Four different closed surfaces, of different shapes and different sizes, are considered. Each one of the four surfaces encloses one and the same single point charge $+q$ (and no other charge). Consider the electric flux through each surface. (a) The flux through the fourth (most irregular) surface is the largest. (b) The flux through the third surface is the least. (c) The flux through the second surface equals that through the third but is smaller than through the fourth. (d) The flux is the same for all four surfaces.

> [!TLDR] > Gauss's law says the net electric flux through a closed surface equals the enclosed charge divided by $\varepsilon_0$ and is completely independent of the surface's shape or size. Since all four surfaces enclose the same single charge $+q$, they all have the same flux. ### Concept Gauss's law: $$\Phi = \oint_S \vec{E}\cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0}.$$ Only the **enclosed** charge matters; the geometry of the surface does not. ### Steps 1. Each of the four surfaces encloses exactly one charge, $+q$. 2. Therefore for each, $q_{enc} = q$. 3. Hence $\Phi = q/\varepsilon_0$ for all four — a common value. ### Why the others fail (a), (b), (c) all assume the flux depends on the size/shape of the surface. It does not — the extra field lines that pierce a larger or more distorted surface enter and leave in equal numbers, leaving the net count fixed by $q_{enc}$ alone. > [!ANSWER] > Option (d): the electric flux is the same for all the figures.

Four different closed surfaces, of different shapes and different sizes, are considered. Each one of the four

NCERT Exemplar · Q3

+q

(a) The flux through the fourth (most irregular) surface is the largest.

(b) The flux through the third surface is the least.

(d) The flux is the same for all four surfaces.

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Concept understanding — Gauss Law

Gauss's Law is a shortcut. Instead of adding up the Coulomb contribution of every charge — a nightmare of vectors and integrals — it lets you read the field straight off the symmetry of a problem. The whole idea rests on one quantity: electric flux.

Flux — field-lines counted through a surface. For a uniform field E crossing a flat area A, the flux is Φ = E·A = EA cosθ, where θ is the angle between the field and the normal to the surface. Picture the field as a bundle of lines; flux is how many pierce the surface. When E is edge-on (θ = 90°) nothing passes through and Φ = 0; when E is ⊥ to the surface (θ = 0) the count is maximal. For a closed surface, Φ is the net number of lines leaving it — lines that enter and exit cancel.

The law itself. Gauss's law states that the net flux out of any closed surface depends only on the charge trapped inside:

Φ = q_enclosed / ε₀.

Two consequences do most of the work. First, only enclosed charge counts — a charge outside sends as many lines in as out, so its net contribution is exactly zero. Second, the surface's shape is irrelevant; move the charge around inside or deform the surface, and Φ never changes.

Why symmetry makes it powerful. By itself Φ = q/ε₀ has E buried in an integral. It becomes a tool only when you pick a Gaussian surface matched to the symmetry — one where E is constant and everywhere either ⊥ to the surface (so Φ = EA) or ∥ to it (contributing nothing). Then E slides out and you solve in one line. This works for exactly three geometries:

1 — Infinite line charge (linear density λ). Use a coaxial cylinder: E = λ / 2πε₀r, falling off as ∝ 1/r.

2 — Infinite sheet (surface density σ). Use a pillbox pierced through the sheet: E = σ / 2ε₀ — uniform and completely independent of distance. The field near a large charged plane simply doesn't weaken as you step back.

3 — Spherical shell / sphere. For a thin shell of charge Q, a Gaussian sphere inside encloses nothing, so E = 0 everywhere within; outside, the charge acts as if concentrated at the centre, E = kQ/r² — indistinguishable from a point charge. For a solid uniformly charged sphere, an interior surface encloses only the charge within radius r, giving E ∝ r (rising linearly from zero at the centre) up to the surface, then 1/r² beyond.

Field just outside a conductor. A charged conductor holds all its charge on the surface with E = 0 inside, so a straddling pillbox gives E = σ / ε₀ just outside — twice the sheet result, because all the flux escapes on the one outer face.

How it's examined. JEE questions test whether you can spot the symmetry, pick the right surface, and recall which result scales as 1/r, which is flat, and which is 1/r². The physics is always the one line Φ = q_enclosed / ε₀, and the skill is knowing that only the enclosed charge — never the far-off one — ever matters.

"Gauss law class 12 physics derivation" and "electric field due to infinite sheet using Gauss law" are heavily searched terms, since this is one of the core results of the Electrostatics chapter in the NCERT/CBSE Class 12 Physics curriculum. Gauss's law applications for spheres, sheets, and line charges are near-guaranteed questions in JEE Main, NEET, and state CETs.

Why this formula?

