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.

Showing the 12 most recent of 16 on this concept.

GUJCET 2026Set x1 markMCQ
$q$
›Reveal solutionSolution
[!TLDR]

The numerator is the derivative of the denominator, so the integral is $\log|e^{x}+e^{-x}|+C$ — option (C).

Concept

Whenever an integrand has the form $\dfrac{f'(x)}{f(x)}$ , the integral is $\log|f(x)|+C$ . Here take $f(x)=e^{x}+e^{-x}$ , whose derivative is exactly $e^{x}-e^{-x}$ .

Solution

Let $u=e^{x}+e^{-x}$ . Then $du=(e^{x}-e^{-x})\,dx$ , and
$\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx=\int \frac{du}{u}=\log|u|+C=\log|e^{x}+e^{-x}|+C.$
The only option that is the correct antiderivative of this standard integrand is (C).

[!ANSWER]

(C) $\log|e^x + e^{-x}|$

Note

This solution was worked out by our team and independently cross-checked by a second solve. The official answer key on record for this question could not be confirmed, so please cross-verify with the official paper where possible.
GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. A point charge of 2.0 microC is at the centre of a cubic Gaussian surface 9.0 cm on edge. The net electric flux through the surface is ___ Nm^2/C.$
(a) 2.2 x 10^-6
(b) 2.2 x 10^5
(c) 2.2 x 10^6
(d) 2.2 x 10^-5
›Reveal solutionSolution

Gauss's law states the net electric flux through any closed surface is q_enclosed / epsilon_0, regardless of the surface's shape or size (as long as it encloses the same charge).

phi = q / epsilon_0

Given q = 2.0 microC = 2.0 x 10^-6 C, epsilon_0 = 8.85 x 10^-12 C^2/(N m^2).

phi = (2.0 x 10^-6) / (8.85 x 10^-12) = 2.26 x 10^5 N m^2/C

(The 9.0 cm edge length of the cube is a distractor - it does not matter for the total flux, only the enclosed charge does.)

✓Final answer
$(b) 2.2 x 10^5.$
GUJCET 2025Set 031 markMCQ
$2q$
›Reveal solutionSolution
[!TLDR]

Using the shell theorem, $E' = 4kq/r^2 = 2E$ .

Concept

A uniformly charged thin spherical shell produces, at any external point, the same field as if all its charge were concentrated at the centre: $E = \dfrac{kQ}{d^2}$ .

Solution

For the point charge: $E = \dfrac{k(2q)}{r^2} = \dfrac{2kq}{r^2}$ .

For the shell of charge $q$ , at distance $r/2$ from the centre (which is $\gg R$ , hence external):
$E' = \frac{kq}{(r/2)^2} = \frac{kq}{r^2/4} = \frac{4kq}{r^2}.$
Compare: $\dfrac{E'}{E} = \dfrac{4kq/r^2}{2kq/r^2} = 2$ , so $E' = 2E$ .

[!ANSWER]

(B) 2E
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. Consider a uniform electric field E = 3 x 10^3 î N/C. What is the flux of this field through a square of 10cm on a side whose plane is parallel to the xy plane?$
(a) 30 Nm^2/C
(b) Zero
(c) 15 Nm^2/C
(d) 60 Nm^2/C
›Reveal solutionSolution

Electric flux Φ = E·A = EA cosθ, where θ is the angle between the field and the surface's normal vector.

E = 3 × 10³ x̂ N/C is directed along x. The square lies in a plane parallel to the xy-plane, so its normal vector is along z — perpendicular to E. Hence θ = 90°, cosθ = 0, and Φ = 0 regardless of the square's area.

✓Final answer
$(b) Zero.$
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. A charge q is placed at the center of one of the faces of a cube. The electric flux linked with the cube is ______.$
(a) q/ε0
(b) q/6ε0
(c) q/2ε0
(d) q/4ε0
›Reveal solutionSolution

Gauss's law gives total enclosed charge → flux, but here the charge sits exactly on a face, not fully inside the cube.

Imagine a second identical cube placed mirror-symmetric on the other side of that face, so together the two cubes fully enclose the charge q. By symmetry, each cube receives exactly half the total flux q/ε0 that q would produce through a fully enclosing surface. So flux through the original cube = (1/2)(q/ε0) = q/2ε0.

✓Final answer
$(c) q/2ε0.$
GUJCET 2024Set 131 markMCQ
$M^1L^3T^{-3}A^1$
›Reveal solutionSolution

Electric field has dimensions $MLT^{-3}A^{-1}$ ; multiplying by area $L^2$ gives electric flux $= M^1L^3T^{-3}A^{-1}$ .

Concept. Electric flux $\Phi_E = E\cdot A$ . Electric field $E=\dfrac{F}{q}$ has dimensions $\dfrac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$ .

