The electric flux emerging out from 1 C charge is (A) $\dfrac{1}{\varepsilon_0}$ (B) $4\pi$ (C) $\dfrac{4\pi}{\varepsilon_0}$ (D) $\varepsilon_0$

The electric flux emerging out from 1 C charge is (A) (1)/(varepsilon₀) (B) 4π (C) (4π)/(varepsilon₀) (D) vare

Question

\dfrac{1}{\varepsilon_0}

CBSECBSE Class XII Board 2020MCQ· 1mImportance★★★★★

✓ Free question

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 tells us that the total electric flux through any closed surface depends only on the enclosed charge: $\Phi = \frac{Q}{\varepsilon_0}$ . For $Q = 1\,\text{C}$ , the flux is $\frac{1}{\varepsilon_0}$ .

The question asks for the total electric flux emerging from a point charge. This is a direct application of Gauss's law, one of the four Maxwell equations that form the foundation of electromagnetism.

Gauss's law states that the total electric flux through any closed surface equals the net charge enclosed divided by the permittivity of free space:

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

The beauty of this law is that it doesn't depend on the shape of the surface you choose, the distance from the charge, or how the field varies. The flux is determined entirely by the enclosed charge.

Why this works: Electric field lines originate from positive charges and terminate on negative charges. Every field line that emerges from a charge must pierce through any closed surface surrounding it exactly once. The total "flow" of field lines (flux) is therefore a property of the source charge alone.

Identify the enclosed charge. We have $Q = 1\,\text{C}$ enclosed by our imaginary surface.
Apply Gauss's law directly. The total flux emerging from this charge is:

\Phi = \frac{Q}{\varepsilon_0} = \frac{1\,\text{C}}{\varepsilon_0}

Recognize what this means physically. The permittivity $\varepsilon_0 \approx 8.85 \times 10^{-12}\,\text{C}^2/\text{N·m}^2$ is a fundamental constant. The flux $\frac{1}{\varepsilon_0}$ is enormous (about $1.13 \times 10^{11}\,\text{N·m}^2/\text{C}$ ), reflecting how strong the electric field is around a 1 coulomb charge.

Watch out

Don't confuse this with the electric field at a specific distance. The field at distance $r$ is $E = \frac{Q}{4\pi\varepsilon_0 r^2}$ , which involves $4\pi$ and depends on $r$ . The total flux, however, is independent of distance and has no $4\pi$ factor.

\Phi = \frac{Q}{\varepsilon_0}

✓Final answer

\dfrac{1}{\varepsilon_0}

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 37 on this concept.

CBSE 2026Set A1 markMCQ
$Q. S.I. unit of electric flux is (A) Vm (B) Vm^2 (C) Jm (D) NC^-1$
›Reveal solutionSolution

Φ = E·A → (V/m)(m²) = V·m.

Electric flux is $\Phi = \vec{E}\cdot\vec{A}$ .

Unit of electric field $E$ = N/C = V/m (volt per metre).

Unit of area $A$ = m².

So $[\Phi] = \dfrac{V}{m}\times m^2 = V\cdot m$ (equivalently N·m²/C).

✓Final answer
$(A) Vm.$
CBSE 2026Set A1 markMCQ
$Q. The surface charge densities on the surface of two conducting spheres of radii r1 and r2 are equal. The ratio of electric field intensities on the surfaces is (A) r1/r2 (B) r1^2/r2^2 (C) r2^2/r1^2 (D) 1 : 1$
›Reveal solutionSolution

Just outside a charged conductor E = σ/ε₀; with equal σ the fields are equal (1:1).

The electric field just outside the surface of a charged conductor is

$E = \frac{\sigma}{\varepsilon_0},$

which depends only on the local surface charge density σ, not on the radius.

Since the two spheres have equal σ, their surface field intensities are equal:

$\frac{E_1}{E_2} = \frac{\sigma/\varepsilon_0}{\sigma/\varepsilon_0} = 1 : 1.$

✓Final answer
$(D) 1 : 1.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Electric flux is a$
(a) scalar quantity
(b) vector quantity
(c) scalar or vector quantity
(d) constant quantity
›Reveal solutionSolution

Electric flux is a scalar quantity, even though it is defined using two vectors.

Electric flux through a surface is defined as

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

Although $\vec{E}$ (electric field) and $d\vec{A}$ (area vector, normal to the surface element) are both vectors, their dot product $\vec{E}\cdot d\vec{A} = E\,dA\cos\theta$ is a single number (magnitude only, with a sign depending on $\theta$ ) — it has no direction of its own. Summing (integrating) scalar quantities over the surface still gives a scalar. Its SI unit is N·m²/C (equivalently V·m).

