Question

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 $\frac{n q}{6 \epsilon_0}$ (in SI units) The value of $n$ is ______

Answer 1

Correct Answer: 3

In this problem, a charge \( q \) is enclosed by a closed surface consisting of an inverted cone and a hemisphere. The objective is to calculate the electric flux through the conical surface using Gauss's law. The electric flux through a surface is related to the charge enclosed within that surface, and we will use Gauss's law to find the value of the flux.

1. Understanding Gauss's Law:
Gauss's law states that the total electric flux \( \Phi_E \) through a closed surface is proportional to the charge \( q \) enclosed within the surface:

\[ \Phi_E = \frac{q_{\text{enc}}}{\epsilon_0} \] where: - \( \Phi_E \) is the electric flux, - \( q_{\text{enc}} \) is the charge enclosed within the surface, - \( \epsilon_0 \) is the permittivity of free space.

2. Setup of the Problem:
The configuration consists of: - An inverted cone of height \( h \) and base radius \( R \), - A hemisphere of radius \( R \), - A charge \( q \) placed at the center of the hemisphere. We are tasked with calculating the electric flux through the conical surface, which is the surface of the cone that excludes the hemispherical part.

3. Using Symmetry:
Because the charge \( q \) is at the center of the hemisphere, the flux through the surface will be symmetric. The flux through the hemisphere is uniform due to symmetry, and the flux through the conical surface can be found by subtracting the flux through the hemisphere from the total flux. Since the total flux through the closed surface is: \[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \] The flux through the hemisphere is \( \frac{q}{2\epsilon_0} \) because half of the total flux will pass through the hemisphere. Therefore, the flux through the conical surface is: \[ \Phi_{\text{cone}} = \Phi_{\text{total}} - \Phi_{\text{hemisphere}} = \frac{q}{\epsilon_0} - \frac{q}{2\epsilon_0} = \frac{q}{2\epsilon_0} \]

4. Final Result:
The electric flux through the conical surface is \( \frac{q}{2 \epsilon_0} \). Comparing this with the given form of the flux \( \frac{nq}{6 \epsilon_0} \), we can conclude that:

\[ n = 3 \]

Final Answer:
The value of \( n \) is 3.