Φ = q/ε₀. Why it works for any closed surface.

Gauss's Law — Electric Flux Through a Sphere

Question

A point charge $q$ is placed at the centre of a spherical surface of radius $R$ . Find the total electric flux through the sphere. Does the answer change if the charge is not at the centre?

Solution — Step by Step

Electric flux $\Phi$ is defined as the surface integral of the electric field over a closed surface:

\Phi = \oint \vec{E} \cdot d\vec{A}

We need this for a sphere with charge $q$ at its centre.

Because $q$ sits exactly at the centre, the electric field at every point on the sphere points radially outward and has the same magnitude. This means $\vec{E}$ and $d\vec{A}$ are always parallel, so $\vec{E} \cdot d\vec{A} = E \, dA$ .

Without this symmetry argument, the integral would be a nightmare. Symmetry is what makes Gauss’s Law powerful.

At distance $R$ from a point charge:

E = \frac{q}{4\pi\varepsilon_0 R^2}

This is constant over the entire spherical surface.

Since $E$ is constant, it comes out of the integral:

\Phi = E \oint dA = \frac{q}{4\pi\varepsilon_0 R^2} \times 4\pi R^2

The $R^2$ cancels. We get:

\boxed{\Phi = \frac{q}{\varepsilon_0}}

The flux is still $q/\varepsilon_0$ . Gauss’s Law says the flux depends only on the enclosed charge, not its position inside the surface. Moving the charge off-centre makes $E$ non-uniform over the surface, but the total flux stays the same.

Why This Works

The $R^2$ in the area formula and the $R^2$ in the denominator of Coulomb’s Law cancel each other — this is not a coincidence. The inverse-square nature of the electric force is precisely why Gauss’s Law holds. If Coulomb’s Law had any other power law, this cancellation wouldn’t happen.

Think of it this way: the charge shoots out a fixed number of “field lines.” No matter how large or small we make the sphere, all those lines must pass through it. A bigger sphere has weaker $E$ but more area — they balance exactly.

This is also why the shape of the surface doesn’t matter. Any closed surface enclosing charge $q$ gives $\Phi = q/\varepsilon_0$ . The sphere is just the easiest shape to calculate with.

Alternative Method

We can state Gauss’s Law directly without going through Coulomb’s Law derivation:

\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}

Here $q_{\text{enc}} = q$ (the only charge inside), so $\Phi = q/\varepsilon_0$ immediately. This is the approach to use in JEE when the question just says “find flux through a closed surface enclosing charge $q$ ” — write the law, substitute, done in two lines.

In JEE Main, several questions give an irregular closed surface (a cube, a cone, a random blob) with a charge inside and ask for flux. The answer is always $q/\varepsilon_0$ regardless of shape. Never try to integrate — just apply Gauss’s Law directly.

Common Mistake

Students often write $\Phi = E \times 4\pi R^2$ and then substitute $E = kq/R^2$ , getting the right answer — but then panic when asked the same question with a cube of side $2R$ instead of a sphere. They try to find $E$ on the cube’s surface (which is non-uniform and has no clean formula). The correct move: Gauss’s Law gives $\Phi = q/\varepsilon_0$ for any closed surface. The shape and size are irrelevant. This exact trap appeared in JEE Main 2024 Shift 1, where a hemispherical surface with charge at the flat face’s centre confused students who forgot the law applies to closed surfaces only — a hemisphere is open.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next