✓Final answer
$(a) scalar quantity.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. A charge Q, is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will$
(a) decrease to half
(b) increase two times
(c) remain unchanged
(d) increase four times
›Reveal solutionSolution

Gauss's law: flux through any closed surface = Q_enclosed / epsilon_0, and this does NOT depend on the surface's size or shape.

Gauss's law states that for any closed (Gaussian) surface,

flux (phi) = Q_enclosed / epsilon_0

Here the same charge Q sits at the centre of the sphere both before and after the radius is doubled - the enclosed charge Q_enclosed is unchanged. Since flux depends ONLY on Q_enclosed and the permittivity of free space epsilon_0 (both unchanged here), the flux does not change even though the surface area (4piR^2) has increased fourfold. Doubling R spreads the same total flux over 4 times the area, so the electric field at the new surface drops to 1/4, but the total flux (field x area, integrated) stays exactly the same.

✓Final answer
$(c) The outward electric flux remains unchanged, since it depends only on the enclosed charge Q, not on the radius R.$
CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) increases as $r$ increases for $r<R$ and for $r>R$
(b) is zero as $r$ increases for $r<R$ and decreases as $r$ increases for $r>R$
(c) is zero as $r$ increases for $r<R$ and increases as $r$ increases for $r>R$
(d) decreases as $r$ increases for $r<R$ and for $r>R$
›Reveal solutionSolution
A charged conducting (hollow metal) sphere carries all its charge on the outer surface. Gauss's law gives $E=0$ inside and $E\propto 1/r^2$ (decreasing) outside.

Setting up Gauss's law

For a hollow, uniformly charged conducting sphere of radius $R$ and total charge $Q$ , all the charge resides on the outer surface (a fundamental property of conductors in electrostatic equilibrium — free charges repel each other and move to the surface where the electric field inside the conducting material is zero).

Take a concentric spherical Gaussian surface of radius $r$ .

Case 1: $r<R$ (inside the shell)

The Gaussian surface of radius $r$ encloses no charge, because all the charge $Q$ lies on the surface at radius $R>r$ .

$\oint \vec E\cdot d\vec A = \frac{Q_{\text{enc}}}{\varepsilon_0} = 0 \implies E=0$

This is true for every $r<R$ — the field is zero throughout the interior, it does not "increase" or "decrease," it is simply zero.

Case 2: $r>R$ (outside the shell)

Now the Gaussian surface encloses the entire charge $Q$ . By spherical symmetry, $\vec E$ is radial and has the same magnitude everywhere on the Gaussian sphere, so

$E(4\pi r^2) = \frac{Q}{\varepsilon_0} \implies E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$

As $r$ increases (for $r>R$ ), $E \propto 1/r^2$ , so $E$ decreases monotonically — exactly as it would for a point charge $Q$ placed at the centre.

Conclusion

Inside ( $r<R$ ): $E=0$ (stays zero — not increasing, not decreasing).

Outside ( $r>R$ ): $E$ decreases as $r$ increases.

This matches option (b).

✓Final answer
$r$
CBSE 2026Set ANNUAL1 mark
$Q. Match the Column A with Column B and write the correct pair. Column A: Electric flux. Column B:$
(i) μ₀nI,
(ii) Volt × second,
(iii) μ₀nI/2,
(iv) Volt × meter,
(v) Volt × meter⁻¹,
(vi) Volt.
›Reveal solutionSolution

Electric flux φ_E = E·A; unit = (V/m)(m²) = V·m, option (iv).

Electric flux through a surface is φ_E = E·A (for a uniform field perpendicular to area A). The SI unit of electric field E is volt per metre (V/m) = N/C, and area is in m². Therefore the unit of electric flux is

(V/m) × m² = V·m (volt × metre), which is also N·m²/C.

This matches Column B entry (iv).

✓Final answer
$(iv) Volt \times meter.$
CBSE 2025Set D1 markMCQ
$Q. Gauss's law states that the electric flux through a closed surface is (A) proportional to the charge enclosed (B) inversely proportional to the charge enclosed (C) zero (D) proportional to the square of the charge enclosed$
›Reveal solutionSolution

Gauss's law states the total electric flux through a closed surface equals the enclosed charge divided by ε₀, so Φ ∝ q_enclosed.

Gauss's law is written as

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

The left side is the total electric flux Φ through the closed (Gaussian) surface. Thus

$\Phi = \frac{q_{enc}}{\varepsilon_0}$

The flux is directly proportional to the net charge enclosed and is independent of the shape of the surface or the location of the charge within it.

