A point charge $+10\,\mu\text{C}$ is a distance $5\,\text{cm}$ directly above the centre of a square of side $10\,\text{cm}$, as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge $10\,\text{cm}$.) ![A point charge 5 cm above the centre of a 10 cm square, used to find the electric flux through the square via Gauss's law](/figures/electric-charges-and-fields/point-charge-cm-above-centre-10-square-fig-1-31.svg)

A point charge +10muC is a distance 5cm directly above the centre of a square of side 10cm, as shown in Fig. 1

Exercises · 1.17

+10\,\mu\text{C}

A point charge 5 cm above the centre of a 10 cm square, used to find the electric flux through the square via Gauss's law — Figure 1.31

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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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Completing the square into a cube of edge $10\,\text{cm}$ places the charge at the cube's centre; Gauss's law gives total flux $q/\varepsilon_0$ , and by symmetry each of the six faces carries $q/6\varepsilon_0=1.88\times10^{5}\,\text{N·m}^2/\text{C}$ .

A single square is an open surface, so Gauss's law cannot be applied to it directly. The hint tells us to complete it into a closed surface.

Step 1 — Build the cube. The charge sits $5\,\text{cm}$ above the centre of the $10\,\text{cm}$ square. Imagine a cube of edge $10\,\text{cm}$ having this square as one face. The centre of such a cube is $5\,\text{cm}$ from each face — exactly where the charge is. So the charge is at the centre of the cube, and the given square is one of its six faces.

Step 2 — Total flux through the cube. The cube is now a closed surface enclosing $q=+10\,\mu\text{C}$ . Gauss's law gives

\Phi_{\text{total}}=\frac{q}{\varepsilon_0},\qquad \varepsilon_0=8.854\times10^{-12}\,\text{C}^2/\text{N·m}^2.

Step 3 — Use symmetry. With the charge at the centre, the six faces are equivalent, so each receives one‑sixth of the total flux: …

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Common Mistakes Students Make with This Gauss Law Problem

Mistake 1: Trying to integrate directly over the square

What students do wrong:

They attempt to compute $\Phi = \int \vec{E} \cdot d\vec{A}$ directly, setting up a double integral over the square's surface. This is messy because the electric field from a point charge varies in both magnitude and direction across the square.

Why it's wrong:

The integration is unnecessarily complex. The electric field is not uniform over the square — its magnitude changes with distance from the charge, and its direction changes relative to the surface normal. This leads to a difficult integral that most students cannot evaluate correctly.

How to avoid:

Use the hint in the problem. Place the square as one face of a cube of side $10\,\text{cm}$ , with the charge at the cube's centre. By Gauss's law, the total flux through the entire cube is:

\Phi_{\text{cube}} = \frac{q_{\text{enc}}}{\varepsilon_0}

Since the charge is at the centre, the flux is equally distributed through all 6 faces. Therefore:

\Phi_{\text{square}} = \frac{1}{6} \cdot \frac{q}{\varepsilon_0}

Mistake 2: Forgetting that the charge is not at the centre of the square

What students do wrong:

They assume the charge is at the centre of the square and use symmetry arguments incorrectly — for example, claiming the flux through the square is $\frac{q}{4\varepsilon_0}$ (as if the square were one face of a tetrahedron).

Why it's wrong:

The charge is 5 cm above the centre, not at the centre of the square itself. The square is only one face of an imaginary cube. The symmetry that works is the cubic symmetry — the charge is at the cube's centre, so all 6 faces are equivalent.

How to avoid:

Visualise the cube clearly. The square is the top face of a cube of side $10\,\text{cm}$ , and the charge is at the cube's centre (5 cm below the top face). This makes all 6 faces symmetric with respect to the charge.

Mistake 3: Using the wrong value of $q$ or units

What students do wrong:

They forget to convert $10\,\mu\text{C}$ to SI units ( $10 \times 10^{-6}\,\text{C}$ ) or use $10\,\text{cm}$ as $10\,\text{m}$ instead of $0.1\,\text{m}$ .

Why it's wrong:

Gauss's law in SI form requires charge in coulombs and distances in metres. Using wrong units gives a numerically incorrect answer.

