Q.When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
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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: …
The key idea is that charge is neither created nor destroyed — it is only transferred. Rubbing transfers electrons from one body to the other, leaving one positively charged and the other negatively charged.
- Before rubbing, both the glass rod and the silk cloth are electrically neutral — the total charge is zero.
- When rubbed, electrons (negative charge) move from the glass rod to the silk cloth.
- The glass rod loses electrons and becomes positively charged; the silk cloth gains those electrons and becomes negatively charged.
- The magnitude of charge on each is exactly equal, so the net charge of the system (rod + cloth) remains zero — unchanged from the initial state. …
Rubbing transfers electrons from one body to the other, creating equal and opposite charges. The total charge before and after remains zero — this is the law of conservation of charge in action.
When you rub a glass rod with silk, the rod becomes positively charged and the silk negatively charged. This isn’t magic — it’s a transfer of electrons. The glass rod loses electrons to the silk, so the rod ends up with a net positive charge (more protons than electrons), and the silk gains those electrons, giving it a net negative charge.
The key point is that charge is never created or destroyed — it’s only moved from one place to another. Before rubbing, both the rod and the silk were electrically neutral. After rubbing, the total charge of the system (rod + silk) is still zero: the positive charge on the rod is exactly equal in magnitude to the negative charge on the silk.
This is exactly what the law of conservation of charge states: the net electric charge of an isolated system remains constant. Here, the system is the rod and silk together — no charge enters or leaves the pair.
Let’s walk through it step by step.
- Initial state: Both the glass rod and the silk cloth are neutral. That means each has an equal number of protons and electrons. The total charge of the system is
Qinitial=0+0=0.
-
During rubbing: The friction causes electrons to be transferred from the glass rod to the silk cloth. Glass holds its electrons less tightly than silk does, so electrons move from glass to silk.
- The glass rod loses, say, n electrons. Each electron has charge −e, so the rod’s charge becomes +ne (since it now has n more protons than electrons).
- The silk gains those n electrons, so its charge becomes −ne.
-
Final state: The rod has charge +ne, the silk has charge −ne. The total charge of the system is
Qfinal=(+ne)+(−ne)=0. …
Method: Law of Conservation of Charge — Explanation via Charge Transfer
This is not a numerical problem but a conceptual explanation. The method below shows how to reason step-by-step using the principle of charge conservation.
Step 1: Identify the initial state
- Before rubbing, both the glass rod and the silk cloth are electrically neutral.
- This means the total charge on each is zero:
Qrod, initial=0,Qsilk, initial=0
- Therefore, the total charge of the system (rod + silk) is also zero.
Step 2: Describe what happens during rubbing
- Rubbing transfers electrons from one material to the other.
- Glass loses electrons → becomes positively charged.
- Silk gains those electrons → becomes negatively charged.
Step 3: Apply the law of conservation of charge
- The total charge of an isolated system remains constant.
- Initially:
Qtotal, initial=0
- After rubbing: Let charge on glass rod = +q …
Here’s a breakdown of the common mistakes students make when connecting the rubbing of a glass rod with silk to the law of conservation of charge, along with how to avoid each.
Mistake 1: Thinking charge is created during rubbing
What students do wrong:
They see that the rod becomes positively charged and the cloth becomes negatively charged, and assume that rubbing “produces” charge out of nowhere.
Why it’s wrong:
Charge is never created or destroyed — it is only transferred. The total charge before rubbing is zero, and after rubbing it is still zero.
How to avoid:
Always remember: Rubbing does not create charge; it separates charge.
- Before rubbing: rod (neutral) + cloth (neutral) → total charge = 0
- After rubbing: rod (+Q) + cloth (−Q) → total charge = 0
Key idea: The net charge of an isolated system remains constant.
Mistake 2: Ignoring the sign of the charges
What students do wrong:
They say “the rod gains positive charge and the cloth gains negative charge” without checking that the magnitudes are equal.
Why it’s wrong:
Conservation of charge requires that the algebraic sum stays zero. If the rod gets +5μC, the cloth must get exactly −5μC — not −4μC or −6μC.
How to avoid:
Always write the equation:
Qrod+Qcloth=0
So if Qrod=+q, then Qcloth=−q.
Mistake 3: Confusing “conservation of charge” with “charge is constant on each body”
What students do wrong:
They think that because the rod ends up positive, it “keeps” that charge forever, and that this is the conservation law.
Why it’s wrong:
Conservation of charge applies to the total charge of an isolated system, not to individual objects. The rod can later lose its charge by touching a conductor — that’s fine, as long as the total charge in the system remains unchanged.
How to avoid:
- Conservation of charge: ∑Qinitial=∑Qfinal for the whole system.
- Individual objects can gain or lose charge — that’s transfer, not violation.
Mistake 4: Not linking the observation to the law explicitly
What students do wrong:
They describe the experiment but never state: “This shows that charge is conserved.”
Why it’s wrong:
The question asks how the observation is consistent — you must explicitly connect the observation to the law.
How to avoid:
End your explanation with a clear sentence:
“Thus, the total charge before and after rubbing is zero, which is consistent with the law of conservation of charge.”
Mistake 5: Using Gauss’s law incorrectly in this context
What students do 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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