A charge $q$ is surrounded by a closed surface consisting of an inverted cone of height $h$ and base radius $R$, and a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{nq}{6{\epsilon }_0}$ (in SI units). The value of $n$ is ______.

The correct answer is 3.

Step 1: Understand the Geometry and Enclosed Charge

As shown in the given figure, the closed Gaussian surface consists of two parts:

1. A hemisphere of radius $R$ at the top.
2. An inverted cone of height $h$ and base radius $R$ at the bottom.

The point charge $q$ is placed precisely at the center of the common circular base of radius $R$, which connects the hemisphere and the inverted cone.

Step 2: Total Electric Flux Using Gauss's Law

According to Gauss's law, the total electric flux ${\Phi }_{\text{total}}$ passing through any closed surface enclosing a net charge $q_{\text{enclosed}}$ is given by:

${\Phi }_{\text{total}}=\frac{q_{\text{enclosed}}}{{\epsilon }_0}=\frac{q}{{\epsilon }_0}$

Step 3: Distribution of Flux

Since the point charge $q$ is located at the center of the circular interface separating the upper hemisphere and the lower conical surface, electric field lines emanate radially outwards symmetrically in all directions (filling a total solid angle of $4\pi$ steradians).

Half of the space (a solid angle of $2\pi$ steradians) lies above the circular base inside the hemisphere, and the other half of the space (a solid angle of $2\pi$ steradians) lies below the circular base inside the inverted cone.

Therefore, by symmetry, exactly half of the total electric flux passes through the hemispherical surface and the remaining half passes through the conical surface:

${\Phi }_{\text{cone}}=\frac{1}{2}{\Phi }_{\text{total}}=\frac{q}{2{\epsilon }_0}$

Step 4: Finding the Value of $n$

We are given that the electric flux through the conical surface is expressed as:

${\Phi }_{\text{cone}}=\frac{nq}{6{\epsilon }_0}$

Equating our derived flux value with the given expression:

$\frac{q}{2{\epsilon }_0}=\frac{nq}{6{\epsilon }_0}$

Canceling $\frac{q}{{\epsilon }_0}$ from both sides:

$\frac{1}{2}=\frac{n}{6}$

Solving for $n$:

$n=\frac{6}{2}=3$

Thus, the value of $n$ is 3.