How to avoid:

Always convert to SI before plugging into formulas:

$q = 10\,\mu\text{C} = 10 \times 10^{-6}\,\text{C} = 1.0 \times 10^{-5}\,\text{C}$
Side of square = $10\,\text{cm} = 0.1\,\text{m}$

Mistake 4: Forgetting $\varepsilon_0$ or using the wrong value

What students do wrong:

They either omit $\varepsilon_0$ entirely or use $\varepsilon_0 = 8.85 \times 10^{-12}$ incorrectly (e.g., forgetting units).

Why it's wrong: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL2 marks
$Q. Classify the following into$
(i) polar, and
(ii) non-polar molecules: CO2, HCl, H2 and H2O. OR Calculate the flux passing through a circular area of radius 5cm placed perpendicular to a uniform electric field E = 200i NC^-1.
›Reveal solutionSolution
Option 1: HCl and H₂O are polar (asymmetric charge distribution, net dipole moment); CO₂ and H₂ are non-polar (symmetric, zero net dipole moment). Option 2: electric flux $\Phi = EA \approx 1.57\,\mathrm{N\,m^2\,C^{-1}}$ .

Option 1 — Polar vs non-polar molecules

A molecule is polar if its centres of positive and negative charge do not coincide, giving it a permanent (net) dipole moment; it is non-polar if they coincide (net dipole moment zero), usually due to symmetry.

CO₂ — linear, symmetric ( $O=C=O$ ); the two C=O bond dipoles are equal and opposite, so they cancel $\Rightarrow$ non-polar.

HCl — different atoms (H, Cl) with unequal electronegativity, bond dipole does not cancel $\Rightarrow$ polar.

H₂ — homonuclear diatomic (identical atoms), no charge asymmetry $\Rightarrow$ non-polar.

H₂O — bent (angular) shape; the two O–H bond dipoles do not cancel $\Rightarrow$ polar.

So: Polar: HCl, H₂O. Non-polar: CO₂, H₂.

Option 2 — Electric flux

…
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AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL2 marks
$Q. A closed spherical surface encloses a charge q at its centre. Show that electric flux through the closed surface is q/ε0. OR A pair of charges +q and -q, separated by a small distance 2a is placed in an electric field E, so that the line joining the charges makes an angle θ with E. Write the expressions for torque τ and also its magnitude |τ|.$
›Reveal solutionSolution

Gauss's law shows the flux out of a sphere enclosing a point charge q is q/ε₀, independent of the sphere's radius; and the torque on a dipole in a field is τ = pE sinθ.

Main question: Consider a point charge $q$ at the centre O of a spherical surface of radius $r$ . By Coulomb's law, the magnitude of the electric field at every point on this sphere (all at distance $r$ from $q$ ) is the same:

$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

and it points radially outward, i.e. parallel to the outward area vector $d\vec A$ at every point on the sphere (by symmetry). So the flux through the whole closed surface is simply $E$ times the total surface area $4\pi r^2$ :

$\Phi_E = \oint \vec E \cdot d\vec A = E \times 4\pi r^2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\times 4\pi r^2 = \frac{q}{\epsilon_0}$

Notice the radius $r$ cancels out completely — the flux depends only on the enclosed charge $q$ , not on the size of the sphere. This is exactly Gauss's law for this special symmetric case.

…

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←Previous question1.16 — Careful measurement of the electric field at the surface of a black bo…→Next question1.18 — A point charge of 2.0muC is at the centre of a cubic Gaussian surface …

Q.A point charge $+10\,\mu\text{C}$ is a distance $5\,\text{cm}$ directly above the centre of a square of side $10\,\text{cm}$ , as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge $10\,\text{cm}$ .)

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

Method: Gauss's Law with Symmetry (Cube Construction)

Why This Method Works

Steps

Common Mistakes Students Make with This Gauss Law Problem

Mistake 1: Trying to integrate directly over the square

Mistake 2: Forgetting that the charge is not at the centre of the square

Mistake 3: Using the wrong value of $q$ or units

Mistake 4: Forgetting $\varepsilon_0$ or using the wrong value

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