Q.A conducting sphere of radius 10cm has an unknown charge. If the electric field 20cm from the centre of the sphere is 1.5×103N/C and points radially inward, what is the net charge on the sphere?
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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 — the net electric flux through a closed surface equals ε0Qenc.
Step 1: Choose a spherical Gaussian surface of radius r=20 cm=0.2 m, concentric with the conducting sphere. The field is radial and uniform over this surface.
Step 2: The electric flux is
Φ=E⋅4πr2
Given E=1.5×103 N/C (inward, so negative if outward is positive), but magnitude is enough — sign will come from direction.
Step 3: By Gauss’s law,
Qenc=ε0Φ=ε0⋅E⋅4πr2
Substitute values:
Q=(8.85×10−12)×(1.5×103)×4π×(0.2)2 …
Using Gauss’s law, the electric field outside a conducting sphere is the same as that of a point charge at the centre. The inward field tells us the charge is negative. The net charge is found to be q=−6.67×10−9C.
The key idea is that for a conducting sphere, any excess charge resides entirely on its surface. Outside the sphere, the electric field behaves exactly as if all that charge were concentrated at the centre. This is a direct consequence of spherical symmetry and Gauss’s law.
Why does this matter? Because it means we can treat the sphere as a point charge when calculating the field at any point outside it. The problem gives us the field at a distance of 20cm from the centre — that’s outside the sphere (radius 10cm), so the point-charge model is valid.
The field points radially inward. That’s a crucial detail: it tells us the charge is negative. A positive charge would produce an outward field.
Now let’s work through the calculation.
-
Identify the relevant distance.
The sphere’s radius is R=10cm=0.10m.
The point where the field is given is r=20cm=0.20m from the centre.
Since r>R, we are outside the sphere.
-
Apply Gauss’s law for a spherical Gaussian surface.
For a spherically symmetric charge distribution, the electric field at distance r from the centre is:
E=4πε01⋅r2∣q∣
where q is the net charge enclosed. For a conducting sphere, all charge is on the surface, so the enclosed charge is just the net charge on the sphere.
- Plug in the known values. We have E=1.5×103N/C and r=0.20m. The constant 4πε01=9×109N⋅m2/C2. So:
1.5×103=(9×109)⋅(0.20)2∣q∣
- Solve for ∣q∣. First, (0.20)2=0.04. Then: ∣q∣=9×1091.5×103×0.04 …
Method: Gauss's Law
Why Gauss's Law?
For a spherically symmetric charge distribution, the electric field at any point depends only on the net charge enclosed inside a spherical Gaussian surface. This lets us find the charge without knowing the internal distribution.
Steps
Step 1: Choose a Gaussian surface
Draw a spherical surface of radius r=20 cm (where the field is given), concentric with the conducting sphere.
Step 2: Apply Gauss's Law
Gauss's Law states:
∮E⋅dA=ε0Qenc
For a spherical surface, E is radial and constant in magnitude over the surface, so:
E⋅(4πr2)=ε0Qenc
Step 3: Identify the enclosed charge
The Gaussian surface at 20 cm encloses the entire conducting sphere (radius 10 cm). So Qenc is the net charge on the sphere — this is what we need.
Step 4: Solve for Q
Q=ε0⋅E⋅4πr2
Step 5: Plug in values
- r=20 cm=0.20 m
- E=1.5×103 N/C (inward direction)
- ε0=8.85×10−12 C2/N⋅m2 …
1. ✗ Mistake: Forgetting the sign of the charge
What students do wrong:
They compute the magnitude of charge correctly but write the answer as positive.
Why it’s wrong:
The electric field points radially inward. For a spherical Gaussian surface, an inward field means the net flux is negative (field lines enter the surface). By Gauss’s law:
ΦE=ε0qenc
If ΦE<0, then qenc<0.
✓ How to avoid:
Always check the direction of the field relative to the outward normal of your Gaussian surface.
- Outward field → positive charge
- Inward field → negative charge
Here, inward field → negative charge.
2. ✗ Mistake: Using the wrong radius for the Gaussian surface
What students do wrong:
They take the Gaussian surface radius as 10cm (the sphere’s radius) instead of 20cm (where the field is given).
Why it’s wrong:
Gauss’s law requires you to draw a closed surface that passes through the point where you know the field. The field is given at 20cm, so your Gaussian sphere must have radius r=20cm=0.2m.
✓ How to avoid:
- The Gaussian surface radius = distance from centre to the point where E is known.
- The sphere’s own radius only matters to decide if the point is inside or outside the conductor.
Here, 20cm>10cm, so the point is outside — the sphere behaves like a point charge at the centre.
3. ✗ Mistake: Using the formula for a point charge incorrectly
What students do wrong:
They write E=r2kq but forget to convert cm to metres, or they use r=10cm.
✓ How to avoid:
Always convert to SI units:
- r=20cm=0.20m
- Use k=9×109 N m2/C2 or ε0=8.85×10−12 C2/N m2
Then:
E=r2k∣q∣⇒∣q∣=kEr2
Plug in:
∣q∣=9×109(1.5×103)(0.20)2=9×109(1.5×103)(0.04)=9×10960=6.67×10−9 C
So q=−6.67×10−9 C (negative because field is inward).
4. ✗ Mistake: Forgetting that the sphere is a conductor
What students do wrong:
They worry about charge distribution inside the sphere.
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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