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:

d\Phi_E = \vec{E} \cdot d\vec{A} = E \, dA \cos\theta

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:

\Phi_E = \oint_S \vec{E} \cdot d\vec{A}

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:

\Phi_E = \oint E \, dA = E \oint dA = \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \right) (4\pi r^2) = \frac{q}{\varepsilon_0}

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:

\Phi_E = \frac{q}{4\pi\varepsilon_0} \cdot 4\pi = \frac{q}{\varepsilon_0}

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:

\vec{E} = \vec{E}_1 + \vec{E}_2 + \dots + \vec{E}_n

Flux is linear:

\oint \vec{E} \cdot d\vec{A} = \oint \vec{E}_1 \cdot d\vec{A} + \oint \vec{E}_2 \cdot d\vec{A} + \dots = \frac{q_1}{\varepsilon_0} + \frac{q_2}{\varepsilon_0} + \dots = \frac{Q_{\text{enc}}}{\varepsilon_0}

Charges outside the surface contribute zero net flux — their field lines enter and exit the surface, cancelling out.

5. The Final Law

\boxed{\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}}

Why it's profound:

It relates a global property (flux through a surface) to a local source (charge inside).
It's true for any closed surface, not just symmetric ones.
It's a direct consequence of Coulomb's inverse-square law — the $1/r^2$ dependence is essential for the cancellation.

6. Quick Exam Tip

Situation	What to remember
Point charge	Flux = $q/\varepsilon_0$ through any enclosing surface
Dipole inside	Net flux = 0 (equal + and -)
Charge outside	Flux contribution = 0
Symmetric surfaces	Use Gauss's law to find $\vec{E}$ easily

Key takeaway: Gauss's law holds because the electric field from a point charge obeys the inverse-square law, making the flux through any closed surface independent of the surface's shape — it depends only on the total charge enclosed.

Gauss's law says the net electric flux through a closed surface equals the enclosed charge divided by $\varepsilon_0$ and is completely independent of the surface's shape or size. Since all four surfaces enclose the same single charge $+q$ , they all have the same flux.

Concept

Gauss's law:

\Phi = \oint_S \vec{E}\cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0}.

Only the enclosed charge matters; the geometry of the surface does not.

Steps

Each of the four surfaces encloses exactly one charge, $+q$ .
Therefore for each, $q_{enc} = q$ .
Hence $\Phi = q/\varepsilon_0$ for all four — a common value.

Why the others fail

,
,
all assume the flux depends on the size/shape of the surface. It does not — the extra field lines that pierce a larger or more distorted surface enter and leave in equal numbers, leaving the net count fixed by $q_{enc}$ alone.
✓Final answer
$Option (d): the electric flux is the same for all the figures.$

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04194 marksMCQ
$cm$
›Reveal solutionSolution

Find the surface charge, then use $E=kQ/r^2$ at the sphere's surface.

Total charge:
$Q=ne=(5\times10^6)(1.6\times10^{-19})=8\times10^{-13}\ \text{C}.$
Field at the surface of a conducting sphere of radius $r=0.1$ m:
$E=\frac{kQ}{r^2}=\frac{(9\times10^9)(8\times10^{-13})}{(0.1)^2}=\frac{7.2\times10^{-3}}{0.01}=0.72\ \text{N C}^{-1}.$
✓Final answer
$The correct option is (E) .$
KEAM 2026Set eng-2026-04214 marksMCQ
$E$
›Reveal solutionSolution

Invert $E=\dfrac{\lambda}{2\pi\epsilon_0 d}$ to get $\lambda=2\pi\epsilon_0 E d$ .

For an infinite line charge, Gauss's law gives $E=\dfrac{\lambda}{2\pi\epsilon_0 d}$ at perpendicular distance $d$ .

Rearranging: $\lambda=2\pi\epsilon_0 E d$ .

✓Final answer
$The correct option is (D) .$
KEAM 2026Set eng-2026-04224 marksMCQ
$8.85\ \mu\text{C cm}^{-2}$
›Reveal solutionSolution

By Gauss's law the flux out of the enclosing sphere is the total charge divided by $\varepsilon_0$ .

The cube (side 1 cm) has 6 faces of $1\,\text{cm}^2$ each, total area $6\,\text{cm}^2$ .

Total charge $Q=\sigma A=8.85\,\mu\text{C cm}^{-2}\times6\,\text{cm}^2=53.1\,\mu\text{C}=53.1\times10^{-6}\,\text{C}$ .

Flux $=\dfrac{Q}{\varepsilon_0}=\dfrac{53.1\times10^{-6}}{8.85\times10^{-12}}=6\times10^{6}\,\text{Nm}^2\text{C}^{-1}$ .

✓Final answer
$The correct option is (E) .$
KEAM 2026Set pha-2026-0420F4 marksMCQ
$Q. Electric-flux through a closed surface depends on the (A) shape of the surface (B) area of the surface (C) volume of the surface (D) electric field outside the surface (E) charge enclosed by the surface$
›Reveal solutionSolution

Gauss's law: $\Phi = q_{enc}/\varepsilon_0$ — only the enclosed charge matters.

