PhysicsCBSE Class 12 · Electric Charges and Fields

Gauss's Law and Its Applications Class 12: Wire, Sheet and Shell

Q: Do I need to prove Gauss's law for CBSE Class 12?

No. The 2026-27 syllabus asks you to state Gauss's law and apply it to an infinitely long straight wire, an infinite plane sheet and a thin spherical shell (inside and outside).

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

The field lines from an infinite sheet stay parallel; they do not spread out as you move away. With the pill-box, the area through which flux passes does not change with distance, so E stays the same.

Q: Why is the field inside a charged spherical shell zero?

Any Gaussian sphere drawn inside the shell encloses no charge. By symmetry E would be the same all over that sphere, so the only way the total flux can be zero is for E to be zero at every point inside.

Q: Can Gauss's law be used to find the field of a finite wire?

Gauss's law is always true, but for a finite wire the field is not the same size all over a cylinder, so the flux cannot be written as E times an area. It is only useful as a shortcut when symmetry is high enough.

Q: What is the unit of electric flux?

N m² C⁻¹, which is the same as V m.

Updated 15 October 2026 · Fact-checked

Gauss's law says the total electric flux through any closed surface equals the charge enclosed divided by ε₀. With a symmetric Gaussian surface it gives the field of an infinitely long wire (λ/2πε₀r), an infinite plane sheet (σ/2ε₀) and a thin spherical shell (kq/r² outside, zero inside), the three cases in the CBSE syllabus.

Only charge inside the closed surface sets the net flux; a charge outside sends lines in and out, so the net flux is zero.

Illustration drawn for QuestionBuddy.

Understand Gauss's Law and Its Applications

Think of field lines as streams of water coming out of a tap (a positive charge) or going down a drain (a negative charge). Wrap any closed bag around the tap and the water passing out through the bag depends only on what is inside, not on the bag's shape. Gauss's law says the same about electric field lines.

Electric flux. For a small flat area $Δ S$ with outward unit normal $\overset{n}{^}$ , the flux is $Δ ϕ = E \cdot Δ S = E Δ S cos θ$ , where $θ$ is the angle between $E$ and the normal. Flux measures how many field lines pass through the area. It is a scalar, its SI unit is N m² C⁻¹ (equivalently V m), it is largest when the field is along the normal and zero when the field skims along the surface. For a closed surface, outward flux is counted positive and inward flux negative.

Statement of Gauss's law. The total flux through any closed surface S is $\oint E \cdot d S = \frac{q _{e n c}}{ε _{0}}$ , where $q_{e n c}$ is the algebraic sum of the charges inside S. Three consequences matter in exams:

Charges outside the surface add nothing to the net flux (their lines enter and leave), although they do change the field $E$ at points of the surface.
If no net charge is enclosed, the net flux is zero (for example a closed surface around a dipole).
The result does not depend on the shape or size of the surface. The surface used is called a Gaussian surface; it must not pass through a point charge. (The CBSE syllabus asks for the statement only, not a proof.)

How to use it. Gauss's law gives the field easily only when symmetry tells you the direction of $E$ and that its size is the same all over part of the Gaussian surface. Then $\oint E \cdot d S$ becomes $E \times$ (area where E is normal). The syllabus has three such cases.

**1. Infinitely long straight wire, charge per length $λ$ .** By symmetry $E$ is radial and depends only on the distance r. Take a coaxial cylinder of radius r and length l. Flux through the flat ends is zero ( $E$ lies along them); through the curved face it is $E \times 2 π r l$ . Enclosed charge is $λ l$ . So $E \cdot 2 π r l = \frac{λ l}{ε _{0}}$ , giving $E = \frac{λ}{2 π ε _{0} r}$ , outward for $λ > 0$ . The field falls as $\frac{1}{r}$ .

**2. Infinite plane sheet, charge per area $σ$ .** $E$ is normal to the sheet, pointing away on both sides for $σ > 0$ , and the same size at equal distances. Take a cylinder (a 'pill-box') of cross-section A crossing the sheet. Flux passes only through the two end faces: $2 E A$ . Charge enclosed is $σ A$ . So $E = \frac{σ}{2 ε _{0}}$ , independent of the distance from the sheet. Two large parallel sheets with $+ σ$ and $- σ$ give $\frac{σ}{ε _{0}}$ between them and zero outside, which is the idea behind the parallel plate capacitor.