Steps.
$[\Phi_E]=[E][A]=(MLT^{-3}A^{-1})(L^2)=M^1L^3T^{-3}A^{-1}.$
✓Final answer
$M^1L^3T^{-3}A^{-1}$

ANSWER: (D)
GUJCET 2024Set 131 markMCQ
$9 \times 10^4$
›Reveal solutionSolution

Using $E=\dfrac{2K\lambda}{r}$ , solve for $\lambda = \dfrac{Er}{2K}=10^{-7}$ C/m $=0.1\,\mu$ C/m.

Concept. The field of an infinite line charge is $E=\dfrac{2K\lambda}{r}=\dfrac{\lambda}{2\pi\varepsilon_0 r}$ .

Steps. With $E=9\times10^4$ N/C, $r=0.02$ m, $K=9\times10^9$ :
$\lambda=\frac{Er}{2K}=\frac{(9\times10^4)(0.02)}{2(9\times10^9)}=\frac{1800}{1.8\times10^{10}}=10^{-7}\ \text{C/m}=0.1\,\mu\text{C/m}.$
✓Final answer
$0.1\,\mu C/m$

ANSWER: (A)
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. If an electric charge 'q' is placed at the centre of a cube, then the flux associated with each surface of the cube is ___.$
(a) q/ε0
(b) q/6ε0
(c) q/4ε0
(d) q/2ε0
›Reveal solutionSolution

By Gauss's law, total flux through a closed surface enclosing charge q is q/ε0; a cube has 6 identical faces symmetric about the centre.

Total flux through the cube (Gauss's law) = q/ε0.

Since the charge is at the centre, by symmetry each of the 6 faces receives an equal share of this flux.

Flux per face = (q/ε0)/6 = q/6ε0.

✓Final answer
$(b) q/6ε0.$
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. If two infinite plane sheets having same surface charge density σ are placed parallel to each other, then the electric field between the two sheets is ___.$
(a) zero
(b) σ/ε0
(c) σ/2ε0
(d) 2σ/ε0
›Reveal solutionSolution

Each infinite charged sheet produces a uniform field of magnitude σ/2ε0 pointing away from it (for positive σ) on both sides.

Between the two sheets, the field due to the left sheet points away from it (rightward, into the gap) with magnitude σ/2ε0, while the field due to the right sheet points away from it (leftward, into the gap) with the same magnitude σ/2ε0. Since both sheets carry the same sign and magnitude of charge density, these two contributions are equal and opposite in the gap, so they cancel exactly.

(Outside the pair, on either side, the two fields add to give σ/ε0.)

✓Final answer
$(a) zero.$
GUJCET 2023Set 091 markMCQ
$\vec{E} = 3 \times 10^3\,\hat{k}$
›Reveal solutionSolution
[!TLDR]

The field is along $\hat{k}$ while the area vector is along $\hat{i}$ , so the flux is zero.

Concept

Electric flux through a flat surface is $\Phi = \vec{E}\cdot\vec{A} = EA\cos\theta$ , where $\theta$ is the angle between the field and the outward normal (area vector).

Solution

The square lies in a plane parallel to the yz-plane, so its normal (area vector) is along the x-axis, $\vec{A} = A\,\hat{i}$ .

The field is $\vec{E} = 3\times10^{3}\,\hat{k}$ N/C.
$\Phi = \vec{E}\cdot\vec{A} = (3\times10^{3}\,\hat{k})\cdot(A\,\hat{i}) = 0,$
since $\hat{k}\cdot\hat{i} = 0$ (the field is parallel to the plane, cutting no field lines through it).

[!ANSWER]

(D) Zero
GUJCET 2023Set 091 markMCQ
$Q. Figure shows the electric field lines of four point charges A, B, C and D. [FIGURE: A has 3 field lines; B and C are joined by many field lines (dipole-like) with C also having outward lines; D has 4 field lines] Which charge has the maximum magnitude? (A) C charge (B) B charge (C) A charge (D) D charge$
›Reveal solutionSolution

Field-line count ∝ $|q|$ ; charge C has the most lines, so the largest magnitude.

Concept — field lines and charge magnitude. The number of electric field lines starting from (or ending on) a charge is proportional to the magnitude of that charge. Counting: A has 3 lines, D has 4 lines, while charges B and C are linked by many lines (a dipole-like pair) with C additionally showing outgoing lines — C is associated with the largest number of field lines.

Therefore charge C has the maximum magnitude.

✓Final answer
$Option (A) C charge$

ANSWER: (A)
GUJCET 2022Set 171 markMCQ
$M^1 L^{-3} T^{-3} A^{-1}$
›Reveal solutionSolution

$\Phi_E=E\cdot A$ ; with $[E]=MLT^{-3}A^{-1}$ and area $L^2$ , flux is $M^1L^3T^{-3}A^{-1}$ .

Concept: Electric field $E=\dfrac{\text{force}}{\text{charge}}=\dfrac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$ .

Electric flux $\Phi_E=E\times\text{area}=MLT^{-3}A^{-1}\times L^2=M^1L^3T^{-3}A^{-1}$ .

✓Final answer
$M^1 L^3 T^{-3} A^{-1}$

ANSWER: (C)

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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.

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

Solution

Concept

Solution

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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