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 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.
So [Φ]=mV×m2=V⋅m (equivalently N·m²/C).
✓Final answer(A) Vm.
- 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.
Since the two spheres have equal σ, their surface field intensities are equal:
E2E1=σ/ε0σ/ε0=1:1.
✓Final answer(D) 1 : 1.
- 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 has no direction of its own. Summing (integrating) scalar quantities over the surface still gives a scalar. Its SI unit is N·m²/C (equivalently V·m).
✓Final answer(a) scalar quantity.
- 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 times the area, so the electric field at the new surface drops to 1/4, but the total flux (field x area, integrated) stays exactly the same.
✓Final answer(c) The outward electric flux remains unchanged, since it depends only on the enclosed charge Q, not on the radius R.
- 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
E(4πr2)=ε0Q⟹E=4πε01r2Q
As r increases (for r>R), E∝1/r2, so E decreases monotonically — exactly as it would for a point charge Q placed at the centre.
Conclusion
- Inside (r<R): E=0 (stays zero — not increasing, not decreasing).
- Outside (r>R): E decreases as r increases.
This matches option (b).
✓Final answer(b) is zero as r increases for r<R, and decreases as r increases for r>R.
- 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
(V/m) × m² = V·m (volt × metre), which is also N·m²/C.
This matches Column B entry (iv).
✓Final answer(iv) Volt × meter.
- 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
The flux is directly proportional to the net charge enclosed and is independent of the shape of the surface or the location of the charge within it.
✓Final answer(A) proportional to the charge enclosed.
- 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.
Therefore Φ = 0/ε₀ = 0. (Field lines from each +q terminate on its own −q inside the surface, so no net flux escapes.)
✓Final answer(D) zero.
- 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 of electrostatic equilibrium (no current flow). This redistribution happens almost instantaneously and is why conductors are used for electrostatic shielding.
✓Final answerTrue.
- 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
So the SI unit of ΦE = N C−1 × m2 = N C−1 m2 (this is also equivalent to V·m, since E can also be expressed in V/m).
✓Final answer(c) N C^-1 m^2.
- 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
This is the same physical unit as Nm2C−1 (order of writing does not matter), but among the given options only (b) is written correctly as NC−1m2; option (a) incorrectly shows C−2.
✓Final answerNC−1m2 — option (b).
- 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 unit (V/m)×m2=V⋅m (this is dimensionally the same as N⋅m2/C).
✓Final answer(a) Vm.
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