A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle

Q: Why can't we just use $\int \mathbf{E} \cdot d\mathbf{A}$ for the curved surface?

You can, but the electric field $\mathbf{E}$ is not constant in magnitude over the curved surface, nor is the angle between $\mathbf{E}$ and the area vector constant. The solid angle method bypasses this complex integration entirely.

Q: Where does the formula $2\pi(1 - \cos\theta)$ come from?

It comes from integrating the solid angle element $d\Omega = \sin\theta d\theta d\phi$ over the azimuthal angle $\phi$ from $0$ to $2\pi$ and the polar angle from $0$ to $\theta$.

Q: Why do we multiply by 2 for the flat faces?

The charge is exactly at the center, so the top and bottom flat faces subtend the exact same solid angle and thus have the exact same flux passing through them.

Q: What if the cylinder was infinitely long?

If the cylinder was infinitely long, $\theta$ would approach $0^\circ$. The flux through the flat faces would approach $0$, and the entire flux $\frac{q}{\varepsilon_0}$ would pass through the curved surface. Our formula $\frac{q}{\varepsilon_0}\cos(0^\circ) = \frac{q}{\varepsilon_0}$ perfectly confirms this!

Q: Does the radius of the cylinder matter?

Interestingly, no! As long as the half-angle $\theta$ is the same, the solid angle is the same, and the flux remains identical regardless of the actual physical dimensions of the cylinder.

Question

A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle $	heta$ at P , as shown in the figure. When $	heta = 30^\circ$ , then the electric flux through the curved surface of the cylinder is $\Phi$ . If $	heta = 60^\circ$ , then the electric flux through the curved surface becomes $\frac{\Phi}{\sqrt{n}}$ , where the value of n is _____________

Sigma Percentile · Accepted Answer

### The Setup: A Charge in a Cylinder

Imagine you are looking at a transparent cylinder, and right at its geometric center, a point charge $q$ is suspended. According to Gauss's Law, we know that this charge acts like a fountain of electric field lines, radiating a total electric flux of $\frac{q}{\varepsilon_0}$ in all directions.

This total flux must escape the cylinder by passing through its boundaries. The cylinder has three distinct boundaries: the top flat circular face, the bottom flat circular face, and the curved lateral surface. Our goal is to find the flux passing **only** through the curved surface.

### The Solid Angle Shortcut

You might initially think about setting up an integral $\int \mathbf{E} \cdot d\mathbf{A}$ over the curved surface. I know this differential equation looks terrifying, but let's take a breath. The electric field is not uniform over the curved surface, and the angle between the field and the area vector constantly changes. Integration would be a nightmare!

Instead, we use a brilliant shortcut: **Solid Angles**.

Think of a solid angle as a 3D cone of vision. The solid angle $\Omega$ subtended by a cone with a half-angle $	heta$ is given by the beautiful geometric formula:
$$ \Omega = 2\pi(1 - \cos	heta) $$

Because the charge is exactly at the center, the top and bottom flat faces of the cylinder act like the bases of two identical cones, each with a half-angle $	heta$.

### Deriving the Master Equation

The total solid angle of a full sphere is $4\pi$ steradians, which corresponds to the total flux $\frac{q}{\varepsilon_0}$. Therefore, the flux passing through *one* flat face (which subtends a solid angle $\Omega$) is simply a proportional fraction:
$$ \phi_{	ext{flat}} = \frac{q}{4\pi\varepsilon_0} 	imes 2\pi(1 - \cos	heta) = \frac{q}{2\varepsilon_0}(1 - \cos	heta) $$

Since there are two identical flat faces (top and bottom), the total flux escaping through the flat ends is double this amount:
$$ \phi_{	ext{flat\_total}} = 2 	imes \frac{q}{2\varepsilon_0}(1 - \cos	heta) = \frac{q}{\varepsilon_0}(1 - \cos	heta) $$

Now, the magic happens. By conservation of flux, whatever doesn't go out the ends must go out the sides! We subtract the flux of the flat faces from the total flux to find the flux through the curved surface:
$$ \phi_{	ext{curved}} = \phi_{	ext{total}} - \phi_{	ext{flat\_total}} $$
$$ \phi_{	ext{curved}} = \frac{q}{\varepsilon_0} - \frac{q}{\varepsilon_0}(1 - \cos	heta) $$

When we expand the bracket, the $\frac{q}{\varepsilon_0}$ terms cancel out perfectly, leaving us with an incredibly elegant master equation:
$$ \mathbf{\phi_{	ext{curved}} = \frac{q}{\varepsilon_0}\cos	heta} $$

### Plugging the Values

Now the physics is done, and it's just a matter of plugging in the numbers.

**Case 1:** The problem states that when $	heta = 30^\circ$, the flux is $\Phi$. Let's substitute this into our master equation:
$$ \Phi = \frac{q}{\varepsilon_0}\cos 30^\circ = \frac{q}{\varepsilon_0} \left(\frac{\sqrt{3}}{2}ight) $$

**Case 2:** When the angle widens to $	heta = 60^\circ$, let's call the new flux $\phi_2$. Substituting again:
$$ \phi_2 = \frac{q}{\varepsilon_0}\cos 60^\circ = \frac{q}{\varepsilon_0} \left(\frac{1}{2}ight) $$

### The Final Reveal

We need to find the relationship between the new flux $\phi_2$ and the original flux $\Phi$. The easiest way to compare them is to take their ratio:
$$ \frac{\phi_2}{\Phi} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} $$

Rearranging this gives us:
$$ \phi_2 = \frac{\Phi}{\sqrt{3}} $$

The problem tells us that the new flux is $\frac{\Phi}{\sqrt{n}}$. By directly comparing our result with the given expression, it is crystal clear that:
$$ \mathbf{n = 3} $$

This problem is a classic example of how recognizing symmetry and utilizing the concept of solid angles can turn a seemingly impossible calculus problem into a straightforward algebraic calculation. Always look for the elegant path!

Enter Numerical Value:

Visualized Solution

Total Flux $ϕ_{total}$

Solid Angle $Ω$

Flux through Flat Faces

Flux through Curved Surface

Case 1: $θ = 3 0^{\circ}$

Case 2: $θ = 6 0^{\circ}$

Calculating Ratio and $n$

The Power of Symmetry

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

The Setup

The Solid Angle Shortcut

Deriving the Master Equation

Plugging the Values

The Final Reveal

Similar Questions

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