PhysicsISC Class 12 · Electric Charges and Fields

Gauss's Theorem: Electric Flux and the Field of a Line, a Sheet and a Shell

Updated 15 October 2026 · Fact-checked

Gauss's theorem states that the total electric flux through any closed surface equals the net charge enclosed divided by ε₀: $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0}$. Choosing a symmetric Gaussian surface gives $E = \frac{\lambda}{2\pi\varepsilon_0 r}$ for a line charge, $\frac{\sigma}{2\varepsilon_0}$ for a plane sheet and $\frac{q}{4\pi\varepsilon_0 r^2}$ outside a thin shell.

Every field line leaving +q crosses both spheres, so flux is the same for any size; lines from an outside charge go in and out, so net flux is zero.

Illustration drawn for QuestionBuddy.

Understand Gauss's Theorem and Its Applications

Electric flux. Flux measures how much electric field passes through a surface. For a small area $d A$ (a vector normal to the surface, outward for a closed surface), $d ϕ = E \cdot d A = E d A cos θ$ . For a flat area in a uniform field, $ϕ = E A cos θ$ , where θ is the angle between $E$ and the normal. Flux is a scalar; its SI unit is $N m^{2} C^{- 1}$ (or V m). Field lines leaving a closed surface give positive flux, lines entering give negative flux.

Statement. The net electric flux through any closed surface is $\frac{1}{ε _{0}}$ times the net charge enclosed:
$ϕ = \oint E \cdot d A = \frac{q _{e n c}}{ε _{0}} .$

Checking it for a point charge. Surround $q$ by a sphere of radius $r$ . On it, $E$ is radial with $E = \frac{q}{4 π ε _{0} r ^{2}}$ , parallel to $d A$ everywhere, so $ϕ = E \cdot 4 π r^{2} = \frac{q}{ε _{0}}$ . The radius cancels: the flux does not depend on the size or shape of the closed surface, only on the charge inside. That was the point most candidates missed in ISC 2024 Q2, where "almost all" thought the flux changes with radius.

Consequences.

A charge outside the surface sends in as many lines as it takes out: zero net flux.
A dipole (net charge zero) inside gives zero net flux, even though the field is not zero.
The field $E$ in the integral is due to all charges, inside and outside; only the flux depends on the enclosed charge alone.
For a charge at the centre of a cube, symmetry gives $\frac{q}{6 ε _{0}}$ through each face.

Gaussian surface. An imaginary closed surface chosen so that on each part $E$ is either constant and along the normal, or perpendicular to the normal (no flux). It must not pass through a point charge.

1. Infinitely long straight line charge (density λ). By symmetry, $E$ is radial and depends only on distance $r$ . Take a coaxial cylinder of radius $r$ and length $l$ . The flat ends have $E$ parallel to the surface, so no flux. The curved surface has $E$ along the normal: $ϕ = E (2 π r l)$ . Enclosed charge $λ l$ . Gauss: $E \cdot 2 π r l = \frac{λ l}{ε _{0}}$ , so
$E = \frac{λ}{2 π ε _{0} r} = \frac{2 k λ}{r} .$
So $E \propto \frac{1}{r}$ , not $\frac{1}{r ^{2}}$ , and $E \propto λ$ (the 2024 line-charge question).

2. Infinite plane thin sheet (surface density σ). $E$ is perpendicular to the sheet, pointing away on both sides (for σ > 0) and equal in magnitude at equal distances. Take a cylinder (pill-box) of cross-section $A$ piercing the sheet. Flux leaves through both flat ends and none through the curved side: $ϕ = 2 E A$ . Enclosed charge $σ A$ . So
$E = \frac{σ}{2 ε _{0}},$
independent of distance: the field is uniform.

Two parallel sheets. Add the fields as vectors. For $+ σ$ and $- σ$ : between the sheets both fields point from + to −, so $E = \frac{σ}{2 ε _{0}} + \frac{σ}{2 ε _{0}} = \frac{σ}{ε _{0}}$ ; outside they cancel. For two sheets of the same sign the reverse happens: zero between, $\frac{σ}{ε _{0}}$ outside. (A thick charged conducting plate has charge on its outer face and a field $\frac{σ}{ε _{0}}$ just outside: the 2025 examiners saw candidates confuse the two cases.)

