Q.A hollow cylinder carries a uniform positive charge spread over its curved outer surface (both circular ends are open). The curved wall is uniformly covered with positive charge all around and along its length. Sketch and describe the electric field lines for this uniformly charged hollow cylinder.

Gauss's Law: Why It Holds

Gauss's Law is one of the four Maxwell's equations and a cornerstone of electromagnetism. Let's build the understanding from the ground up — not just the formula, but the why.

1. The Core Idea: Flux as "Flow" of Field

Imagine an electric field $\vec{E}$ passing through a small patch of area $dA$. The electric flux through that patch is:

where $\theta$ is the angle between $\vec{E}$ and the outward normal to the surface.

Why this definition?

• If $\vec{E}$ is perpendicular to the surface ($\theta = 0$), maximum field "flows through".
• If $\vec{E}$ is parallel ($\theta = 90^\circ$), no flux — the field just slides along the surface.

Total flux through a closed surface $S$ is:

2. The Key Insight: Flux Depends Only on Enclosed Charge

Consider a single point charge $+q$ at the centre of a spherical surface of radius $r$.

• By Coulomb's law, at every point on the sphere: $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$, radially outward.
• The area vector $d\vec{A}$ is also radially outward.
• So $\vec{E} \cdot d\vec{A} = E \, dA$ everywhere.

The total flux:

Notice: The $r^2$ cancels! The flux is independent of the radius.

3. Why Shape Doesn't Matter

Now, what if the surface is not spherical but any closed shape enclosing the charge?

• Draw a small cone from the charge to the surface.
• The flux through a patch $dA$ at distance $r$ is $d\Phi = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \cos\theta \, dA$.
• But $\frac{\cos\theta \, dA}{r^2}$ is exactly the solid angle $d\Omega$ subtended by $dA$ at the charge.

So $d\Phi = \frac{q}{4\pi\varepsilon_0} d\Omega$.

Integrating over the entire closed surface: $\oint d\Omega = 4\pi$ (total solid angle around a point).

Hence:

Result: For any closed surface enclosing $q$, the flux is $\frac{q}{\varepsilon_0}$.

4. Multiple Charges: Superposition

If there are many charges $q_1, q_2, \dots, q_n$ inside the surface, the total electric field is the vector sum of individual fields:

Flux is linear: …