Step-by-Step Solution

1. Understanding the relationship between Electric Flux and Solid Angle The electric flux dΦ through a small area element $d\vec{A}$ due to a point charge q is given by $dΦ = \vec{E} \cdot d\vec{A}$. For a point charge, the electric field is $\vec{E} = \frac{q}{4\pi\epsilon_0 r^2} \hat{r}$. The total flux Φ through a surface is the integral of dΦ over that surface. This integral simplifies when using the concept of solid angle. The electric flux Φ through a surface subtending a solid angle Ω at the location of the point charge q is given by: $\Phi = \frac{q}{4\pi\epsilon_0} \Omega$

2. Calculating the Solid Angle The point charge q is at the center P. The entire space around P corresponds to a total solid angle of $4\pi$ steradians. This total solid angle can be conceptually divided into the solid angle subtended by the two circular bases ($Ω_{bases}$) and the solid angle subtended by the curved surface ($Ω_{curved}$). $Ω_{total} = Ω_{bases} + Ω_{curved} = 4\pi$

3. Solid Angle of the Circular Bases The solid angle subtended by a circular cap (forming a cone) with a semi-vertical angle θ at its apex is given by the formula $Ω_{cap} = 2\pi(1 - \cos\theta)$. In our problem, there are two such identical circular bases, one on each side of the point charge P, both subtending the same half-angle θ. Therefore, the total solid angle subtended by both bases is: $Ω_{bases} = 2 \times Ω_{cap} = 2 \times [2\pi(1 - \cos\theta)] = 4\pi(1 - \cos\theta)$

4. Solid Angle of the Curved Surface We can find the solid angle subtended by the curved surface by subtracting the solid angle of the bases from the total solid angle: $Ω_{curved} = Ω_{total} - Ω_{bases} = 4\pi - 4\pi(1 - \cos\theta) = 4\pi - 4\pi + 4\pi\cos\theta$ $Ω_{curved} = 4\pi\cos\theta$

5. Flux through the Curved Surface Now we can calculate the electric flux through the curved surface, $Φ_{curved}$, using the solid angle $Ω_{curved}$: $Φ_{curved} = \frac{q}{4\pi\epsilon_0} Ω_{curved} = \frac{q}{4\pi\epsilon_0} (4\pi\cos\theta)$ $Φ_{curved} = \frac{q}{\epsilon_0} \cos\theta$

6. Applying the Given Conditions We are given two cases:

   • Case 1: When θ = 30°, the flux is Φ. $\Phi = \frac{q}{\epsilon_0} \cos(30°) = \frac{q}{\epsilon_0} \frac{\sqrt{3}}{2}$
   • Case 2: When θ = 60°, the flux is Φ'. Let's calculate Φ'. $Φ' = \frac{q}{\epsilon_0} \cos(60°) = \frac{q}{\epsilon_0} \frac{1}{2}$

7. Finding the value of n We are told that the new flux in Case 2 is $Φ' = Φ / \sqrt{n}$. Let's substitute the expressions we found for Φ and Φ': $\frac{q}{2\epsilon_0} = \frac{1}{\sqrt{n}} \left( \frac{q}{\epsilon_0} \frac{\sqrt{3}}{2} \right)$ We can cancel the common terms $\frac{q}{2\epsilon_0}$ from both sides: $1 = \frac{\sqrt{3}}{\sqrt{n}}$ $\sqrt{n} = \sqrt{3}$ Squaring both sides, we get: $n = 3$

Thus, the value of n is 3.