3. Uniformly charged thin spherical shell (charge q, radius R). By symmetry $E$ is radial.

Outside ( $r > R$ ): a concentric Gaussian sphere encloses $q$ , so $E \cdot 4 π r^{2} = \frac{q}{ε _{0}}$ , $E = \frac{q}{4 π ε _{0} r ^{2}}$ : as if all the charge were at the centre.
On the surface: $E = \frac{q}{4 π ε _{0} R ^{2}} = \frac{σ}{ε _{0}}$ (since $q = 4 π R^{2} σ$ ).
Inside ( $r < R$ ): the Gaussian sphere encloses no charge, so $E = 0$ .

The E–r graph is zero up to R, jumps to its maximum at R and falls as $\frac{1}{r ^{2}}$ .

Gauss's Theorem and Its Applications: formulas and key facts

Flux through a flat area: $ϕ = E A cos θ$; θ between E and the normal; unit N m² C⁻¹.
Gauss's theorem: $\oint E \cdot d A = \frac{q _{e n c}}{ε _{0}}$; Independent of the size and shape of the surface.
Line charge: $E = \frac{λ}{2 π ε _{0} r} = \frac{2 k λ}{r}$; Radial; $E \propto \frac{1}{r}$ .
Thin plane sheet: $E = \frac{σ}{2 ε _{0}}$; Uniform, independent of distance.
Two opposite sheets: $E_{b e tw ee n} = \frac{σ}{ε _{0}}, E_{o u t s i d e} = 0$; Same-sign sheets: 0 between, σ/ε₀ outside.
Thin spherical shell: $E = \frac{q}{4 π ε _{0} r ^{2}} (r \geq R), E = 0 (r < R)$; On the surface $E = \frac{σ}{ε _{0}}$ .
Charge at cube centre: $ϕ_{f a ce} = \frac{q}{6 ε _{0}}$; By symmetry.

How to solve Gauss's Theorem and Its Applications questions

Gauss questions are either "what is the flux?" or "find E using a symmetric surface".

1For flux through a closed surface, add up only the charges inside, with signs, and divide by ε₀.
2For flux through part of a closed surface, use symmetry (equal faces share equally).
3To find E, identify the symmetry: line (cylinder), plane (pill-box), sphere (concentric sphere).
4Write the flux through the chosen surface in terms of E: $E (2 π r l)$ , $2 E A$ or $E (4 π r^{2})$ .
5Write the enclosed charge ( $λ l$ , $σ A$ or q) and equate flux to $\frac{q _{e n c}}{ε _{0}}$ .
6For several sheets or shells, find each field separately and add as vectors, stating direction.

Gauss's Theorem and Its Applications in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Field of a long line charge

Suppose a long plastic conveyor pipe in a factory becomes charged by friction to a linear charge density of $5 \times 1 0^{- 8}$ C m⁻¹. Air breaks down (sparks) when the field reaches about $3 \times 1 0^{6}$ N C⁻¹.

Find the field 1 cm from the pipe. Will there be sparking there?

Try it first, then show the solutionSolution

$E = \frac{2 k λ}{r} = \frac{2 \times 9 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 8}}{0.01}$ .
$E = \frac{900}{0.01} = 9 \times 1 0^{4}$ N C⁻¹.
This is far below $3 \times 1 0^{6}$ N C⁻¹.

Answer: $9 \times 1 0^{4}$ N C⁻¹: no sparking at 1 cm.

Memory hook: Line: one over r. Point: one over r squared.

Zero field inside a charged shell

Imagine a hollow metal sphere of radius 0.5 m on an insulating stand carrying a charge of $2 μ$ C, with a small sensor placed inside it.

What field does the sensor read inside, and what is the field just outside the surface?

Try it first, then show the solutionSolution

Inside: a Gaussian sphere encloses no charge, so $E = 0$ .
Just outside: $E = \frac{k q}{R ^{2}} = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 6}}{0.25} = 7.2 \times 1 0^{4}$ N C⁻¹.

Answer: Zero inside; $7.2 \times 1 0^{4}$ N C⁻¹ just outside. This is why a closed metal enclosure shields what is inside.

Memory hook: No charge inside the Gaussian surface, no field inside the shell.

Tricks to remember Gauss's Theorem and Its Applications

Short memory aids. Say each one aloud once, then use it on a question the same day.

