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.

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

$\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx=\int \frac{du}{u}=\log|u|+C=\log|e^{x}+e^{-x}|+C.$ …
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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

…

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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: $E = \dfrac{k(2q)}{r^2} = \dfrac{2kq}{r^2}$ . …
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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 — perpe …

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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 …

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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 $\dfrac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$ . …
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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, $K=9\times10^9$ : …
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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.

…

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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 d …

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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. …
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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 outg …
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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 $E=\dfrac{\text{force}}{\text{charge}}=\dfrac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$ . …
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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

Solution

Concept

Solution

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

Concept understanding

Testing each option

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