E = λ / (2πε₀ r), where λ is linear charge density and r is perpendicular distance. Decreases as 1/r.
-- NCERT Class 12 Physics, Ch. 1, p. 34Field Long Wire Gauss
8 MCQs1 revision card9-step worked example
Source: NCERT ElectrostaticsOfficial key: NTA-verifiedLast updated: 21 Sep 2026
Field Long Wire Gauss, explained for NEET
The trap: treating a long charged wire like a point charge and writing a 1/r² field. For an infinitely long wire the field falls as 1/r, not 1/r². Double the distance and the field halves; it does not drop to a quarter. A second slip is dropping the factor 2 and writing λ/(4πε₀r).
The result. For an infinitely long thin straight wire with uniform linear charge density λ, the field at perpendicular distance r is
E = λ / (2πε₀ r), directed along the radial unit vector (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.32, page 34)
The field points radially outward if λ is positive and inward if λ is negative (page 34).
Why Gauss's law gives it quickly (Section 1.14.1, pages 33–34).
- Symmetry: the field is radial and its magnitude depends only on r.
- Take a cylindrical Gaussian surface of radius r and length l, coaxial with the wire.
- The field is parallel to the two flat ends, so the flux through them is zero. On the curved part, E is normal to the surface and constant, so flux = E × 2πrl.
- Charge enclosed = λl. Gauss's law: E × 2πrl = λl/ε₀, so E = λ/(2πε₀r). The length l cancels.
Two NCERT fine points (page 34). Only the enclosed charge λl enters Gauss's law, but the field E is due to the charge on the entire wire. And the infinite-length assumption is essential: without it E cannot be taken normal to the curved surface. For a real long wire, Eq. 1.32 is approximately true near its central portion, where end effects can be ignored.
Bridge to NEET. Useful form: 1/(2πε₀) = 2 × 1/(4πε₀) ≈ 1.8 × 10¹⁰ N m² C⁻², taking 1/(4πε₀) ≈ 9 × 10⁹ N m² C⁻² (page 7). Typical questions give λ and r and ask for E, ask for λ from E, or ask how E changes when r changes.
Watch out: in ratio questions use E₁/E₂ = r₂/r₁ (inverse first power), never the square.
Can you answer these Field Long Wire Gauss MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The magnitude of the electric field at perpendicular distance r from an infinitely long thin straight wire with uniform linear charge density λ is:
Show answer and why every option is right or wrong
Answer: C. Gauss's law with a coaxial cylindrical surface gives E × 2πrl = λl/ε₀, so E = λ/(2πε₀r) (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.32, page 34).
Why A is wrong: A is wrong because it is the point-charge form with λ in place of q; a line charge gives a 1/r field, and the constant is 1/(2πε₀), not 1/(4πε₀).
Why B is wrong: B is wrong because it keeps the correct constant but uses the point-charge 1/r² dependence; for an infinite wire the curved area 2πrl grows as r, so E falls only as 1/r.
Why D is wrong: D is wrong because it puts r in the numerator, which would make the field grow with distance from the wire; dividing flux by the curved area 2πrl puts r in the denominator.
MCQ 2Easy RecallPractice
In NCERT's Gauss's-law derivation for an infinitely long charged wire, a cylindrical Gaussian surface coaxial with the wire is used. The electric flux through its two flat end faces is:
Show answer and why every option is right or wrong
Answer: D. The field is everywhere radial, so it lies along the plane of each end face and the flux through both ends is zero; all the flux, E × 2πrl, passes through the curved part (NCERT Class 12 Physics Part I, Chapter 1, Section 1.14.1, page 33).
Why A is wrong: A is wrong because 2πrl is the area of the curved surface, not of the end faces; the flux E × 2πrl belongs to the curved part.
Why B is wrong: B is wrong because E × πr² would apply only if the field were normal to the end faces; a radial field is parallel to them, so E·ΔS = 0 there.
Why C is wrong: C is wrong because it swaps the roles of the surfaces: λl/ε₀ is the total flux, and it passes entirely through the curved surface, not the ends.
MCQ 3Easy RecallPractice
An infinitely long straight wire carries a uniform negative linear charge density (λ < 0). The electric field at a nearby point is directed:
Show answer and why every option is right or wrong
Answer: C. NCERT states that E is directed outward if λ is positive and inward if λ is negative; the direction is along the radial unit vector in the plane normal to the wire (NCERT Class 12 Physics Part I, Chapter 1, page 34).
Why A is wrong: A is wrong because outward is the direction for positive λ; for negative λ the field reverses and points towards the wire.
Why B is wrong: B is wrong because the components along the wire cancel for every pair of line elements; since the wire is infinite, the field has no component along its length.
