An arbitrary surface encloses a dipole. What is the electric flux through this surface?

NCERT Exemplar · Q14

Q. An arbitrary surface encloses a dipole. What is the electric flux through this 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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The electric flux through any closed surface enclosing an electric dipole is zero because the net charge enclosed by the surface is zero, as per Gauss's Law.

To determine the electric flux through a surface enclosing an electric dipole, we turn to Gauss's Law, a fundamental principle in electrostatics. Gauss's Law provides a powerful way to calculate electric flux, especially when dealing with closed surfaces and charge distributions.

The core idea behind Gauss's Law is that the total electric flux out of any closed surface is directly proportional to the total electric charge enclosed within that surface. It doesn't matter how the charge is distributed inside, or what the shape of the surface is; only the net charge matters.

\Phi_E

Let's apply this concept step-by-step to the given problem.

Understand an Electric Dipole:

An electric dipole consists of two equal and opposite point charges, typically $+q$ and $-q$ , separated by a small fixed distance $d$ . These two charges together form the dipole.
Identify Charges Enclosed by the Surface:

The problem states that "an arbitrary surface encloses a dipole." This means that both the positive charge ( $+q$ ) and the negative charge ( $-q$ ) that constitute the dipole are located inside this closed surface.
Calculate the Net Charge Enclosed ( $Q_{enc}$ ):

Since both charges of the dipole are inside the surface, the total (net) charge enclosed by the surface is the algebraic sum of these charges:

Q_{enc} = (+q) + (-q)

$$ Q_{enc} = 0 $$ …

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Here’s a breakdown of the common mistakes students make on this question, along with how to avoid each.

The Core Concept

The question tests Gauss’s Law:

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

The key is that only the net charge enclosed matters. A dipole consists of two equal and opposite charges ( $+q$ and $-q$ ). Therefore, the net charge enclosed by any surface that completely contains the dipole is:

Q_{\text{enc}} = (+q) + (-q) = 0

Hence, the electric flux through the surface is zero.

Common Mistake #1: Thinking the Flux is Non-Zero Because There is an Electric Field

The Mistake: Students see that the dipole creates an electric field everywhere (including on the surface). They assume that because $\vec{E}$ is not zero on the surface, the flux cannot be zero.
Why it’s wrong: Flux depends on the net flow of field lines through the surface. For a dipole, field lines leave the positive charge, loop around, and enter the negative charge. Every field line that exits the surface re-enters it. The inward and outward contributions cancel exactly.
How to avoid: Remember Gauss’s Law is about net enclosed charge, not the presence of a field. Draw the field lines of a dipole. Count the lines entering vs. leaving the surface — they are equal.

Common Mistake #2: Forgetting the Dipole is a Single System

The Mistake: A student might treat the dipole as two separate charges and try to calculate flux for each, then add them. They might forget that the surface encloses both charges.
Why it’s wrong: If you enclose only one charge of the dipole, the flux is non-zero. But the question says the surface encloses the entire dipole. The net enclosed charge is zero.
How to avoid: Always ask: “What is the total charge inside the closed surface?” For a dipole, the answer is always zero. Do not split the problem into two separate Gauss’s law calculations unless the surface only encloses one charge.

Common Mistake #3: Confusing Flux with Field Strength

The Mistake: Students think that if the electric field is strong at some points on the surface, the flux must be large.
Why it’s wrong: Flux is the dot product $\vec{E} \cdot d\vec{A}$ . Even if $\vec{E}$ is large, the angle between $\vec{E}$ and the surface normal can make the contribution small or negative. For a dipole, the field points outward on one side and inward on the other, leading to cancellation. …

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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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Q.An arbitrary surface encloses a dipole. What is the electric flux through this 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

Concept: Gauss’s Law and Net Charge Enclosed

Method: Net Charge Argument

Steps

The Core Concept

Common Mistake #1: Thinking the Flux is Non-Zero Because There is an Electric Field

Common Mistake #2: Forgetting the Dipole is a Single System

Common Mistake #3: Confusing Flux with Field Strength

Concept

Solution

Concept

Solution

Concept

Solution

Q.An arbitrary surface encloses a dipole. What is the electric flux through this surface?

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