coaxial cylindrical Gaussian surface of radius $R$ and length $L$ centered around the line charge, the total electric flux passing through its curved surface of length $L$ is given by Gauss's Law:

$\Phi_{\text{total}} = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{\lambda L}{\varepsilon_0}$

Since the electric field has cylindrical symmetry, the flux through any longitudinal section of length $L$ that subtends an angle $\theta$ at the line charge is directly proportional to the fraction of the total $2\pi$ radians it subtends:

$\Phi = \left(\frac{\theta}{2\pi}\right) \Phi_{\text{total}}$

Substituting $\theta = \frac{\pi}{3}$ and $\Phi_{\text{total}} = \frac{\lambda L}{\varepsilon_0}$:

$\Phi = \left(\frac{\pi/3}{2\pi}\right) \frac{\lambda L}{\varepsilon_0} = \frac{\lambda L}{6\varepsilon_0}$

4. Comparison and Conclusion

Comparing the derived expression for the flux with the given expression:

$\frac{\lambda L}{6\varepsilon_0} = \frac{\lambda L}{n\varepsilon_0}$

We find: $n = 6$