✓Final answer
$(A) proportional to the charge enclosed.$
CBSE 2025Set D1 markMCQ
$Q. Inside a closed surface n electric dipoles are situated. The electric flux coming out from the closed surface will be (A) q/ε0 (B) 2q/ε0 (C) nq/ε0 (D) zero$
›Reveal solutionSolution

A dipole has zero net charge, so n dipoles enclose zero charge and the net flux is zero.

Gauss's law states the net electric flux out of a closed surface is Φ = q_enclosed/ε₀.

Each electric dipole consists of +q and −q; its net charge is +q + (−q) = 0. With n dipoles inside, the total enclosed charge is n × 0 = 0.

Therefore Φ = 0/ε₀ = 0. (Field lines from each +q terminate on its own −q inside the surface, so no net flux escapes.)

✓Final answer
$(D) zero.$
CBSE 2025Set A1 mark
$Q. Write True or False: Inside a conductor, electrostatic field is zero.$
›Reveal solutionSolution

The statement is True: the electrostatic field inside a conductor is zero in equilibrium.

In electrostatic equilibrium, free charges in a conductor redistribute themselves on the surface such that the electric field inside the body of the conductor is exactly zero. If there were a residual field inside, it would exert a force on the free electrons, causing them to keep moving — contradicting the assumption of electrostatic equilibrium (no current flow). This redistribution happens almost instantaneously and is why conductors are used for electrostatic shielding.

✓Final answer
$True.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The SI unit of electric flux is$
(a) N C m^2
(b) N C^-1 m^-2
(c) N C^-1 m^2
(d) N^2 C^-1 m^2
›Reveal solutionSolution
Electric flux is the field strength times the perpendicular area through which it passes, so its unit is simply the product of the units of $E$ and area.

Electric flux through a surface is defined as $\Phi_E = \vec{E}\cdot\vec{A}$ (or $\int \vec{E}\cdot d\vec{A}$ for a general surface).

SI unit of electric field $E$ = N C $^{-1}$

SI unit of area $A$ = m $^2$

So the SI unit of $\Phi_E$ = N C $^{-1}$ $\times$ m $^2$ = N C $^{-1}$ m $^2$ (this is also equivalent to V·m, since $E$ can also be expressed in V/m).

✓Final answer
$(c) N C^-1 m^2.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The SI unit of electric flux is:$
(a) $Nm^2C^{-2}$
(b) $NC^{-1}m^2$
(c) $CN^2m^{-1}$
(d) $C^2N^{-1}m^{-2}$
›Reveal solutionSolution

Electric flux $\Phi_E = \vec{E}\cdot\vec{A}$ , so its unit is the unit of $E$ (N C⁻¹) times the unit of area (m²).

Electric flux through a surface is defined as $\Phi_E = \oint \vec{E}\cdot d\vec{A}$ , i.e. the product of the electric field and the area component perpendicular to it.

Since the SI unit of electric field $E$ is newton per coulomb ( $NC^{-1}$ ) and the unit of area $A$ is square metre ( $m^2$ ), the unit of electric flux is:

$NC^{-1} \times m^2 = NC^{-1}m^2$

This is the same physical unit as $Nm^2C^{-1}$ (order of writing does not matter), but among the given options only (b) is written correctly as $NC^{-1}m^2$ ; option (a) incorrectly shows $C^{-2}$ .

✓Final answer
$NC^{-1}m^2$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The SI unit of the surface integral of electric field is -$
(a) Vm
(b) V
(c) $NC^{-1}$
(d) $Cm^{-3}$
›Reveal solutionSolution

Electric flux $\Phi_E=\oint\vec{E}\cdot d\vec{A}$ has SI unit V·m (equivalent to N·m $^2$ /C).

The surface integral of the electric field, $\Phi_E = \oint \vec{E}\cdot d\vec{A}$ , is the electric flux. Since $E$ has SI unit V/m (or equivalently N/C) and area has unit m $^2$ , the flux has unit $(\text{V/m})\times\text{m}^2 = \text{V·m}$ (this is dimensionally the same as $\text{N·m}^2/\text{C}$ ).

✓Final answer
$(a) Vm.$

Q.The electric flux emerging out from 1 C charge is (A) 1ε0\dfrac{1}{\varepsilon_0}ε0​1​ (B) 4π4\pi4π (C) 4πε0\dfrac{4\pi}{\varepsilon_0}ε0​4π​ (D) ε0\varepsilon_0ε0​

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Q.The electric flux emerging out from 1 C charge is (A) $\dfrac{1}{\varepsilon_0}$ (B) $4\pi$ (C) $\dfrac{4\pi}{\varepsilon_0}$ (D) $\varepsilon_0$

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