Q.The electric field components in Fig. 1.24 are Ex=αx1/2, Ey=Ez=0, in which α=800N/C m1/2. Calculate
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.
3 — Spherical shell / sphere. For a thin shell of charge Q, a Gaussian sphere inside encloses nothing, so E = 0 everywhere within; outside, the charge acts as if concentrated at the centre, E = kQ/r² — indistinguishable from a point charge. For a solid uniformly charged sphere, an interior surface encloses only the charge within radius r, giving E ∝ r (rising linearly from zero at the centre) up to the surface, then 1/r² beyond.
Field just outside a conductor. A charged conductor holds all its charge on the surface with E = 0 inside, so a straddling pillbox gives E = σ / ε₀ just outside — twice the sheet result, because all the flux escapes on the one outer face.
How it's examined. JEE questions test whether you can spot the symmetry, pick the right surface, and recall which result scales as 1/r, which is flat, and which is 1/r². The physics is always the one line Φ = q_enclosed / ε₀, and the skill is knowing that only the enclosed charge — never the far-off one — ever matters.
"Gauss law class 12 physics derivation" and "electric field due to infinite sheet using Gauss law" are heavily searched terms, since this is one of the core results of the Electrostatics chapter in the NCERT/CBSE Class 12 Physics curriculum. Gauss's law applications for spheres, sheets, and line charges are near-guaranteed questions in JEE Main, NEET, and state CETs.
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:
∮E⋅dA=∮E1⋅dA+∮E2⋅dA+⋯=ε0q1+ε0q2+⋯=ε0Qenc
Charges outside the surface contribute zero net flux — their field lines enter and exit the surface, cancelling out.
5. The Final Law
∮SE⋅dA=ε0Qenc
Why it's profound:
- It relates a global property (flux through a surface) to a local source (charge inside).
- It's true for any closed surface, not just symmetric ones.
- It's a direct consequence of Coulomb's inverse-square law — the 1/r2 dependence is essential for the cancellation.
6. Quick Exam Tip
| Situation | What to remember |
|---|---|
| Point charge | Flux = q/ε0 through any enclosing surface |
| Dipole inside | Net flux = 0 (equal + and -) |
| Charge outside | Flux contribution = 0 |
| Symmetric surfaces | Use Gauss's law to find E easily |
Key takeaway: Gauss's law holds because the electric field from a point charge obeys the inverse-square law, making the flux through any closed surface independent of the surface's shape — it depends only on the total charge enclosed.
Only the two faces perpendicular to the x‑axis carry flux, since Ey=Ez=0 and Ex=αx1/2 is constant on each such face.
Left face (x=a, outward normal −x^): ΦL=−αa1/2a2=−αa5/2.
Right face (x=2a, outward normal +x^): ΦR=+α(2a)1/2a2=2αa5/2.
- Net flux:
With α=800N/C⋅m1/2 and a=0.1m, a5/2=3.16×10−3:
Φ=ΦL+ΦR=αa5/2(2−1).
Φ=800(3.16×10−3)(0.414)≈1.05N⋅m2/C.
- Enclosed charge (Gauss's law):
q=ε0Φ=(8.854×10−12)(1.05)≈9.27×10−12C.
✓Final answerNet flux Φ≈1.05N⋅m2/C; enclosed charge q≈9.27×10−12C.
Only the two faces ⊥ to the x‑axis carry flux; with Ex=αx1/2 the net flux is Φ=αa5/2(2−1)≈1.05N⋅m2/C, and by Gauss's law the enclosed charge is q=ε0Φ≈9.27×10−12C.
The field points only along x, so flux passes only through faces whose normal has an x‑component. For the axis‑aligned cube, those are the left face at x=a and the right face at x=2a; the four faces parallel to the x‑axis contribute nothing because E⋅dA=0 there.
Left face (x=a). Here Ex=αa1/2 is uniform over the face of area a2, and the outward normal points in −x^:
ΦL=−(αa1/2)a2=−αa5/2.
Right face (x=2a). Now Ex=α(2a)1/2=2αa1/2, and the outward normal points in +x^:
ΦR=+(2αa1/2)a2=2αa5/2.
- Net flux.
With α=800N/C⋅m1/2 and a=0.1m,
Φ=ΦL+ΦR=αa5/2(2−1).
a5/2=(0.1)5/2=3.162×10−3,2−1=0.4142,
Φ=800×3.162×10−3×0.4142≈1.05N⋅m2/C.