3. Thin spherical shell of radius R with charge q. $E$ is radial and the same everywhere on a concentric sphere. Outside ( $r > R$ ): $E \cdot 4 π r^{2} = \frac{q}{ε _{0}}$ , so $E = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}$ , as if all the charge were at the centre. At the surface, $E = \frac{σ}{ε _{0}}$ (since $q = σ \cdot 4 π R^{2}$ ). Inside ( $r < R$ ): the Gaussian sphere encloses no charge, so $E = 0$ everywhere inside the shell. The E versus r graph is zero up to R, jumps to its largest value at R, and then falls as $\frac{1}{r ^{2}}$ .

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

Electric flux through an area: $ϕ = E \cdot S = E S cos θ$; θ is the angle between $E$ and the normal to the area. Unit N m² C⁻¹.
Gauss's law: $\oint E \cdot d S = \frac{q _{e n c}}{ε _{0}}$; Only charge inside the closed surface counts for net flux.
Infinitely long straight wire: $E = \frac{λ}{2 π ε _{0} r} = \frac{2 k λ}{r}$; Radial; falls as 1/r.
Infinite plane sheet: $E = \frac{σ}{2 ε _{0}}$; Normal to the sheet, independent of distance.
Two oppositely charged parallel sheets: $E_{b e tw ee n} = \frac{σ}{ε _{0}}, E_{o u t s i d e} = 0$; Uniform field between the plates.
Thin spherical shell: $E = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}} (r \geq R), E = 0 (r < R)$; At the surface $E = \frac{σ}{ε _{0}}$ .
Flux of a point charge at the centre of a cube: $ϕ_{f a ce} = \frac{q}{6 ε _{0}}$; By symmetry the total $\frac{q}{ε _{0}}$ shares equally between six faces.

How to solve Gauss's Law and Its Applications questions

Use this method for flux questions and for finding fields with Gauss's law.

1For flux through a flat area, find the angle between $E$ and the normal (not the surface), then use $E S cos θ$ .
2For flux through a closed surface, add only the charges inside (with signs) and divide by $ε_{0}$ ; ignore charges outside.
3For a field, identify the symmetry: line (cylinder), plane (pill-box) or sphere (concentric sphere).
4Draw the Gaussian surface through the point where E is wanted, so that E is either normal and constant or parallel to each part of it.
5Write the flux as E times the area where E is normal; write $q_{e n c}$ in terms of $λ$ , $σ$ or q.
6Equate and solve for E; state its direction and, for a shell, give the inside and outside cases separately.
7For a sketch, plot E against r and mark the special points (the surface of the shell, or the constant value for a sheet).

Gauss's Law 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 charged spherical shell

Suppose a hollow metal dome of radius 15 cm on an insulating stand is given a charge of 3 μC, which spreads evenly over its outer surface. Assume air starts to conduct (and sparks jump) when the field reaches about $3 \times 1 0^{6}$ N/C.

(a) Find the field just outside the dome and at 30 cm from its centre. (b) What is the field inside the dome? (c) What is the largest charge the dome can hold before sparking starts?

Try it first, then show the solutionSolution

Just outside: $E = \frac{k Q}{R ^{2}} = \frac{9 \times 1 0 ^{9} \times 3 \times 1 0 ^{- 6}}{( 0.15 ) ^{2}} = 1.2 \times 1 0^{6}$ N/C.
At r = 30 cm (twice R): $E = \frac{1.2 \times 1 0 ^{6}}{4} = 3 \times 1 0^{5}$ N/C.
Inside: a Gaussian sphere inside the shell encloses no charge, so E = 0.
Largest charge: $Q_{ma x} = \frac{E _{ma x} R ^{2}}{k} = \frac{3 \times 1 0 ^{6} \times 0.0225}{9 \times 1 0 ^{9}} = 7.5 \times 1 0^{- 6}$ C.

Answer: (a) $1.2 \times 1 0^{6}$ N/C and $3 \times 1 0^{5}$ N/C; (b) zero; (c) about 7.5 μC.

Memory hook: A charged shell is a point charge from outside and nothing at all from inside.

Field of a long line of charge

Imagine a long, thin plastic pipe that has been rubbed so that it carries a uniform charge of 3 nC per metre along its length.

(a) Find the field at 5 cm and at 50 cm from the pipe (far from its ends). (b) By what factor does the field fall when the distance is made 10 times larger, and how does this compare with a point charge?

Try it first, then show the solutionSolution

$E = \frac{2 k λ}{r}$ . At 5 cm: $E = \frac{2 \times 9 \times 1 0 ^{9} \times 3 \times 1 0 ^{- 9}}{0.05} = 1080$ N/C.
At 50 cm: $E = \frac{54}{0.5} = 108$ N/C.
The field falls by a factor of 10 (as 1/r). For a point charge, a tenfold distance would cut the field 100 times (1/r²).