Flux counts charge, not size
Big sphere, small sphere, cube: same charge inside, same flux.
Use it when: "How does the flux change if…" questions.
1, 0, 2 powers of r
Line: $\frac{1}{r}$ ; sheet: $r^{0}$ (constant); point or outside a shell: $\frac{1}{r ^{2}}$ .
Use it when: Predicting how E changes with distance.
Half for a sheet, whole between plates
One sheet: $\frac{σ}{2 ε _{0}}$ . Two opposite sheets: $\frac{σ}{ε _{0}}$ between.
Use it when: Sheet questions.
Pick the surface to match the charge
Line → cylinder, sheet → pill-box, ball → sphere.
Use it when: Choosing a Gaussian surface.

Gauss's Theorem and Its Applications: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Flux through one face · 1 mark · MCQBoard level

A point charge q is placed at the centre of a cube. The flux through one face is:
(a) $\frac{q}{ε _{0}}$
(b) $\frac{q}{6 ε _{0}}$
(c) $\frac{q}{8 ε _{0}}$
(d) zero

Try it first, then show the solutionSolution

Total $\frac{q}{ε _{0}}$ shared equally by six identical faces.

Answer: (b).

Example 2: Line charge field · 2 marks · Short answerBoard level

An infinitely long wire carries $2 μ$ C m⁻¹. Find the electric field 9 cm from it. What would it be at 18 cm? ( $k = 9 \times 1 0^{9}$ N m² C⁻²)

Try it first, then show the solutionSolution

$E = \frac{2 k λ}{r} = \frac{2 \times 9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 6}}{0.09} = 4 \times 1 0^{5}$ N C⁻¹, radially outward. (1)
$E \propto \frac{1}{r}$ , so at 18 cm it halves: $2 \times 1 0^{5}$ N C⁻¹. (1)

Answer: $4 \times 1 0^{5}$ N C⁻¹; $2 \times 1 0^{5}$ N C⁻¹.

Example 3: Two sheets of the same sign · 3 marks · Short answerHarder board level

Two large, thin, parallel sheets each carry $+ 1.77 \times 1 0^{- 6}$ C m⁻². Find the field (a) between them and (b) outside them. ( $ε_{0} = 8.85 \times 1 0^{- 12}$ C² N⁻¹ m⁻²)

Try it first, then show the solutionSolution

Each sheet: $\frac{σ}{2 ε _{0}} = \frac{1.77 \times 1 0 ^{- 6}}{2 \times 8.85 \times 1 0 ^{- 12}} = 1.0 \times 1 0^{5}$ N C⁻¹, pointing away from the sheet. (1)
Between: the two fields point in opposite directions and cancel: $E = 0$ . (1)
Outside: both point away from the pair, so $E = 2.0 \times 1 0^{5}$ N C⁻¹ directed outward. (1)

Answer: (a) 0; (b) $2.0 \times 1 0^{5}$ N C⁻¹ outward.

Example 4: Concentric shells · 3 marks · Short answerHarder board level

Two concentric thin shells have radii 10 cm and 20 cm and charges +4 nC and −6 nC. Find the field at 5 cm, 15 cm and 30 cm from the centre.

Try it first, then show the solutionSolution

5 cm: no charge enclosed, $E = 0$ . (1)
15 cm: encloses +4 nC: $E = \frac{9 \times 1 0 ^{9} \times 4 \times 1 0 ^{- 9}}{0.1 5 ^{2}} = 1600$ N C⁻¹, outward. (1)
30 cm: encloses −2 nC: $E = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 9}}{0.09} = 200$ N C⁻¹, inward. (1)

Answer: 0; 1600 N C⁻¹ outward; 200 N C⁻¹ inward.

Example 5: Derivation: infinite line charge · 5 marks · Long answerHarder board level

Use Gauss's theorem to derive the electric field at a distance r from an infinitely long, thin, straight wire with uniform linear charge density λ. Sketch E against r.

Try it first, then show the solutionSolution

By symmetry, $E$ is radial (perpendicular to the wire) and has the same magnitude at all points at distance r. (1)
Gaussian surface: a coaxial cylinder of radius r and length l. On the flat ends $E ⊥ d A$ , so their flux is zero. (1)
On the curved surface $E ∥ d A$ and E is constant: $ϕ = E \cdot 2 π r l$ . (1)
Charge enclosed: $λ l$ . Gauss: $E \cdot 2 π r l = \frac{λ l}{ε _{0}}$ , so $E = \frac{λ}{2 π ε _{0} r}$ . (1)
Graph: a rectangular hyperbola, E falling as $\frac{1}{r}$ ; the field points away from the wire for λ > 0. (1)

Answer: $E = \frac{λ}{2 π ε _{0} r}$ , radially outward for positive λ.