Watch outNote a5/2=a2a, not a3/2 — the extra a2 is the face area. Keep the minus sign on the left face, or the two contributions wrongly add.
- Enclosed charge. Gauss's law Φ=q/ε0 gives
q=ε0Φ=(8.854×10−12)(1.05)≈9.27×10−12C.
✓Final answer- Net flux Φ≈1.05N⋅m2/C.
- Enclosed charge q≈9.27×10−12C.
Method: Gauss's Law Flux Calculation via Surface Integration
This problem uses Gauss's Law in integral form:
ΦE=∮E⋅dA=ε0qenc
Step 1: Identify the non-zero field contribution
Given:
- Ex=αx1/2, where α=800 N/C m1/2
- Ey=Ez=0
- Cube side length a=0.1 m, placed with one corner at origin
Since only Ex is non-zero, flux only passes through faces perpendicular to the x-axis — the left face (at x=0) and the right face (at x=a).
Step 2: Calculate flux through each x-face
Right face (x=a=0.1 m):
- Area vector: dA=i^dydz (outward normal is +i^)
- Field at this face: Ex=αa1/2 (constant over the face)
- Flux:
Φright=∫ExdA=αa1/2×a2=αa5/2
Left face (x=0):
- Area vector: dA=−i^dydz (outward normal is −i^)
- Field at this face: Ex=α(0)1/2=0
- Flux: Φleft=0
Step 3: Total flux through the cube
ΦE=Φright+Φleft=αa5/2+0
Substitute values:
ΦE=800×(0.1)5/2=800×(0.1)2×(0.1)1/2
Since (0.1)1/2=0.1≈0.3162:
ΦE=800×0.01×0.3162=8×0.3162
ΦE=2.53 N m2/C
Step 4: Find enclosed charge using Gauss's Law
qenc=ε0ΦE
ε0=8.85×10−12 C2/N m2
qenc=(8.85×10−12)×2.53
qenc=2.24×10−11 C
Final Answer Summary
| Quantity | Value |
|---|---|
| Flux through cube | 2.53 N m2/C |
| Charge inside cube | 2.24×10−11 C |
Common Mistakes Students Make with This Gauss Law Problem
Mistake 1: Forgetting That Flux Depends Only on the Perpendicular Component
The error: Students often try to integrate Ex over all six faces of the cube, including faces where the field is parallel to the surface.
Why it's wrong: Flux through a surface is Φ=∫E⋅dA. Since Ey=Ez=0, only the two faces perpendicular to the x-axis contribute. The four side faces (parallel to the x-axis) have zero flux because E⋅dA=0 there.
How to avoid: Always check which field components are non-zero. If Ey=Ez=0, only faces with normals along i^ matter. Draw the cube and label each face's normal vector.
Mistake 2: Using the Same x Value for Both Faces
The error: Plugging x=a into Ex=αx1/2 for both the left and right faces.
Why it's wrong: The left face is at x=0, the right face is at x=a. The field strength is different at each location:
- Left face: Ex(0)=α⋅01/2=0
- Right face: Ex(a)=αa1/2
How to avoid: Write the coordinates explicitly:
- Left face: x=0, area vector dA=−dAi^
- Right face: x=a, area vector dA=+dAi^
Then compute each flux separately.
Mistake 3: Ignoring the Direction of the Area Vector
The error: Treating both faces as having +dAi^ and getting zero net flux.
Why it's wrong: By convention, the area vector points outward from the closed surface:
- Left face: outward normal is −i^, so dA=−dAi^
- Right face: outward normal is +i^, so dA=+dAi^
The flux through the left face is:
Φleft=∫E⋅dA=∫(Exi^)⋅(−dAi^)=−∫ExdA
How to avoid: Always draw outward normals on each face before computing dot products.
Mistake 4: Forgetting That Ex Varies Over the Face
The error: Treating Ex as constant over the entire right face and writing Φ=Ex⋅A.
Why it's wrong: Ex=αx1/2 depends on x. On the right face, x=a is constant, so this actually works here — but only because the face is perpendicular to the x-axis. Students often carry this habit to problems where the field varies across the face.
How to avoid: Check if the field component is constant over the surface. Here, since the right face is at fixed x=a, Ex is uniform across it. But be cautious — this is a special case.
