12 min readConcept sheet3 or 5-mark derivations with a labelled Gaussian surface and graph, and 1-mark MCQ on the field inside a shell; also a NEET favourite

Gauss's theorem: field of a long wire, a plane sheet and a spherical shell

Choose a Gaussian surface that matches the symmetry. A wire gives E = λ/(2πε₀r), a plane sheet gives E = σ/(2ε₀), and a shell gives zero field inside.

Infinitely long straight wireE = λ/(2πε₀r), radially outward for λ > 0 (λ = charge per unit length)

Infinite plane sheetE = σ/(2ε₀), the same at every distance (σ = charge per unit area)

Thin spherical shell (radius R, charge q)Outside (r ≥ R): E = q/(4πε₀r²); inside (r < R): E = 0

Derivation 1: infinitely long straight wire

A thin, infinitely long straight wire carries a uniform linear charge density λ. Find E at a distance r from it.

1
Symmetry: E is radial, at right angles to the wire, and has the same size at all points at distance r.
2
Gaussian surface: a cylinder of radius r and length l, with the wire along its axis.
3
Flat ends: E runs along these faces (at 90° to their normal), so their flux is zero.
4
Curved surface: E is along the normal everywhere, so the flux = E × 2πrl.
5
Charge enclosed = λl. Gauss's theorem: E × 2πrl = λl/ε₀.
6★
So, E = λ/(2πε₀r), radially outward for λ > 0 (inward for λ < 0). E ∝ 1/r.

★ key step

Derivation 2: uniformly charged infinite plane sheet

A thin, infinite plane sheet carries a uniform surface charge density σ. Find E at a point near it.

1
Symmetry: E is at right angles to the sheet, away from it on both sides (for σ > 0), and the same at equal distances.
2
Gaussian surface: a small cylinder of cross-section A that crosses the sheet, with one flat end on each side.
3
Curved surface: E runs along it, so its flux is zero.
4
Each flat end: E is along the normal, so the flux is EA. Both ends together give 2EA.
5
Charge enclosed = σA. Gauss's theorem: 2EA = σA/ε₀.
6★
So, E = σ/(2ε₀). It does not depend on the distance from the sheet.

★ key step

Derivation 3: uniformly charged thin spherical shell

A thin shell of radius R carries a charge q spread evenly over it (σ = q/4πR²).

1
Symmetry: E is radial and has the same size at all points at distance r from the centre.
2
Outside (r > R): take a concentric Gaussian sphere of radius r. Flux = E × 4πr².
3
It encloses the whole charge q. So E × 4πr² = q/ε₀, giving E = q/(4πε₀r²).
4
Outside, the shell acts as if all its charge were at the centre. In terms of σ: E = σR²/(ε₀r²).
5
Inside (r < R): a concentric Gaussian sphere of radius r encloses no charge. So E × 4πr² = 0.
6★
So, E = 0 everywhere inside the shell.

★ key step

How to draw the figures and the graph

Wire: a vertical line with + signs; a dashed cylinder round it; label r and l; arrows for E pointing radially out from the curved surface; mark the normals on the flat ends.
Sheet: a vertical plane with + signs; a small dashed cylinder crossing it; label A on both ends; E arrows pointing away on both sides.
Shell: a circle of radius R with + signs on it; one dashed Gaussian circle outside (r > R) and one inside (r < R); E arrows radially out from the outer one.
E versus r for a shell: E = 0 from r = 0 to R; a jump to q/(4πε₀R²) at r = R; then a fall as 1/r².

Compare the results

Charge distribution	Gaussian surface	Field E	How E varies with r
Point charge q	Sphere centred on q	q/(4πε₀r²)	∝ 1/r²
Infinite line (λ)	Coaxial cylinder	λ/(2πε₀r)	∝ 1/r
Infinite plane sheet (σ)	Cylinder crossing the sheet	σ/(2ε₀)	Does not change
Two parallel sheets, +σ and −σ	Add the two sheet fields	σ/ε₀ between them, zero outside	Does not change
Thin spherical shell (q)	Concentric sphere	q/(4πε₀r²) outside, zero inside	∝ 1/r² outside

Worked example: three quick fields

Problem

(a) A long straight wire has λ = 3 × 10⁻⁸ C m⁻¹. Find E 2 cm from it. (b) A thin spherical shell of radius 10 cm carries 4 nC. Find E at 5 cm and at 20 cm from its centre. (c) A large sheet has σ = 8.85 × 10⁻⁸ C m⁻². Find E near it. (Take 1/4πε₀ = 9 × 10⁹ N m² C⁻² and ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².)

Working

(a) E = λ/(2πε₀r) = 2kλ/r = 2 × 9 × 10⁹ × 3 × 10⁻⁸ ÷ 0.02 = 540 ÷ 0.02 = 2.7 × 10⁴ N C⁻¹.
(b) At 5 cm the point is inside the shell, so E = 0.
At 20 cm: E = kq/r² = 9 × 10⁹ × 4 × 10⁻⁹ ÷ (0.20)² = 36 ÷ 0.04 = 900 N C⁻¹.
(c) E = σ/(2ε₀) = 8.85 × 10⁻⁸ ÷ (2 × 8.85 × 10⁻¹²) = 5 × 10³ N C⁻¹.

Answer

(a) 2.7 × 10⁴ N C⁻¹; (b) zero at 5 cm, 900 N C⁻¹ at 20 cm; (c) 5 × 10³ N C⁻¹

In real life

The field near a thin charged wire is very strong because E ∝ 1/r. Electrostatic precipitators in Indian thermal power plants and cement factories use thin high-voltage wires to charge fly ash, so that it is collected instead of escaping through the chimney.

Exam angle

5 marks: “State Gauss's theorem. Use it to derive the electric field due to an infinitely long, uniformly charged straight wire. Draw the graph of E against r.” Gaussian cylinder of radius r and length l; flat ends give no flux; the curved surface gives E × 2πrl = λl/ε₀; so E = λ/(2πε₀r). Graph: a curve falling as 1/r.

3 marks: “Using Gauss's theorem, find the field due to a uniformly charged infinite plane sheet. Does it depend on the distance from the sheet?” A cylinder of area A crossing the sheet: 2EA = σA/ε₀, so E = σ/(2ε₀), the same at every distance.

3 marks: “Find the electric field inside and outside a uniformly charged thin spherical shell. Draw the graph of E against r.” Outside: E × 4πr² = q/ε₀, so E = q/(4πε₀r²). Inside: no charge is enclosed, so E = 0. Graph: zero up to R, then a 1/r² fall.

Remember it

Wire 1/r, sheet r-free, shell 1/r² outside and zero inside. Match the surface to the shape: cylinder, small cylinder across the sheet, sphere.