A charge q is surrounded by a closed surface consisting of an inverted cone of height h and base rad

Q: Why does exactly half the flux go through the hemisphere?

Because the charge is located exactly on the flat circular boundary. The electric field lines radiate uniformly in all directions (a full $4\pi$ steradians). The flat boundary divides this 3D space into two perfectly equal halves ($2\pi$ steradians each). Thus, exactly half the lines go up, and half go down.

Q: Does the height $h$ of the cone affect the flux passing through it?

Not at all! As long as the cone completely encloses the lower half of the space and the charge remains at the center of its base, the flux through it will always be $q/(2\epsilon_0)$. Gauss's Law depends only on the enclosed charge, not the shape or size of the bounding surface.

Q: What if the charge was placed at the apex of the cone instead?

If the charge were at the apex, the solid angle subtended by the cone would be different. You would need to calculate the specific solid angle using the formula $\Omega = 2\pi(1 - \cos\theta)$, where $\theta$ is the semi-vertical angle of the cone.

Q: Why didn't we need to integrate the electric field over the surface?

While you *could* integrate $\int \mathbf{E} \cdot d\mathbf{A}$, it would be mathematically tedious. The beauty of Gauss's Law combined with symmetry is that it allows us to bypass complex surface integrals entirely.

Q: Is the electric field uniform across the conical surface?

No, the electric field is not uniform. Points on the cone closer to the charge will experience a stronger electric field than points further away (like the apex). However, the *total flux* (the integral of the field over the area) remains constant.

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 $\frac{nq}{6\epsilon_0}$ (in SI units). The value of $n$ is ________.

Sigma Percentile · Accepted Answer

Welcome, future engineers and physicists, to a beautiful exploration of electrostatics! Today, we are going to tackle a problem from JEE Advanced 2022 that perfectly encapsulates the elegance of Gauss's Law and the profound power of symmetry. I know that dealing with 3D geometries like cones and hemispheres can sometimes feel intimidating, but let's take a breath. Once you see the hidden symmetry in this setup, the math will unravel beautifully.

### The Setup: Visualizing the Geometry

Imagine you are looking at a unique closed container. The top half is a perfect hemisphere of radius $R$, and the bottom half is an inverted cone of height $h$ and base radius $R$. These two shapes are glued together at their flat circular bases, creating a single, continuous closed surface.

Right at the exact center of this circular boundary—the origin of our setup—lies a point charge $q$. Think of this charge as a glowing lightbulb, radiating electric field lines uniformly in all directions across the entire 3D space.

### The Master Equation: Gauss's Law

Whenever we are asked to find the electric flux through a closed surface, our immediate instinct should be to invoke Gauss's Law. Gauss's Law is like an accounting system for electric field lines. It states that the total electric flux $\phi_{	ext{total}}$ passing through *any* closed surface is simply the total enclosed charge divided by the permittivity of free space, $\epsilon_0$.

$$
\phi_{	ext{total}} = \frac{q_{	ext{enclosed}}}{\epsilon_0}
$$

In our specific problem, the closed surface is composed of two distinct parts: the upper hemispherical surface and the lower conical surface. Therefore, the total flux is the sum of the flux through these two regions:

$$
\phi_{	ext{hemisphere}} + \phi_{	ext{cone}} = \frac{q}{\epsilon_0}
$$

### The Power of Symmetry

Here is where the magic happens. Notice the strategic placement of the charge $q$. It is not floating randomly inside the volume; it is pinned exactly on the flat circular plane that separates the hemisphere from the cone.

Because the charge radiates field lines uniformly in all directions (a full $4\pi$ steradians of solid angle), and because it sits exactly on the dividing plane, that plane slices the radiating field perfectly in half. Exactly half of the electric field lines will shoot upwards into the hemisphere, and the exact other half will shoot downwards into the cone.

### Calculating the Flux

Thanks to this pure symmetry, we don't need to perform any complex surface integrals. We can confidently state that the flux through the upper hemisphere is exactly half of the total flux:

$$
\phi_{	ext{hemisphere}} = \frac{1}{2} \left( \frac{q}{\epsilon_0} ight) = \frac{q}{2\epsilon_0}
$$

Consequently, the flux passing through the lower conical surface must account for the remaining half:

$$
\phi_{	ext{cone}} = \frac{1}{2} \left( \frac{q}{\epsilon_0} ight) = \frac{q}{2\epsilon_0}
$$

### Final Calculation

We are almost at the finish line! The problem states that the electric flux through the conical surface is given by the expression $\frac{nq}{6\epsilon_0}$. All we have to do now is equate our derived flux to this given expression:

$$
\frac{nq}{6\epsilon_0} = \frac{q}{2\epsilon_0}
$$

Notice how beautifully the physics constants cancel out. The charge $q$ and the permittivity $\epsilon_0$ vanish from both sides, leaving us with a simple algebraic equation:

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

Multiplying both sides by 6, we get:

$$
**n = 3**
$$

And there we have it! By trusting in Gauss's Law and leveraging the geometric symmetry of the setup, we bypassed tedious calculus and arrived at the elegant integer answer of 3. Always remember to look for symmetry in physics problems—it is often the key to unlocking the simplest and most beautiful solutions.

Enter Numerical Value:

Visualized Solution

System Overview

Gauss's Law

Symmetry Analysis

Flux through Hemisphere

Flux through Cone

Equating Flux

Solving for $n$

Final Answer

The Way Forward

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

The Setup

The Master Equation

The Power of Symmetry

Calculating the Flux

Final Calculation

Similar Questions

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