Answer: (a) 1080 N/C and 108 N/C; (b) 10 times weaker, against 100 times weaker for a point charge.

Memory hook: Point: 1/r². Line: 1/r. Sheet: no r at all.

Tricks to remember Gauss's Law and Its Applications

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

Point, line, sheet: 2, 1, 0
The power of r in the denominator drops by one each time: point charge 1/r², line 1/r, sheet 1/r⁰ (constant).
Use it when: Comparing how the three fields vary with distance.
Half for one sheet, full for two
One sheet: σ/2ε₀. Two opposite sheets between them: σ/ε₀. Shell surface: σ/ε₀ too.
Use it when: Remembering where the 2 goes.
Shapes for symmetry
Line → can (cylinder), sheet → pill-box (cylinder across it), ball → ball (concentric sphere).
Use it when: Choosing the Gaussian surface.
Inside out
Charge outside: changes E but not net flux. Charge inside: decides net flux.
Use it when: Flux MCQs with charges both inside and outside.

Gauss's Law and Its Applications: solved examples

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

Example 1: Flux through a cube · 2 marks · Very short answerBoard level

A point charge of 10 μC is at the centre of a cube of side 20 cm. Find (a) the total flux through the cube, and (b) the flux through one face. (c) Would the answers change if the cube's side were doubled?

Try it first, then show the solutionSolution

(a) $ϕ = \frac{q}{ε _{0}} = \frac{1 0 ^{- 5}}{8.854 \times 1 0 ^{- 12}} \approx 1.13 \times 1 0^{6}$ N m² C⁻¹. (½)
(b) By symmetry each of the six faces gets an equal share: $\frac{1.13 \times 1 0 ^{6}}{6} \approx 1.88 \times 1 0^{5}$ N m² C⁻¹. (1)
(c) No: flux depends only on the enclosed charge, not on the size of the surface. (½)

Answer: (a) about $1.13 \times 1 0^{6}$ N m² C⁻¹; (b) about $1.88 \times 1 0^{5}$ N m² C⁻¹; (c) no change.

Example 2: Derivation: field of an infinitely long wire · 3 marks · Short answerBoard level

Using Gauss's law, derive the electric field at a distance r from an infinitely long straight wire with uniform linear charge density λ. Draw the Gaussian surface.

Try it first, then show the solutionSolution

By symmetry, $E$ is radial (perpendicular to the wire) and has the same size at all points at distance r. Choose a coaxial cylinder of radius r and length l as the Gaussian surface (draw it around the wire). (1)
Flux through the two flat ends is zero since $E$ is parallel to them; through the curved surface $E$ is along the normal: $ϕ = E \times 2 π r l$ . (1)
Enclosed charge $= λ l$ . Gauss's law: $E \cdot 2 π r l = \frac{λ l}{ε _{0}}$ , so $E = \frac{λ}{2 π ε _{0} r}$ , radially outward for positive λ. (1)

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

Example 3: Derivation: thin spherical shell, inside and outside · 5 marks · Long answerBoard level

(a) Using Gauss's law, find the electric field due to a uniformly charged thin spherical shell of radius R and charge q at a point (i) outside and (ii) inside the shell. (3)
(b) Sketch the variation of E with distance r from the centre. (2)

Try it first, then show the solutionSolution

By symmetry $E$ is radial and has the same size at every point of a concentric sphere. (½)
(i) Gaussian sphere of radius r > R: $ϕ = E \cdot 4 π r^{2}$ and $q_{e n c} = q$ . So $E = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}$ , radially outward for q > 0, as if q were at the centre. (1½)
(ii) Gaussian sphere of radius r < R: $q_{e n c} = 0$ , so $E \cdot 4 π r^{2} = 0$ and E = 0 at every point inside. (1)
(b) Graph: E = 0 from r = 0 to r = R; at r = R it jumps to $\frac{q}{4 π ε _{0} R ^{2}}$ $(= \frac{σ}{ε _{0}})$ ; for r > R it falls off as $\frac{1}{r ^{2}}$ . Label axes, R and the peak value. (2)

Answer: Outside: $E = \frac{q}{4 π ε _{0} r ^{2}}$ ; inside: E = 0; the graph is zero up to R, peaks at R and falls as 1/r².

