A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle  at P, as shown in the figure. When  = 30∘ , then the electric flux through the curved surface of the cylinder is Φ. If  = 60∘ , then the electric flux through the curved surface becomes Φ/√𝑛, where the value of n is______.

 To solve this problem, we need to analyze the electric flux through the curved surface of a cylindrical region when a charge is placed at its central point 

$P$. The flux depends on the angle $\theta$ subtended by the edges of the cylinder at $P$.

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Step 1: Electric Flux Through a Closed Surface

The total electric flux ${\mathrm{\Phi }}_{\text{total}}$ through a closed surface enclosing a charge $q$ is given by Gauss's law:

$${\mathrm{\Phi }}_{\text{total}}=\frac{q}{{ϵ}_0}$$

where ${ϵ}_0$ is the permittivity of free space.

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Step 2: Flux Through the Curved Surface

The cylindrical region has two flat circular faces and a curved surface. The flux through the curved surface depends on the angle $\theta$. The total flux ${\mathrm{\Phi }}_{\text{total}}$ is distributed between the two flat faces and the curved surface.

The flux through each flat face is proportional to the solid angle subtended by the face at $P$. The solid angle $\mathrm{\Omega }$ subtended by a cone of half-angle $\theta$ is:

$$\mathrm{\Omega }=2\pi (1-\cos \theta )$$

The flux through one flat face is:

$${\mathrm{\Phi }}_{\text{flat}}=\frac{\mathrm{\Omega }}{4\pi }\cdot {\mathrm{\Phi }}_{\text{total}}=\frac{2\pi (1-\cos \theta )}{4\pi }\cdot \frac{q}{{ϵ}_0}=\frac{1-\cos \theta }{2}\cdot \frac{q}{{ϵ}_0}$$

Since there are two flat faces, the total flux through the flat faces is:

$${\mathrm{\Phi }}_{\text{flat, total}}=2\cdot \frac{1-\cos \theta }{2}\cdot \frac{q}{{ϵ}_0}=(1-\cos \theta )\cdot \frac{q}{{ϵ}_0}$$

The flux through the curved surface is the remaining flux:

$${\mathrm{\Phi }}_{\text{curved}}={\mathrm{\Phi }}_{\text{total}}-{\mathrm{\Phi }}_{\text{flat, total}}=\frac{q}{{ϵ}_0}-(1-\cos \theta )\cdot \frac{q}{{ϵ}_0}=\cos \theta \cdot \frac{q}{{ϵ}_0}$$

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Step 3: Flux for $\theta =30^{\circ }$ and $\theta =60^{\circ }$

For $\theta =30^{\circ }$:

$${\mathrm{\Phi }}_{\text{curved}}=\cos 30^{\circ }\cdot \frac{q}{{ϵ}_0}=\frac{\sqrt{3}}{2}\cdot \frac{q}{{ϵ}_0}$$

This is given as $\mathrm{\Phi }$:

$$\mathrm{\Phi }=\frac{\sqrt{3}}{2}\cdot \frac{q}{{ϵ}_0}$$

For $\theta =60^{\circ }$:

$${\mathrm{\Phi }}_{\text{curved}}=\cos 60^{\circ }\cdot \frac{q}{{ϵ}_0}=\frac{1}{2}\cdot \frac{q}{{ϵ}_0}$$

This is given as $\frac{\mathrm{\Phi }}{\sqrt{n}}$:

$$\frac{\mathrm{\Phi }}{\sqrt{n}}=\frac{1}{2}\cdot \frac{q}{{ϵ}_0}$$

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Step 4: Solve for $n$

From the expression for $\mathrm{\Phi }$:

$$\mathrm{\Phi }=\frac{\sqrt{3}}{2}\cdot \frac{q}{{ϵ}_0}$$

Substitute into the equation for $\theta =60^{\circ }$:

$$\frac{\frac{\sqrt{3}}{2}\cdot \frac{q}{{ϵ}_0}}{\sqrt{n}}=\frac{1}{2}\cdot \frac{q}{{ϵ}_0}$$

Cancel $\frac{q}{{ϵ}_0}$ and $\frac{1}{2}$:

$$\frac{\sqrt{3}}{\sqrt{n}}=1$$

Square both sides:

$$\frac{3}{n}=1⟹n=3$$

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Final Answer:

The value of $n$ is:

$$\boxed{{\displaystyle 3}}$$