If there were only one type of charge in the universe, then (a) ointS vecE· dvecS ≠ 0 on any surface. (b) oint

NCERT Exemplar · Q10

Q. If there were only one type of charge in the universe, then

(a)

\oint_S \vec{E}\cdot d\vec{S} \neq 0

on any surface.

(b)

\oint_S \vec{E}\cdot d\vec{S} = 0

if the charge is outside the surface.

(c)

\oint_S \vec{E}\cdot d\vec{S}

could not be defined.

(d)

\oint_S \vec{E}\cdot d\vec{S} = \dfrac{q}{\varepsilon_0}

if charges of magnitude

q

were inside the surface.

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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. …

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: …

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Gauss's law $\oint_S \vec{E}\cdot d\vec{S}=q_{enc}/\varepsilon_0$ ties the net flux only to the enclosed charge, and holds regardless of how many kinds of charge exist. A charge outside gives zero net flux; a charge $q$ inside gives $q/\varepsilon_0$ . Correct: (b) and (d).

Concept understanding

Gauss's law is a geometric consequence of the inverse-square Coulomb field. For any closed surface,

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

Nothing in its derivation requires the existence of both positive and negative charge — it depends only on the $1/r^2$ field and the solid-angle geometry of a closed surface.

Testing each option

(a) claims the flux is never zero. This is false: the field lines from a charge lying outside the surface enter one side and leave the other, so their contributions cancel and the net flux vanishes.
(b) If the charge is outside the surface, $q_{enc}=0$ , so $\oint_S \vec{E}\cdot d\vec{S}=0$ . True. …

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Here are the common mistakes students make when reasoning about a universe with only one type of charge, along with how to avoid each.

Mistake 1: Assuming Electric Fields Still Exist as We Know Them

The mistake: Students often continue to draw field lines starting from the single charge type and ending at infinity, or they assume that field lines can still "terminate" somewhere. They forget that field lines only begin on positive charges and end on negative charges.
How to avoid: Recall the definition of an electric field line: it starts at a positive charge and ends at a negative charge. If only one type exists, field lines cannot terminate. They must either go to infinity or form closed loops. In a universe with only one charge type, the net flux through any closed surface is non-zero (by Gauss's law), so the field cannot be zero everywhere. The correct picture is that field lines radiate outward (if the charge is positive) and never end — they extend to infinity.

Mistake 2: Forgetting Gauss’s Law Implications

The mistake: Students think that if there is only one charge, the electric field must be zero everywhere because there is no opposite charge to "balance" it. They ignore that Gauss’s law ( $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$ ) still holds. A single charge produces a non-zero field.
How to avoid: Apply Gauss’s law directly. For a single point charge $q$ , the field at distance $r$ is $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$ , regardless of whether other charges exist. The field does not vanish — it simply diverges from the charge. The only way to have zero field everywhere is if the total charge in the universe is zero.

Mistake 3: Confusing "One Type of Charge" with "One Charge"

The mistake: Students interpret "one type of charge" as meaning there is only a single charge particle in the entire universe. They then try to apply Coulomb’s law between two identical charges, but there is no second charge to interact with.
How to avoid: "One type of charge" means all charges have the same sign (all positive or all negative). There can be many such charges. The key point is that unlike charges do not exist, so there is no attraction — only repulsion. The universe would be filled with repelling charges, and no stable neutral matter could form.

Mistake 4: Assuming Electric Forces Can Still Cancel

The mistake: Students think that multiple charges of the same sign can arrange themselves so that the net force on a test charge is zero (like in a neutral atom). They forget that without opposite charges, there is no possibility of cancellation — all forces are repulsive.
How to avoid: Realize that in our universe, cancellation happens because positive and negative charges can coexist. With only one sign, any configuration of charges will produce a net repulsive force on a test charge of the same sign. The only way to have zero net force is if the test charge is infinitely far away or if the charges are arranged in a symmetric pattern (e.g., at the center of a uniformly charged sphere), but even then, the field inside a conductor would be zero only if charges can move — and they would all repel to the surface. …

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

$\Phi_E = \oint \vec{E}\cdot d\vec{A} = \frac{q_{enclosed}}{\varepsilon_0}.$ …
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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

$q_{enc} = \varepsilon_0\,\phi_{net} = -3\times10^4\,\varepsilon_0\ \mathrm{C}.$ …
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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 …
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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 $r$ : …
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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 …
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←Previous questionQ9 — The electric field at a point is (a) always continuous. (b) continuous…→Next questionQ11 — Consider a region inside which there are various types of charges but …

Q.If there were only one type of charge in the universe, then

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

Concept understanding

Testing each option

Concept: Electric Field of a Point Charge

Method: Coulomb’s Law + Superposition (Single Charge Case)

Steps:

Mistake 1: Assuming Electric Fields Still Exist as We Know Them

Mistake 2: Forgetting Gauss’s Law Implications

Mistake 3: Confusing "One Type of Charge" with "One Charge"

Mistake 4: Assuming Electric Forces Can Still Cancel

Concept

Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Q.If there were only one type of charge in the universe, then

Concept understanding

Testing each option

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