Q.A point charge +10μC is a distance 5cm directly above the centre of a square of side 10cm, 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 10cm.)
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🔒 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 E passing through a small patch of area dA. The electric flux through that patch is:
dΦE=E⋅dA=EdAcosθ
where θ is the angle between E and the outward normal to the surface.
Why this definition?
- If E is perpendicular to the surface (θ=0), maximum field "flows through".
- If E is parallel (θ=90∘), no flux — the field just slides along the surface.
Total flux through a closed surface S is:
ΦE=∮SE⋅dA
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=4πε01r2q, radially outward.
- The area vector dA is also radially outward.
- So E⋅dA=EdA everywhere.
The total flux:
ΦE=∮EdA=E∮dA=(4πε01r2q)(4πr2)=ε0q
Notice: The r2 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 dΦ=4πε01r2qcosθdA.
- But r2cosθdA is exactly the solid angle dΩ subtended by dA at the charge.
So dΦ=4πε0qdΩ.
Integrating over the entire closed surface: ∮dΩ=4π (total solid angle around a point).
Hence:
ΦE=4πε0q⋅4π=ε0q
Result: For any closed surface enclosing q, the flux is ε0q.
4. Multiple Charges: Superposition
If there are many charges q1,q2,…,qn inside the surface, the total electric field is the vector sum of individual fields:
E=E1+E2+⋯+En
Flux is linear: …
By the hint, treat the square as one face of a cube of edge 10cm. Because the charge is 5cm (half an edge) above the square's centre, it lies exactly at the cube's centre.
Gauss's law gives the total flux through the closed cube:
Φtotal=ε0q.
By symmetry the six faces share this equally, so the flux through one face is …
Completing the square into a cube of edge 10cm places the charge at the cube's centre; Gauss's law gives total flux q/ε0, and by symmetry each of the six faces carries q/6ε0=1.88×105N⋅m2/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 5cm above the centre of the 10cm square. Imagine a cube of edge 10cm having this square as one face. The centre of such a cube is 5cm 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=+10μC. Gauss's law gives
Φtotal=ε0q,ε0=8.854×10−12C2/N⋅m2.
Step 3 — Use symmetry. With the charge at the centre, the six faces are equivalent, so each receives one‑sixth of the total flux: …
Method: Gauss's Law with Symmetry (Cube Construction)
Why This Method Works
The hint suggests a powerful symmetry trick. A point charge above the centre of a square has no simple symmetry by itself — but if we imagine the square as one face of a cube with the charge at its centre, the full cube has perfect symmetry.
Steps
Step 1: Construct an imaginary cube
Place the +10μC charge at the exact centre of a cube of side 10cm. The given square becomes the top face of this cube.
Step 2: Apply Gauss's Law to the entire cube
Gauss's Law states:
Φcube=ε0Qenclosed
Here, Qenclosed=+10μC=10×10−6C.
So:
Φcube=8.85×10−1210×10−6≈1.13×106N⋅m2/C
Step 3: Use symmetry to find flux through one face
The charge is at the cube's centre. By symmetry, the total flux is divided equally among all 6 faces of the cube.
Therefore: …
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 Φ=∫E⋅dA 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 10cm, with the charge at the cube's centre. By Gauss's law, the total flux through the entire cube is:
Φcube=ε0qenc
Since the charge is at the centre, the flux is equally distributed through all 6 faces. Therefore:
Φsquare=61⋅ε0q
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 4ε0q (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 10cm, 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 10μC to SI units (10×10−6C) or use 10cm as 10m instead of 0.1m.
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=10μC=10×10−6C=1.0×10−5C
- Side of square = 10cm=0.1m
Mistake 4: Forgetting ε0 or using the wrong value
What students do wrong:
They either omit ε0 entirely or use ε0=8.85×10−12 incorrectly (e.g., forgetting units).
Why it's wrong: …
Showing the 12 most recent of 37 on this concept.
- CBSE 2026Set A1 markMCQQ.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 Φ=E⋅A.
Unit of electric field E = N/C = V/m (volt per metre).
Unit of area A = m².
…
- CBSE 2026Set A1 markMCQQ.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=ε0σ,
which depends only on the local surface charge density σ, not on the radius.
…
- CBSE 2026Set ANNUAL1 markMCQQ.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
Φ=∮E⋅dA
Although E (electric field) and dA (area vector, normal to the surface element) are both vectors, their dot product E⋅dA=EdAcosθ is a single number (magnitude only, with a sign depending on θ) — it …
- CBSE 2026Set ANNUAL1 markMCQQ.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 …
- CBSE 2026Set ANNUAL1 markMCQQ.A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre(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∝1/r2 (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.
∮E⋅dA=ε0Qenc=0⟹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, E is radial and has the same magnitude everywhere on the Gaussian sphere, so
…
- CBSE 2026Set ANNUAL1 markQ.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
…
- CBSE 2025Set D1 markMCQQ.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
∮E⋅dA=ε0qenc
The left side is the total electric flux Φ through the closed (Gaussian) surface. Thus
Φ=ε0qenc
…
- CBSE 2025Set D1 markMCQQ.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.
…
- CBSE 2025Set A1 markQ.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 …
- CBSE 2025Set ANNUAL1 markMCQQ.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 ΦE=E⋅A (or ∫E⋅dA for a general surface).
- SI unit of electric field E = N C−1
- SI unit of area A = m2 …
- CBSE 2025Set ANNUAL1 markMCQQ.The SI unit of electric flux is:(a) Nm2C−2(b) NC−1m2(c) CN2m−1(d) C2N−1m−2
›Reveal solutionSolution
Electric flux ΦE=E⋅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 ΦE=∮E⋅dA, 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 (m2), the unit of electric flux is:
NC−1×m2=NC−1m2
…
- CBSE 2025Set ANNUAL1 markMCQQ.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 ΦE=∮E⋅dA has SI unit V·m (equivalent to N·m2/C).
The surface integral of the electric field, ΦE=∮E⋅dA, is the electric flux. Since E has SI unit V/m (or equivalently N/C) and area has unit m2, the flux has uni …
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