A cube of side $0.1\ \text{m}$ is placed, as shown in the figure, in a region where the electric field $\vec{E} = 500\,x\,\hat{i}$ exists. Here $x$ is in metres and $E$ in $\text{NC}^{-1}$. Calculate : (a) the flux passing through the cube, and (b) the charge within the cube. ![Figure: cube in the field E = 500x i](/figures/pyq/2024-55-1-1/fig-q22.svg)

A cube of side 0.1m is placed, as shown in the figure, in a region where the electric field vecE = 500xhati ex

Question

0.1\ \text{m}

(a) the flux passing through the cube, and

(b) the charge within the cube.

Figure: cube in the field E = 500x i — Figure

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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 field is non-uniform ( $E \propto x$ ), so the flux through opposite faces does not cancel. Only the two faces perpendicular to the $x$ -axis contribute. Net flux $\Phi = 0.5\ \text{N m}^2/\text{C}$ , and the enclosed charge $q = \varepsilon_0 \Phi \approx 4.43 \times 10^{-12}\ \text{C}$ .

The key insight here is that Gauss's law relates the net electric flux through a closed surface to the charge enclosed. But when the field is not uniform, you cannot just multiply $E$ by area — you must sum the flux through each face separately, paying attention to the direction of the field relative to the outward normal.

The field $\vec{E} = 500x\,\hat{i}$ depends only on $x$ . That means on any face of the cube, the field is constant over that face (since $x$ is fixed on that face), but it changes from one face to another. This is the classic "non-uniform field through a cube" problem.

Let's work through it systematically.

Identify which faces contribute.

The flux through a surface is $\Phi = \int \vec{E} \cdot d\vec{A}$ . Since $\vec{E}$ points purely along $\hat{i}$ , only faces whose area vector has an $x$ -component can have non-zero flux. For a cube aligned with the axes, these are the two faces perpendicular to the $x$ -axis: the left face (at $x = 0.1\ \text{m}$ ) and the right face (at $x = 0.2\ \text{m}$ ).

The four faces parallel to the $x$ -axis have $d\vec{A}$ perpendicular to $\hat{i}$ , so $\vec{E} \cdot d\vec{A} = 0$ on them. They contribute nothing.
Flux through the left face ( $x = 0.1\ \text{m}$ ).

On this face, $x$ is constant at $0.1\ \text{m}$ , so $E = 500 \times 0.1 = 50\ \text{N/C}$ . The outward normal for the left face points in the $-\hat{i}$ direction (out of the cube). So $d\vec{A} = -dA\,\hat{i}$ .

The dot product: $\vec{E} \cdot d\vec{A} = (50\,\hat{i}) \cdot (-dA\,\hat{i}) = -50\,dA$ .

Since $E$ is constant over the face, the flux is simply $E \times \text{area}$ with the sign:

\Phi_{\text{left}} = -50 \times (0.1)^2 = -50 \times 0.01 = -0.5\ \text{N m}^2/\text{C}.

Flux through the right face ( $x = 0.2\ \text{m}$ ). Here $E = 500 \times 0.2 = 100\ \text{N/C}$ . The outward normal points in $+\hat{i}$ , so $d\vec{A} = +dA\,\hat{i}$ .

\vec{E} \cdot d\vec{A} = (100\,\hat{i}) \cdot (+dA\,\hat{i}) = +100\,dA.

\Phi_{\text{right}} = +100 \times 0.01 = +1.0\ \text{N m}^2/\text{C}.

Net flux through the cube. The other four faces contribute zero, so: …

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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 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.

…

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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.

…

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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 …

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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 ti …

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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

…

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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

…

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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

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

…

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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.

…

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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 …

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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 $^2$ …
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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:

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

…

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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 uni …

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Q.A cube of side 0.1 m0.1\ \text{m}0.1 m is placed, as shown in the figure, in a region where the electric field E⃗=500 x i^\vec{E} = 500\,x\,\hat{i}E=500xi^ exists. Here xxx is in metres and EEE in NC−1\text{NC}^{-1}NC−1. Calculate :

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Q.A cube of side $0.1\ \text{m}$ is placed, as shown in the figure, in a region where the electric field $\vec{E} = 500\,x\,\hat{i}$ exists. Here $x$ is in metres and $E$ in $\text{NC}^{-1}$ . Calculate :

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