A cube has side length $a$. A point charge $q$ is placed, in four separate cases, at the following positions relative to the cube: (a) at point A, which is a corner (vertex) of the cube; (b) at point B, the mid-point of one edge of the cube; (c) at point C, the centre of one face of the cube; (d) at point D, the mid-point of the straight segment joining B and C (so D lies on that face of the cube). In each case find the total electric flux through all the faces of the cube.

A cube has side length a. A point charge q is placed, in four separate cases, at the following positions relat

NCERT Exemplar · Q19

a

(a) at point A, which is a corner (vertex) of the cube;

(b) at point B, the mid-point of one edge of the cube;

(d) at point D, the mid-point of the straight segment joining B and C (so D lies on that face of the cube). In each case find the total electric flux through all the faces of the cube.

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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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By Gauss's law the flux from a fully-enclosed charge is $q/\varepsilon_0$ . When the charge sits on a symmetry point shared by several identical cubes, that total flux divides equally among them: a corner is shared by 8 cubes, an edge-midpoint by 4, and any point lying on a face by 2. This directly gives the four answers.

Concept

Gauss's law: a charge fully enclosed gives total flux $\Phi_{tot}=q/\varepsilon_0$ . If the charge lies on a boundary shared by $n$ identical cubes tiling the space around it, symmetry splits the flux equally, so each cube gets $\Phi = \dfrac{q}{n\,\varepsilon_0}$ .

(a) Charge at a corner A

A cube corner is shared by $2\times2\times2 = 8$ cubes.

\Phi_A = \frac{q}{8\varepsilon_0}.

(b) Charge at the mid-point of an edge B

An edge is shared by $4$ cubes meeting around it.

\Phi_B = \frac{q}{4\varepsilon_0}.

(c) Charge at the centre of a face C

A face is shared by exactly $2$ cubes. …

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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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Q.A cube has side length $a$ . A point charge $q$ is placed, in four separate cases, at the following positions relative to the cube:

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

(a) Charge at a corner A

(b) Charge at the mid-point of an edge B

(c) Charge at the centre of a face C

Method: The Symmetry-Sharing Trick for Flux Through a Partial Enclosure

Steps

Concept

Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Q.A cube has side length aaa. A point charge qqq is placed, in four separate cases, at the following positions relative to the cube:

Concept

(a) Charge at a corner A

(b) Charge at the mid-point of an edge B

(c) Charge at the centre of a face C

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Q.A cube has side length $a$ . A point charge $q$ is placed, in four separate cases, at the following positions relative to the cube:

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