A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle $\mathrm{\theta <!--\; \theta \; -->}$ at P, as shown in the figure. When $\mathrm{\theta <!--\; \theta \; -->}$ = 30^∘ , then the electric flux through the curved surface of the cylinder is Φ. If $\mathrm{\theta <!--\; \theta \; -->}$ = 60^∘ , then the electric flux through the curved surface becomes Φ/√𝑛, where the value of n is______.

The correct answer is 3.

Step-by-step Explanation:

Let a point charge $q$ be placed at the central point $P$ inside the cylinder. According to Gauss's Law, the total electric flux passing through the entire closed surface of the cylinder is given by:
${\Phi }_{\text{total}}=\frac{q}{{\epsilon }_0}$
where ${\epsilon }_0$ is the permittivity of free space.

The total surface of the cylinder consists of three parts:
1. The top flat circular face.
2. The bottom flat circular face.
3. The curved lateral surface.

As shown in the figure, the circular edges of the cylinder subtend a half-angle $\theta$ at the central point $P$. The solid angle $\Omega$ subtended by a single circular flat face at $P$ is:
$\Omega =2\pi (1-\cos \theta )$

The electric flux passing through one of the flat circular faces is proportional to the fraction of the total solid angle ($\mathrm{4<mi>\pi </mi>}$ steradians) it subtends:
${\Phi }_{\text{flat}}={\Phi }_{\text{total}}\cdot \frac{\Omega }{4\pi }=\frac{q}{{\epsilon }_0}$

Since the cylinder is symmetric about the central point $P$, the total flux passing through both flat circular faces (top and bottom) combined is:
${\Phi }_{\text{flat, total}}=2{\Phi }_{\text{flat}}=\frac{q}{{\epsilon }_0}$

Subtracting the flux through the flat faces from the total flux gives the flux through the curved surface of the cylinder:
${\Phi }_{\text{curved}}={\Phi }_{\text{total}}-{\Phi }_{\text{flat, total}}$
${\Phi }_{\text{curved}}=\frac{q}{{\epsilon }_0}$

Let's use the given values for the angle:
Case 1: When $\theta ={30}^{\circ }$, the flux through the curved surface is $\Phi$:
$\Phi =\frac{q}{{\epsilon }_0}$

Case 2: When $\theta ={60}^{\circ }$, the flux through the curved surface becomes ${\Phi }^{\prime }$:
${\Phi }^{\prime }=\frac{q}{{\epsilon }_0}$

We are given that ${\Phi }^{\prime }=\frac{\Phi }{\sqrt{n}}$. Substituting the expressions for $\Phi$ and ${\Phi }^{\prime }$:
$\frac{q}{{\epsilon }_0}$

Dividing both sides by $\frac{q}{2{\epsilon }_0}$ simplifies the equation to:
$1=\frac{\sqrt{3}}{\sqrt{n}}$

Squaring both sides of the equation yields:
$n=3$