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.

Puducherry CbseShort· 3mImportance★★★★★

81% · 54/67 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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: …

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ18 — A hollow cylinder carries a uniform positive charge spread over its cu…→Next questionQ20 — A paisa coin is made up of Al-Mg alloy and weighs 0.75 g. It has a squ…

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

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

🎓Unlock everything free for 14 days

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:

Unlock everything free for 14 days