Q.Point charges are arranged along a straight horizontal line: a charge $Q$, a charge $5Q$ located a distance $R$ to the left of $Q$, and a charge $-2Q$ located a distance $R/2$ from $Q$ (just below it). A spherical Gaussian surface of radius $R$ is centred on the charge $Q$. Thus $Q$ (at the centre) and $-2Q$ (at distance $R/2 < R$) lie inside the sphere, while $5Q$ lies at distance $R$, on/outside the closed spherical surface. Which of the following statements is/are correct?

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: …