Gauss's law states
$\Phi_E = \oint \vec{E}\cdot d\vec{A} = \frac{q_{enclosed}}{\varepsilon_0}.$
The net electric flux through a closed surface depends only on the total charge enclosed by it — not on the shape, area or volume of the surface, nor on charges lying outside it. Hence (E).

✓Final answer
$The correct option is (E) .$
KEAM 2025Set eng-2025-04234 marksMCQ
$6\times10^4\ NM^2C^{-1}$
›Reveal solutionSolution
[!TLDR]

Using Gauss's law with net flux $=\phi_{out}-\phi_{in}=-3\times10^4$ gives an enclosed charge of $-3\times10^4\,\varepsilon_0$ C, which matches option (E).

Concept

Gauss's law states that the net electric flux through a closed surface equals the enclosed charge divided by the permittivity of free space: $\phi_{net} = \dfrac{q_{enc}}{\varepsilon_0}$ . Outward flux is taken positive and inward flux negative — the standard NCERT/CBSE electrostatics convention the KEAM syllabus is aligned with.

Solution

The net flux through the surface is the outward flux minus the inward flux:
$\phi_{net} = \phi_{out} - \phi_{in} = 3\times10^4 - 6\times10^4 = -3\times10^4\ \mathrm{N\,m^2\,C^{-1}}.$
By Gauss's law the enclosed charge is
$q_{enc} = \varepsilon_0\,\phi_{net} = -3\times10^4\,\varepsilon_0\ \mathrm{C}.$
The negative sign correctly reflects that more flux enters than leaves, so the enclosed charge is negative. This value, $-3\times10^4\,\varepsilon_0$ , is not any of the printed options (A)–(D) ( $-6\times10^4\varepsilon_0,\ 6\times10^4\varepsilon_0,\ 3\times10^4\varepsilon_0,\ 9\times10^4\varepsilon_0$ ), so by elimination it corresponds to option (E).

[!ANSWER]

(E) $q_{enc} = -3\times10^4\,\varepsilon_0$ C
KEAM 2025Set eng-2025-04274 marksMCQ
$R^{2}$
›Reveal solutionSolution

A Gaussian surface inside a uniformly charged spherical shell encloses no charge, so $E=0$ everywhere inside.

For a uniformly charged spherical shell, apply Gauss's law to a concentric spherical surface of radius $r<R$ . It encloses zero net charge, so
$\oint \vec E\cdot d\vec A=\frac{q_{enc}}{\varepsilon_0}=0\;\Rightarrow\;E=0.$
The electric field is zero at every point inside the shell.

✓Final answer
$The correct option is (D) .$
KEAM 2024Set eng-2024-06094 marksMCQ
$\lambda$
›Reveal solutionSolution

By Gauss's law the field of an infinite line charge is $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$ , i.e. proportional to $\lambda/r$ .

Using a cylindrical Gaussian surface of radius $r$ :
$E=\frac{\lambda}{2\pi\varepsilon_0 r}\;\propto\;\frac{\lambda}{r}.$
✓Final answer
$The correct option is (B) .$
KEAM 2023Set eng-2023-P1-A14 marksMCQ
$+q$
›Reveal solutionSolution
The net electric flux through the sphere is zero.

Concept and Intuition

By Gauss's law the flux depends only on the enclosed net charge. A dipole encloses $+q$ and $-q$ , whose sum is zero.

Step-by-Step Solution

Enclosed charge $=+q+(-q)=0$ .

Gauss's law: $\Phi=\dfrac{Q_{enc}}{\varepsilon_0}=\dfrac{0}{\varepsilon_0}$ .

$\Phi=0$ .

Common Mistakes

Thinking the strong dipole field near the surface implies non-zero flux.

Adding the magnitudes of the charges instead of their signed sum.

✓Final answer
$The correct option is (D) — zero.$

ANSWER: D

←Previous questionQ2 — A positive point charge +q is brought near an isolated (uncharged) con…→Next questionQ4 — Five point charges q₁, q₂, q₃, q₄ and q₅ are fixed at their positions …

Q.Four different closed surfaces, of different shapes and different sizes, are considered. Each one of the four surfaces encloses one and the same single point charge $+q$ (and no other charge). Consider the electric flux through each surface.

Concept understanding — Gauss Law

Why this formula?

Gauss's Law: Why It Holds

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

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

3. Why Shape Doesn't Matter

4. Multiple Charges: Superposition

5. The Final Law

6. Quick Exam Tip

Concept

Steps

Why the others fail

Method: Using Gauss's Law to Compare Flux Through Different Surfaces

Steps

Concept

Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes

Q.Four different closed surfaces, of different shapes and different sizes, are considered. Each one of the four surfaces encloses one and the same single point charge +q+q+q (and no other charge). Consider the electric flux through each surface.

Concept

Steps

Why the others fail

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Q.Four different closed surfaces, of different shapes and different sizes, are considered. Each one of the four surfaces encloses one and the same single point charge $+q$ (and no other charge). Consider the electric flux through each surface.

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