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 16 on this concept.
- GUJCET 2026Set x1 markMCQQ.If charge q is placed on one of the vertex of a cube, then total electric flux passing through the cube is ______. (A) ε0q (B) 8ε0q (C) 4ε0q (D) 24ε0q
›Reveal solutionSolution
[!TLDR]
The numerator is the derivative of the denominator, so the integral is log∣ex+e−x∣+C — option (C).
Concept
Whenever an integrand has the form f(x)f′(x), the integral is log∣f(x)∣+C. Here take f(x)=ex+e−x, whose derivative is exactly ex−e−x.
Solution
Let u=ex+e−x. Then du=(ex−e−x)dx, and
∫ex+e−xex−e−xdx=∫udu=log∣u∣+C=log∣ex+e−x∣+C. …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.A point charge of 2.0 microC is at the centre of a cubic Gaussian surface 9.0 cm on edge. The net electric flux through the surface is ___ Nm^2/C.(a) 2.2 x 10^-6(b) 2.2 x 10^5(c) 2.2 x 10^6(d) 2.2 x 10^-5
›Reveal solutionSolution
Gauss's law states the net electric flux through any closed surface is q_enclosed / epsilon_0, regardless of the surface's shape or size (as long as it encloses the same charge).
phi = q / epsilon_0
Given q = 2.0 microC = 2.0 x 10^-6 C, epsilon_0 = 8.85 x 10^-12 C^2/(N m^2).
phi = (2.0 x 10^-6) / (8.85 x 10^-12) = 2.26 x 10^5 N m^2/C
…
- GUJCET 2025Set 031 markMCQQ.The electric field due to point charge 2q at a distance r is E. Now, charge q is uniformly distributed over a thin spherical shell of radius R, the electric field at a distance 2r (r≫R) from the centre of the thin spherical shell is E′= ______. (A) 4E (B) 2E (C) E (D) 2E
›Reveal solutionSolution
[!TLDR]
Using the shell theorem, E′=4kq/r2=2E.
Concept
A uniformly charged thin spherical shell produces, at any external point, the same field as if all its charge were concentrated at the centre: E=d2kQ.
Solution
For the point charge: E=r2k(2q)=r22kq. …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.Consider a uniform electric field E = 3 x 10^3 î N/C. What is the flux of this field through a square of 10cm on a side whose plane is parallel to the xy plane?(a) 30 Nm^2/C(b) Zero(c) 15 Nm^2/C(d) 60 Nm^2/C
›Reveal solutionSolution
Electric flux Φ = E·A = EA cosθ, where θ is the angle between the field and the surface's normal vector.
E = 3 × 10³ x̂ N/C is directed along x. The square lies in a plane parallel to the xy-plane, so its normal vector is along z — perpe …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.A charge q is placed at the center of one of the faces of a cube. The electric flux linked with the cube is ______.(a) q/ε0(b) q/6ε0(c) q/2ε0(d) q/4ε0
›Reveal solutionSolution
Gauss's law gives total enclosed charge → flux, but here the charge sits exactly on a face, not fully inside the cube.
Imagine a second identical cube placed mirror-symmetric on the other side of that face, so together the two cubes fully enclose the charge q. By symmetry, each cube receives exactly half the total flux …
- GUJCET 2024Set 131 markMCQQ.The Dimensional formula for Electric Flux is ________. (A) M1L3T−3A1 (B) M1L1T−3A−1 (C) M−1L−3T3A1 (D) M1L3T−3A−1
›Reveal solutionSolution
Electric field has dimensions MLT−3A−1; multiplying by area L2 gives electric flux =M1L3T−3A−1.
Concept. Electric flux ΦE=E⋅A. Electric field E=qF has dimensions ATMLT−2=MLT−3A−1. …
- GUJCET 2024Set 131 markMCQQ.An infinite line charge produces an electric field of 9×104 N/C at a distance of 2 cm. Then the linear charge density will be ________. (K=9×109 Nm2/C2) (A) 0.1μC/m (B) 10μC/m (C) 0.01μC/m (D) 1μC/m
›Reveal solutionSolution
Using E=r2Kλ, solve for λ=2KEr=10−7 C/m =0.1μC/m.
Concept. The field of an infinite line charge is E=r2Kλ=2πε0rλ.
Steps. With E=9×104 N/C, r=0.02 m, K=9×109: …
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.If an electric charge 'q' is placed at the centre of a cube, then the flux associated with each surface of the cube is ___.(a) q/ε0(b) q/6ε0(c) q/4ε0(d) q/2ε0
›Reveal solutionSolution
By Gauss's law, total flux through a closed surface enclosing charge q is q/ε0; a cube has 6 identical faces symmetric about the centre.
Total flux through the cube (Gauss's law) = q/ε0.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.If two infinite plane sheets having same surface charge density σ are placed parallel to each other, then the electric field between the two sheets is ___.(a) zero(b) σ/ε0(c) σ/2ε0(d) 2σ/ε0
›Reveal solutionSolution
Each infinite charged sheet produces a uniform field of magnitude σ/2ε0 pointing away from it (for positive σ) on both sides.
Between the two sheets, the field due to the left sheet points away from it (rightward, into the gap) with magnitude σ/2ε0, while the field due to the right sheet points away from it (leftward, into the gap) with the same magnitude σ/2ε0. Since both sheets carry the same sign and magnitude of charge d …
- GUJCET 2023Set 091 markMCQQ.Consider a uniform electric field E=3×103k^ N/C. The electric flux of this field through a square of 20 cm on a side whose plane is parallel to yz plane is ______ Nm2/C. (A) 90 (B) 120 (C) 60 (D) Zero
›Reveal solutionSolution
[!TLDR]
The field is along k^ while the area vector is along i^, so the flux is zero.
Concept
Electric flux through a flat surface is Φ=E⋅A=EAcosθ, where θ is the angle between the field and the outward normal (area vector).
Solution
The square lies in a plane parallel to the yz-plane, so its normal (area vector) is along the x-axis, A=Ai^.
The field is E=3×103k^ N/C. …
- GUJCET 2023Set 091 markMCQQ.Figure shows the electric field lines of four point charges A, B, C and D. [FIGURE: A has 3 field lines; B and C are joined by many field lines (dipole-like) with C also having outward lines; D has 4 field lines] Which charge has the maximum magnitude? (A) C charge (B) B charge (C) A charge (D) D charge
›Reveal solutionSolution
Field-line count ∝ ∣q∣; charge C has the most lines, so the largest magnitude.
Concept — field lines and charge magnitude. The number of electric field lines starting from (or ending on) a charge is proportional to the magnitude of that charge. Counting: A has 3 lines, D has 4 lines, while charges B and C are linked by many lines (a dipole-like pair) with C additionally showing outg …
- GUJCET 2022Set 171 markMCQQ.Dimensional formula of Electric flux = ________. (A) M1L−3T−3A−1 (B) M1L3T3A−1 (C) M1L3T−3A−1 (D) M−1L3T−3A−1
›Reveal solutionSolution
ΦE=E⋅A; with [E]=MLT−3A−1 and area L2, flux is M1L3T−3A−1.
Concept: Electric field E=chargeforce=ATMLT−2=MLT−3A−1. …
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