Step-by-Step Solution

1. Understand the Setup and Electric Field

• An infinitely long line charge with linear charge density $\lambda$ is parallel to the y-axis, located at $x=0$ and $z_0 = \frac{\sqrt{3}}{2}a$.
• A rectangular surface ABCD lies in the x-y plane ($z=0$), centered at the origin. Its vertices are not explicitly given, but the problem description and figure (typically associated with this problem) imply its sides are parallel to the x and y axes. Let the length along the x-axis be $L$ (from $x=-L/2$ to $x=L/2$) and the width along the y-axis be $a$ (from $y=-a/2$ to $y=a/2$).
• The electric field $\vec{E}$ due to the infinite line charge at a perpendicular distance $r$ is given by $E = \frac{\lambda}{2\pi\varepsilon_0 r}$. The field lines are radial in the x-z plane.
• For a point $(x, y, 0)$ on the rectangle, the perpendicular distance to the line charge is $r = \sqrt{(x-0)^2 + (0-z_0)^2} = \sqrt{x^2 + z_0^2}$.
• The electric field vector at this point is $\vec{E}(x, y, 0) = \frac{\lambda}{2\pi\varepsilon_0(x^2 + z_0^2)}(x\hat{i} - z_0\hat{k})$.

2. Calculate the Electric Flux

• The electric flux $\Phi$ through the surface is given by the integral $\Phi = \int_S \vec{E} \cdot d\vec{A}$.
• The area element $d\vec{A}$ for the surface in the x-y plane is $d\vec{A} = dx dy \hat{k}$. We are interested in the magnitude of the flux, so the direction of $d\vec{A}$ (up or down) only affects the sign.
• The dot product is $\vec{E} \cdot d\vec{A} = \left( \frac{\lambda(x\hat{i} - z_0\hat{k})}{2\pi\varepsilon_0(x^2 + z_0^2)} \right) \cdot (dx dy \hat{k}) = -\frac{\lambda z_0}{2\pi\varepsilon_0(x^2 + z_0^2)} dx dy$.
• The magnitude of the flux is the integral over the surface area: $|\Phi| = \int_{y=-a/2}^{a/2} \int_{x=-L/2}^{L/2} \frac{\lambda z_0}{2\pi\varepsilon_0(x^2 + z_0^2)} dx dy$

3. Evaluate the Integral

• The integrand is independent of $y$, so we can integrate over $y$ first: $\int_{-a/2}^{a/2} dy = y \Big|_{-a/2}^{a/2} = \frac{a}{2} - (-\frac{a}{2}) = a$
• The flux expression becomes: $|\Phi| = a \cdot \frac{\lambda z_0}{2\pi\varepsilon_0} \int_{-L/2}^{L/2} \frac{dx}{x^2 + z_0^2}$
• The remaining integral is a standard form: $\int \frac{dx}{x^2 + c^2} = \frac{1}{c} \arctan(\frac{x}{c})$. $\int_{-L/2}^{L/2} \frac{dx}{x^2 + z_0^2} = \left[ \frac{1}{z_0} \arctan\left(\frac{x}{z_0}\right) \right]_{-L/2}^{L/2} = \frac{1}{z_0} \left[ \arctan\left(\frac{L}{2z_0}\right) - \arctan\left(\frac{-L}{2z_0}\right) \right]$
• Using the property $\arctan(-u) = -\arctan(u)$, the integral evaluates to: $\frac{1}{z_0} \left[ 2 \arctan\left(\frac{L}{2z_0}\right) \right]$
• Substituting this back into the flux expression: $|\Phi| = a \cdot \frac{\lambda z_0}{2\pi\varepsilon_0} \cdot \frac{2}{z_0} \arctan\left(\frac{L}{2z_0}\right) = \frac{a\lambda}{\pi\varepsilon_0} \arctan\left(\frac{L}{2z_0}\right)$

4. Substitute Known Values

• We are given $z_0 = \frac{\sqrt{3}}{2}a$. Substituting this into the expression for flux: $|\Phi| = \frac{a\lambda}{\pi\varepsilon_0} \arctan\left(\frac{L}{2(\frac{\sqrt{3}}{2}a)}\right) = \frac{a\lambda}{\pi\varepsilon_0} \arctan\left(\frac{L}{\sqrt{3}a}\right)$

5. Solve for n

• The problem states that the flux is $|\Phi| = \frac{\lambda L}{n\varepsilon_0}$.
• Equating the derived expression and the given expression gives: $\frac{a\lambda}{\pi\varepsilon_0} \arctan\left(\frac{L}{\sqrt{3}a}\right) = \frac{\lambda L}{n\varepsilon_0}$
• This equation cannot be solved for a constant integer $n$ for arbitrary $L$ and $a$. This suggests that there is a specific relationship between $L$ and $a$, or there is a typo in the question. In such physics problems designed for integer answers, it is a common convention that a "rectangle" might be a square unless specified otherwise, or the dimensions are chosen to give a simple result.
• Let's assume the rectangle is a square, so $L=a$. This is the most plausible assumption to obtain the integer answer provided.
• Substituting $L=a$ into our flux expression: $|\Phi| = \frac{a\lambda}{\pi\varepsilon_0} \arctan\left(\frac{a}{\sqrt{3}a}\right) = \frac{a\lambda}{\pi\varepsilon_0} \arctan\left(\frac{1}{\sqrt{3}}\right)$
• The value of $\arctan(\frac{1}{\sqrt{3}})$ is $\frac{\pi}{6}$ radians. $|\Phi| = \frac{a\lambda}{\pi\varepsilon_0} \left(\frac{\pi}{6}\right) = \frac{\lambda a}{6\varepsilon_0}$
• Now, we compare this with the given expression for flux, also setting $L=a$: $|\Phi| = \frac{\lambda a}{n\varepsilon_0}$
• By comparing the two expressions, we get: $\frac{\lambda a}{6\varepsilon_0} = \frac{\lambda a}{n\varepsilon_0} \implies n = 6$

Therefore, the value of n is 6.