Example 6: A coaxial cable · 4 marks · Case-basedHOTS / competency

Suppose a long coaxial cable has a thin inner wire carrying $+ 3 \times 1 0^{- 8}$ C m⁻¹ and a thin outer cylindrical sheath of radius 4 mm carrying $- 3 \times 1 0^{- 8}$ C m⁻¹.
(i) Find the field 2 mm from the axis. (1 mark)
(ii) Find the field 1 cm from the axis. (1 mark)
(iii) Find the flux through a closed cylinder of radius 2 mm and length 50 cm, coaxial with the cable. (2 marks)

Try it first, then show the solutionSolution

(i) Only the inner wire is enclosed: $E = \frac{2 k λ}{r} = \frac{2 \times 9 \times 1 0 ^{9} \times 3 \times 1 0 ^{- 8}}{0.002} = 2.7 \times 1 0^{5}$ N C⁻¹, outward.
(ii) A cylinder of radius 1 cm encloses $+ λ - λ = 0$ , so $E = 0$ .
(iii) $q_{e n c} = λ L = 3 \times 1 0^{- 8} \times 0.5 = 1.5 \times 1 0^{- 8}$ C.
$ϕ = \frac{1.5 \times 1 0 ^{- 8}}{8.85 \times 1 0 ^{- 12}} \approx 1.7 \times 1 0^{3}$ N m² C⁻¹.

Answer: (i) $2.7 \times 1 0^{5}$ N C⁻¹; (ii) 0; (iii) about $1.7 \times 1 0^{3}$ N m² C⁻¹.

Where students lose marks in Gauss's Theorem and Its Applications

The slips examiners see most often, and what to write instead.

Mistake: Thinking flux depends on the surface size
In ISC 2024 almost all candidates thought the flux from a charge inside a hollow sphere changes with its radius; some confused flux with field.
Do this: Flux = enclosed charge ÷ ε₀, whatever the size or shape of the surface.
Usually costs: 1 to 2 marks.
Mistake: Dipole inside gives non-zero flux
Also seen in 2024: a dipole has zero net charge, so zero net flux.
Do this: Add the enclosed charges with their signs.
Usually costs: 1 mark.
Mistake: Treating a line charge as a point charge
2024 candidates used inverse-square dependence for an infinite line charge and answered "four times".
Wrong working
Halving r makes E four times.
Right working
$E \propto \frac{1}{r}$ : halving r doubles E.
Do this: Quote the working formula $E = \frac{λ}{2 π ε _{0} r}$ first.
Usually costs: 1 mark each part.
Mistake: Mishandling two sheets
In 2025 candidates confused a thin sheet with a thick conducting plate, subtracted fields that act in the same direction, and could not state the direction of the resultant.
Do this: Draw an arrow for each sheet's field at the point, then add or subtract; state the direction (towards the negative sheet).
Usually costs: 1 to 2 marks.

Exam tips

In derivations, draw the Gaussian surface and say why the flux through each part is zero or EA.
Always state the direction of E as well as its magnitude.
Keep $ε_{0} = 8.85 \times 1 0^{- 12}$ and $k = 9 \times 1 0^{9}$ in SI units; convert cm to m and nC to C.
For "how does E change" questions, write the proportionality first ( $E \propto λ$ , $E \propto \frac{1}{r}$ ).