Mistake 5: Incorrectly Computing the Net Flux
The error: Adding magnitudes without signs, e.g., Φnet=αa1/2⋅a2+0=αa5/2.
Why it's wrong: The left face contributes negative flux because E points inward there (field enters the cube). The correct calculation:
Φnet=Φleft+Φright=−α(0)1/2⋅a2+αa1/2⋅a2=αa5/2
The left face has Ex=0, so its flux is zero. The net flux is just from the right face: Φnet=αa5/2.
How to avoid: Compute each face's flux with its correct sign, then sum. Don't shortcut.
Mistake 6: Using the Wrong Formula for Charge from Flux
The error: Writing q=Φ⋅ε0 instead of q=Φε0.
Why it's wrong: Gauss's law states:
Φ=ε0qenc⇒qenc=Φε0
How to avoid: Memorize the exact form: flux = charge enclosed divided by epsilon-zero. Rearrange carefully.
Mistake 7: Unit Errors in the Final Answer
The error: Reporting flux in N/C or charge in C without checking dimensions.
Why it's wrong: Flux has units N⋅m2/C. With α=800N/C⋅m1/2 and a=0.1m:
Φ=αa5/2=800⋅(0.1)5/2=800⋅(0.1)2⋅(0.1)1/2=800⋅0.01⋅0.316=2.53N⋅m2/C
Then q=Φε0=2.53×8.85×10−12=2.24×10−11C.
How to avoid: Track units at every step. Write the unit of each quantity before substituting numbers.
Quick Checklist to Avoid These Mistakes
| Step | What to Check |
|---|---|
| 1 | Which field components are non-zero? |
| 2 | Which faces have flux? (Only those with E⊥ face) |
| 3 | What is the x-coordinate of each contributing face? |
| 4 | What is the outward normal direction for each face? |
| 5 | Is the field constant over the face? |
| 6 | Did I include the correct sign in the dot product? |
| 7 | Did I use q=Φε0 correctly? |
| 8 | Are the final units consistent? |
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.
The only option that is the correct antiderivative of this standard integrand is (C).
[!ANSWER]
(C) log∣ex+e−x∣
NoteThis solution was worked out by our team and independently cross-checked by a second solve. The official answer key on record for this question could not be confirmed, so please cross-verify with the official paper where possible.
- 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
(The 9.0 cm edge length of the cube is a distractor - it does not matter for the total flux, only the enclosed charge does.)
✓Final answer(b) 2.2 x 10^5.
- 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.
For the shell of charge q, at distance r/2 from the centre (which is ≫R, hence external):
E′=(r/2)2kq=r2/4kq=r24kq.
Compare: EE′=2kq/r24kq/r2=2, so E′=2E.
[!ANSWER]
(B) 2E
- 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 — perpendicular to E. Hence θ = 90°, cosθ = 0, and Φ = 0 regardless of the square's area.
✓Final answer(b) Zero.
- 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 q/ε0 that q would produce through a fully enclosing surface. So flux through the original cube = (1/2)(q/ε0) = q/2ε0.
✓Final answer(c) q/2ε0.
- 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.
Steps.
[ΦE]=[E][A]=(MLT−3A−1)(L2)=M1L3T−3A−1.
✓Final answer(D) M1L3T−3A−1
ANSWER: (D)
- 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:
λ=2KEr=2(9×109)(9×104)(0.02)=1.8×10101800=10−7 C/m=0.1μC/m.
✓Final answer(A) 0.1μC/m
ANSWER: (A)
- 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.
Since the charge is at the centre, by symmetry each of the 6 faces receives an equal share of this flux.
Flux per face = (q/ε0)/6 = q/6ε0.
✓Final answer(b) q/6ε0.
- 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 density, these two contributions are equal and opposite in the gap, so they cancel exactly.
(Outside the pair, on either side, the two fields add to give σ/ε0.)
✓Final answer(a) zero.
- 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.
Φ=E⋅A=(3×103k^)⋅(Ai^)=0,
since k^⋅i^=0 (the field is parallel to the plane, cutting no field lines through it).
[!ANSWER]
(D) Zero
- 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 outgoing lines — C is associated with the largest number of field lines.
Therefore charge C has the maximum magnitude.
✓Final answerOption (A) C charge
ANSWER: (A)
- 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.
Electric flux ΦE=E×area=MLT−3A−1×L2=M1L3T−3A−1.
✓Final answer(C) M1L3T−3A−1
ANSWER: (C)
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