Why D is wrong: D is wrong because a circular (tangential) direction is not an electrostatic field of a line charge; NCERT shows the field is radial in the plane normal to the wire.
MCQ 4Direct ApplicationPractice
A long straight thin wire has a uniform linear charge density of 4.0 × 10⁻⁷ C/m. Taking 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻², the magnitude of the electric field at a perpendicular distance of 0.20 m from the wire is:
Show answer and why every option is right or wrong
Answer: A. E = λ/(2πε₀r) = 2 × (9.0 × 10⁹) × (4.0 × 10⁻⁷)/0.20 = 7.2 × 10³/0.20 = 3.6 × 10⁴ N/C (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.32, page 34).
Why B is wrong: B is wrong because 1.8 × 10⁴ N/C comes from (9.0 × 10⁹)(4.0 × 10⁻⁷)/0.20, dropping the factor 2, i.e. using 1/(4πε₀) instead of 1/(2πε₀).
Why C is wrong: C is wrong because 9.0 × 10⁴ N/C comes from (9.0 × 10⁹)(4.0 × 10⁻⁷)/(0.20)², the point-charge form with a 1/r² dependence and no factor 2.
Why D is wrong: D is wrong because 1.8 × 10⁵ N/C comes from 2(9.0 × 10⁹)(4.0 × 10⁻⁷)/(0.20)², keeping the factor 2 but squaring r.
MCQ 5Direct ApplicationPractice
The electric field due to a long uniformly charged straight wire is 6.0 × 10⁴ N/C at a perpendicular distance of 5.0 cm. What is the field magnitude at a perpendicular distance of 15 cm from the same wire?
Show answer and why every option is right or wrong
Answer: B. E is proportional to 1/r, so E₂ = E₁ × r₁/r₂ = 6.0 × 10⁴ × (5.0/15) = 2.0 × 10⁴ N/C (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.32, page 34).
Why A is wrong: A is wrong because 6.7 × 10³ N/C comes from 6.0 × 10⁴ × (5.0/15)², using the point-charge 1/r² dependence instead of 1/r.
Why C is wrong: C is wrong because 1.8 × 10⁵ N/C comes from 6.0 × 10⁴ × 3, making the field grow with distance; tripling r divides E by 3.
Why D is wrong: D is wrong because it assumes the field does not depend on distance; for a line charge E = λ/(2πε₀r) clearly contains r.
MCQ 6Direct ApplicationPractice
The electric field at a perpendicular distance of 0.10 m from a long straight uniformly charged wire is 9.0 × 10⁴ N/C, directed away from the wire. Taking 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻², the linear charge density of the wire is:
Show answer and why every option is right or wrong
Answer: A. λ = 2πε₀rE = rE/[2 × 1/(4πε₀)] = (0.10 × 9.0 × 10⁴)/(1.8 × 10¹⁰) = 9.0 × 10³/1.8 × 10¹⁰ = 5.0 × 10⁻⁷ C/m; it is positive because the field points outward (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.32, page 34).
Why B is wrong: B is wrong because 1.0 × 10⁻⁶ C/m comes from rE/(9.0 × 10⁹) = 9.0 × 10³/9.0 × 10⁹, dropping the factor 2 in 1/(2πε₀).
Why C is wrong: C is wrong because 1.0 × 10⁻⁷ C/m comes from r²E/(9.0 × 10⁹) = 9.0 × 10²/9.0 × 10⁹, treating the wire with the point-charge 1/r² form and no factor 2.
Why D is wrong: D is wrong because 5.0 × 10⁻⁸ C/m comes from r²E/(1.8 × 10¹⁰) = 9.0 × 10²/1.8 × 10¹⁰, keeping the factor 2 but squaring r.
MCQ 7CalculationPractice
A closed cylindrical Gaussian surface of radius r and length l is coaxial with an infinitely long uniformly charged wire. It is replaced by another coaxial cylinder of radius 2r and length l/2. Compared with the first cylinder, the total flux through the new cylinder and the field magnitude on its curved surface are:
Show answer and why every option is right or wrong
Answer: B. Total flux = enclosed charge/ε₀ = λl/ε₀; the enclosed length halves, so the flux halves. The field on the curved surface is λ/(2πε₀r), which depends only on radial distance; doubling r halves E (NCERT Class 12 Physics Part I, Chapter 1, Section 1.14.1 and Eq. 1.32, page 34).
Why A is wrong: A is wrong because it reasons from the curved area 2π(2r)(l/2) = 2πrl, which is unchanged; but flux is fixed by the enclosed charge λ(l/2), which halves, and E depends on r, which doubled.
Why C is wrong: C is wrong because it gets the flux right but applies a point-charge 1/r² dependence; for a line charge, doubling r only halves the field.
