Q.Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0×10−22C/m2. What is E:
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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: …
Concept: Gauss’s law for parallel plate capacitors — fields from each plate superpose.
Reasoning:
-
Each large plate alone produces a field of magnitude 2ε0σ on either side, directed away from positive charge and toward negative charge.
-
For two plates with equal and opposite surface charge densities σ=17.0×10−22C/m2, the fields add in the region between the plates and cancel in the outer regions.
-
Using ε0=8.85×10−12C2/N⋅m2:
- Outer region of first plate: fields from both plates are equal and opposite → net E=0.
- Outer region of second plate: same cancellation → net E=0. …
For two oppositely charged parallel plates the field is zero outside both plates and E=σ/ε0≈1.9×10−10 N/C between them (directed from the positive to the negative plate).
Field of the pair. Each plate produces a field σ/2ε0. Outside the pair the two fields are opposite and cancel; between the plates they add.
- Outer region of first plate: E=0.
- Outer region of second plate: E=0.
- Between the plates: the fields add, …
Method: Gauss's Law for Parallel Plate Capacitors
This problem uses Gauss's Law applied to the superposition principle for infinite charged sheets.
Key Concept
For a single infinite thin sheet with surface charge density σ, the electric field is:
E=2ε0σ
directed away from the sheet if σ>0, and toward the sheet if σ<0.
Given Data
- Plate 1 inner face: +σ=+17.0×10−22C/m2
- Plate 2 inner face: −σ=−17.0×10−22C/m2
- ε0=8.85×10−12C2/N⋅m2
Steps
Step 1: Identify the three regions
- Region (a): Outer side of plate 1 (left of plate 1)
- Region (b): Outer side of plate 2 (right of plate 2)
- Region (c): Between the plates
Step 2: Apply superposition
Each plate produces its own field. The net field is the vector sum of fields from both plates.
Step 3: Calculate field magnitude
2ε0σ=2×8.85×10−1217.0×10−22=9.6×10−11N/C
Step 4: Determine direction and net field in each region
- (a) Outer region of first plate: Fields from both plates point left (away from + plate, toward – plate). They cancel. Ea=0 …
Here are the common mistakes students make when solving this classic Gauss’s law problem, along with how to avoid each.
Mistake 1: Forgetting that the field due to a single plate is 2ε0σ, not ε0σ
Why it happens
Students often memorise the formula for an infinite sheet as ε0σ, but that’s the total field on one side when using a Gaussian pillbox that encloses both surfaces. For a single thin conducting plate, the field on one side is half that: 2ε0σ.
How to avoid
- Draw a Gaussian pillbox that cuts through the plate. The flux goes out both sides, so 2EA=ε0σA → E=2ε0σ.
- Remember: one plate → half the flux per side.
Mistake 2: Adding fields without considering direction (sign convention)
Why it happens
Students treat all fields as positive magnitudes and add them algebraically, ignoring that fields from opposite charges point in opposite directions.
How to avoid
- Always draw a diagram. Mark the direction of E from each plate (away from positive, toward negative).
- Use a sign convention (e.g., right = positive). Then add vectors, not magnitudes.
- For this problem:
- Outer region of plate 1: fields from both plates point away from plate 1 → they add.
- Between plates: fields point in opposite directions → they subtract.
Mistake 3: Using the wrong value of σ for each plate
Why it happens
The problem gives one magnitude 17.0×10−22C/m2, but the plates have opposite signs. Students sometimes use +σ for both.
How to avoid
- Label plate 1 with +σ and plate 2 with −σ (or vice versa).
- The magnitude is the same, but the sign matters for direction.
- In the formula E=2ε0∣σ∣, the sign only tells you the direction — keep the magnitude positive, then assign direction from the diagram.
Mistake 4: Forgetting that the field is zero inside a conductor in electrostatic equilibrium
Why it happens
Students try to compute the field inside the metal plates using the same formulas, not realising that charges rearrange to cancel the internal field.
How to avoid
- Recall: Inside a conductor in equilibrium, E=0.
- The given surface charges are on the inner faces only. The field inside the metal is zero — that’s why all charge sits on the surfaces.
Mistake 5: Not using superposition correctly for the three regions
Why it happens …
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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