Example 4: Two parallel charged sheets · 3 marks · Short answerHarder board level

Two large thin parallel plates carry surface charge densities $+ 17.7$ nC/m² and $- 17.7$ nC/m². Find (a) the field due to one plate alone, (b) the field between the plates, (c) the field outside them, and (d) the force on an electron between the plates.

Try it first, then show the solutionSolution

(a) $E_{1} = \frac{σ}{2 ε _{0}} = \frac{17.7 \times 1 0 ^{- 9}}{2 \times 8.85 \times 1 0 ^{- 12}} = 1000$ N/C. (½)
(b) Between the plates, both fields point from the positive to the negative plate and add: $E = \frac{σ}{ε _{0}} = 2000$ N/C. (1)
(c) Outside, the two fields are opposite and equal: E = 0. (½)
(d) $F = e E = 1.6 \times 1 0^{- 19} \times 2000 = 3.2 \times 1 0^{- 16}$ N, towards the positive plate. (1)

Answer: (a) 1000 N/C; (b) 2000 N/C, from + to − plate; (c) zero; (d) $3.2 \times 1 0^{- 16}$ N towards the positive plate.

Example 5: Concentric shells · 3 marks · Short answerHOTS / competency

Two concentric thin shells have radii 5 cm and 15 cm. The inner shell carries +3 μC and the outer shell −5 μC. Find the electric field at distances of (a) 2 cm, (b) 10 cm and (c) 20 cm from the centre, with directions.

Try it first, then show the solutionSolution

(a) r = 2 cm: inside both shells, $q_{e n c} = 0$ , so E = 0. (1)
(b) r = 10 cm: only the inner shell is enclosed: $E = \frac{9 \times 1 0 ^{9} \times 3 \times 1 0 ^{- 6}}{( 0.1 ) ^{2}} = 2.7 \times 1 0^{6}$ N/C, radially outward. (1)
(c) r = 20 cm: $q_{e n c} = 3 - 5 = - 2$ μC: $E = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 6}}{( 0.2 ) ^{2}} = 4.5 \times 1 0^{5}$ N/C, radially inward. (1)

Answer: (a) zero; (b) $2.7 \times 1 0^{6}$ N/C outward; (c) $4.5 \times 1 0^{5}$ N/C inward.

Example 6: Flux in a uniform field · 4 marks · Case-basedHarder board level

Read the passage and answer.

A uniform field $E = 2 \times 1 0^{3} \hat{i}$ N/C exists in a region. A square sheet of side 10 cm is placed in it.
(a) Find the flux through the sheet when its normal is along the x-axis. (1)
(b) Find the flux when the sheet is turned so that its normal makes 60° with the x-axis. (1)
(c) A closed cylinder is placed in this field with its axis along x. What is the net flux through it, and why? (1)
(d) A closed surface contains charges of +3 μC and −5 μC, and a charge of +2 μC lies just outside. Find the net flux through it. (1)

Try it first, then show the solutionSolution

(a) $ϕ = E S = 2 \times 1 0^{3} \times (0.1)^{2} = 20$ N m² C⁻¹. (1)
(b) $ϕ = E S cos 60° = 10$ N m² C⁻¹. (1)
(c) Zero: the field enters one flat face and leaves the other with equal flux, and the curved face has no flux; no charge is enclosed. (1)
(d) $ϕ = \frac{( 3 - 5 ) \times 1 0 ^{- 6}}{8.854 \times 1 0 ^{- 12}} \approx - 2.26 \times 1 0^{5}$ N m² C⁻¹ (net inward); the outside charge does not count. (1)

Answer: (a) 20; (b) 10; (c) zero; (d) about $- 2.26 \times 1 0^{5}$ N m² C⁻¹.

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

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

Mistake: Using the angle with the surface instead of with the normal in φ = ES cos θ.
Questions often give the angle between the field and the plane of the sheet.
Wrong working
Field at 30° to the plane: φ = ES cos 30°
Right working
Field at 30° to the plane means 60° to the normal: φ = ES cos 60°
Do this: Always convert to the angle between E and the normal.
Usually costs: 1 mark.
Mistake: Writing σ/ε₀ for a single sheet.
Mixing up one sheet with a pair of plates or a conductor's surface.
Wrong working
Infinite sheet: E = σ/ε₀
Right working
Infinite sheet: E = σ/2ε₀ (flux through both faces of the pill-box)
Do this: Count the faces of the pill-box through which flux passes.
Usually costs: 1 mark in the derivation.
Mistake: No Gaussian surface drawn or named.
Students jump straight to the formula.
Do this: Draw and name the surface (coaxial cylinder, pill-box, concentric sphere) and say why E is constant on it.
Usually costs: ½ to 1 mark in every derivation.
Mistake: Including charges outside the surface in q_enc.
Thinking every charge in the question contributes to flux.
Do this: Circle the charges inside the surface before adding.
Usually costs: The full mark of a flux MCQ.
Mistake: Saying the field inside a charged shell is zero because charges cancel.
The real reason is that a Gaussian sphere inside encloses no charge.
Do this: Write: 'q_enc = 0 for a Gaussian sphere of radius r < R, so E = 0'.
Usually costs: ½ mark for the reason.