Why D is wrong: D is wrong because it gets the field right but ties the flux to the unchanged curved area; with only half the wire length inside, the enclosed charge and hence the flux are halved.
MCQ 8Concept TrapPractice
A student writes: "In the Gauss's-law derivation for a long charged wire only the charge λl inside the cylinder appears, so the field on the curved surface is produced only by that enclosed piece of wire, and the result E = λ/(2πε₀r) is exact for a wire of any length." Which statement is correct according to NCERT?
Show answer and why every option is right or wrong
Answer: D. NCERT notes that although only the enclosed charge λl was used, E is due to the charge on the entire wire, and that the infinite-length assumption is crucial because otherwise E cannot be taken normal to the curved surface; Eq. 1.32 is approximately true around the central portions of a long wire (NCERT Class 12 Physics Part I, Chapter 1, page 34).
Why A is wrong: A is wrong on both counts: the enclosed charge fixes the flux, but the field itself is produced by the whole wire, and the formula depends on the infinite-length assumption.
Why B is wrong: B is wrong because Gauss's law is always true, but using it to get E needs the symmetry that makes E normal to the curved surface; that symmetry fails near the ends of a finite wire.
Why C is wrong: C is wrong because it keeps the first error: Gauss's law counts only enclosed charge for the flux, yet the field at the surface is due to all the charge on the wire.
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Field Long Wire Gauss: quick recall before you leave
How do you solve a Field Long Wire Gauss question? A worked example
Pattern: none of the dossier's PYQ patterns is in scope for this topic; this is an original problem.
- 1
Given
An infinitely long thin straight wire carries a uniform linear charge density λ = +2.0 × 10⁻⁶ C/m. Take 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻².
- 2
Required
(a) The magnitude and direction of the electric field at a perpendicular distance of 0.12 m from the wire. (b) The field magnitude at a perpendicular distance of 0.36 m.
- 3
Concept
By symmetry the field of an infinite line charge is radial and depends only on r. Gauss's law with a coaxial cylinder gives a field that falls as 1/r. A positive λ gives an outward field.
- 4
Formula
E = λ / (2πε₀ r) = 2 × [1/(4πε₀)] × λ / r
- 5
Substitution
(a) E = 2 × (9.0 × 10⁹) × (2.0 × 10⁻⁶) / 0.12
(b) E = 2 × (9.0 × 10⁹) × (2.0 × 10⁻⁶) / 0.36 - 6
Calculation
2 × (9.0 × 10⁹) × (2.0 × 10⁻⁶) = 3.6 × 10⁴ N m C⁻¹
(a) E = 3.6 × 10⁴ / 0.12 = 3.0 × 10⁵ N/C
(b) E = 3.6 × 10⁴ / 0.36 = 1.0 × 10⁵ N/C
Check for (b): the distance is 3 times larger, so E should be 3 times smaller: 3.0 × 10⁵ / 3 = 1.0 × 10⁵ N/C.
The factor 2 and π in 2πε₀ are exact mathematical constants and do not limit significant figures; λ, r and the given value of 1/(4πε₀) each have two significant figures, so the answers are given to two significant figures. - 7
Final answer
(a) E = 3.0 × 10⁵ N/C, directed radially outward from the wire; (b) E = 1.0 × 10⁵ N/C
- 8
Common trap
Using the point-charge form (9.0 × 10⁹)λ/r² gives 1.3 × 10⁶ N/C for part (a), and dropping only the factor 2 gives 1.5 × 10⁵ N/C. In part (b), scaling by 1/r² instead of 1/r gives 3.3 × 10⁴ N/C instead of 1.0 × 10⁵ N/C.
- 9
Similar NEET-style question
"A long straight wire carries a uniform linear charge density of −1.0 × 10⁻⁶ C/m. At what perpendicular distance from the wire is the field magnitude 4.5 × 10⁴ N/C, and which way does the field point there? Take 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻²."
Strategy: r = 2 × [1/(4πε₀)] × |λ| / E = (1.8 × 10¹⁰ × 1.0 × 10⁻⁶)/(4.5 × 10⁴) = 0.40 m. Because λ is negative, the field points radially inward, towards the wire.
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What to remember before solving Field Long Wire Gauss questions
Which Field Long Wire Gauss formulas do you need for NEET?
1 formula — click to collapse
Field of long line charge (Gauss)
Radial field at perpendicular distance r from infinite line of linear charge density lambda.
| Symbol | Quantity | SI Unit |
|---|---|---|
| E | field | V/m |
| lambda | linear charge density | C/m |
| r | perp distance | m |
Valid when
- Infinite line (or much longer than r)
- Uniform charge density
More in Electrostatics: 3 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
Field Long Wire Gauss questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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