Practice questions on Gauss's Theorem and Its Applications

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A closed surface encloses $+ 3 μ$ C and $- 5 μ$ C; a charge of $+ 7 μ$ C lies outside. The net flux through the surface is:
(a) $\frac{5 \times 1 0 ^{- 6}}{ε _{0}}$
(b) $\frac{- 2 \times 1 0 ^{- 6}}{ε _{0}}$
(c) $\frac{1 0 ^{- 6}}{ε _{0}}$
(d) zero
Hint
Only enclosed charge counts.
Answer
(b), about $- 2.26 \times 1 0^{5}$ N m² C⁻¹.
Q2 · 1 mark · Very short answerBoard level
A point charge at the centre of a sphere sends a flux φ through it. What is the flux if the charge is moved off-centre (still inside) and the radius of the sphere is halved? (A variation on ISC 2024.)
Hint
Flux depends only on the enclosed charge.
Answer
Still φ: the flux depends only on the charge enclosed, not on the size of the surface or where the charge sits inside it.
Q3 · 2 marks · Short answerHarder board level
A uniform field $E = 3 \times 1 0^{3} \hat{i}$ N C⁻¹ passes through a square of side 10 cm whose plane makes 60° with the field. Find the flux.
Hint
The angle with the normal is 30°.
Answer
$ϕ = E A cos 30° = 3000 \times 0.01 \times 0.866 \approx 26$ N m² C⁻¹.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The field inside a uniformly charged thin spherical shell is zero.
Reason (R): A concentric Gaussian sphere inside the shell encloses no charge and, by symmetry, E is the same at all its points.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Zero flux plus symmetry gives zero field.
Answer
(a).
Q5 · 1 mark · Very short answerHarder board level
A large thin sheet produces a field of $4.5 \times 1 0^{4}$ N C⁻¹ near it. Find its surface charge density.
Hint
$σ = 2 ε_{0} E$ .
Answer
$σ = 2 \times 8.85 \times 1 0^{- 12} \times 4.5 \times 1 0^{4} \approx 8.0 \times 1 0^{- 7}$ C m⁻².
Q6 · 3 marks · Short answerHOTS / competency
A small ball of mass 2 g carrying $2 \times 1 0^{- 8}$ C hangs by a silk thread near a large vertical charged sheet. The thread makes 30° with the vertical. Find the surface charge density of the sheet. (g = 9.8 m s⁻²)
Hint
$tan θ = \frac{q E}{m g}$ .
Answer
$E = \frac{m g t a n 30°}{q} = \frac{0.002 \times 9.8 \times 0.577}{2 \times 1 0 ^{- 8}} \approx 5.66 \times 1 0^{5}$ N C⁻¹; $σ = 2 ε_{0} E \approx 1.0 \times 1 0^{- 5}$ C m⁻².
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a charge of 8.85 nC is placed inside a closed cubical box. ( $ε_{0} = 8.85 \times 1 0^{- 12}$ C² N⁻¹ m⁻²)
(i) Find the total flux through the box. (1 mark)
(ii) If the charge is at the centre, find the flux through one face. (1 mark)
(iii) The charge is now moved to one corner of a cube (placed exactly at the corner). Using eight identical cubes around that corner, find the total flux through this cube. (1 mark)
(iv) Hence find the flux through each of the three faces not touching that corner. (1 mark)
Hint
At a corner the charge is shared by 8 cubes; the 3 faces through the corner carry no flux.
Answer
(i) $\frac{8.85 \times 1 0 ^{- 9}}{8.85 \times 1 0 ^{- 12}} = 1000$ N m² C⁻¹. (ii) $\approx 167$ N m² C⁻¹. (iii) $\frac{1000}{8} = 125$ N m² C⁻¹. (iv) $\frac{125}{3} \approx 41.7$ N m² C⁻¹ each.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Gauss's theorem.Show answer
The total flux through a closed surface is $\frac{1}{ε _{0}}$ times the net charge enclosed.
2.Field of an infinite thin sheet?Show answer
$\frac{σ}{2 ε _{0}}$ , independent of distance.
3.Field inside a uniformly charged thin spherical shell?Show answer
Zero.
4.SI unit of electric flux?Show answer
N m² C⁻¹ (or V m).
5.Flux through a closed surface around a 1 nC charge? (ε₀ = 8.85 × 10⁻¹² SI)Show answer
About 113 N m² C⁻¹.

Gauss's Theorem and Its Applications: frequently asked questions

Does the field inside the Gaussian surface come only from the enclosed charge?

No. The field at each point is due to all charges, inside and outside. Only the total flux through the closed surface depends on the enclosed charge alone.

Why is the field of an infinite sheet independent of distance?

The field lines from an infinite sheet are parallel and do not spread out, so their density, and hence E, stays the same at every distance.

What makes a good Gaussian surface?

A closed surface that matches the symmetry of the charge, so that E is constant and normal to the surface on some parts and parallel to it (zero flux) on the rest.

Which applications of Gauss's theorem are in the ISC 2027 syllabus?

The field of an infinitely long straight wire, a uniformly charged infinite plane sheet and a uniformly charged thin spherical shell.

How many marks does this carry?

It was the most tested part of the chapter: 4 marks in 2024 (flux and line charge) and 3 marks in 2025 (two charged sheets).