Exam tips

The syllabus asks for the statement of Gauss's law, not its proof; spend your time on the three applications.
Each derivation earns marks for: symmetry argument, Gaussian surface (drawn), flux expression, enclosed charge, final result with direction.
When a question gives λ, σ or q with signs, state the direction of E (outward or inward).
Board papers have set the shell (inside and outside), the plane sheet and the line charge as 3- and 5-mark parts (for example 2024 55/5/1 Q31(b) and 2026 55/1/1 Q22(a)), and electric flux as 3-mark items (2024 55/1/1 Q22 and 55/5/1 Q23).
Practise sketching E versus r for the shell; a jump at r = R and a 1/r² fall must be visible.

Practice questions on Gauss's Law 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 an electric dipole. The net electric flux through the surface is
(a) $\frac{q}{ε _{0}}$ (b) $\frac{2 q}{ε _{0}}$ (c) zero (d) $\frac{p}{ε _{0}}$
Hint
Add the enclosed charges.
Answer
(c) zero, because the enclosed charge is +q − q = 0.
Q2 · 2 marks · Very short answerBoard level
Write the SI unit of electric flux. State Gauss's law in words.
Hint
Flux = E × area.
Answer
N m² C⁻¹ (or V m). Gauss's law: the total electric flux through any closed surface is $\frac{1}{ε _{0}}$ times the net charge enclosed by it.
Q3 · 2 marks · Very short answerHarder board level
A point charge q is surrounded by a Gaussian sphere of radius r. What happens to (a) the total flux and (b) the field at the surface if the radius is doubled?
Hint
Flux depends only on q; field depends on r.
Answer
(a) Unchanged, still $\frac{q}{ε _{0}}$ . (b) The field becomes one-quarter, since $E \propto \frac{1}{r ^{2}}$ ; the area becomes four times, so the product stays the same.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The electric field at a point on a Gaussian surface is due only to the charges enclosed by the surface.
Reason (R): The net flux through a closed surface depends only on the charge enclosed.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Field and flux are different things.
Answer
(d). The field at any point is due to all charges, inside and outside; only the net flux depends on the enclosed charge alone.
Q5 · 3 marks · Short answerHarder board level
The field at 20 cm from the centre of a thin charged spherical shell of radius 10 cm is $1.8 \times 1 0^{5}$ N/C, pointing outward. Find (a) the charge on the shell, (b) its surface charge density, and (c) the field just outside the surface.
Hint
Outside, the shell acts as a point charge at its centre.
Answer
(a) $Q = \frac{E r ^{2}}{k} = \frac{1.8 \times 1 0 ^{5} \times 0.04}{9 \times 1 0 ^{9}} = 8 \times 1 0^{- 7}$ C. (b) $σ = \frac{Q}{4 π R ^{2}} = \frac{8 \times 1 0 ^{- 7}}{4 π \times 0.01} \approx 6.37 \times 1 0^{- 6}$ C/m². (c) $E = \frac{k Q}{R ^{2}} = 7.2 \times 1 0^{5}$ N/C (four times the value at 20 cm).
Q6 · 2 marks · Very short answerHOTS / competency
A point charge q is placed at one corner of a cube. Find the flux through the cube.
Hint
Imagine 8 identical cubes around the corner.
Answer
Eight such cubes, meeting at the corner, would enclose the charge completely, so they share $\frac{q}{ε _{0}}$ equally: flux through one cube $= \frac{q}{8 ε _{0}}$ .
Q7 · 3 marks · Short answerHOTS / competency
Using the result for one infinite sheet, show that the field just outside a charged conductor of surface charge density σ is $\frac{σ}{ε _{0}}$ , not $\frac{σ}{2 ε _{0}}$ . (Hint: the field inside a conductor is zero.)
Hint
Use a pill-box with one face inside the conductor.
Answer
Take a pill-box of area A with one face just outside and one inside the conductor. Inside, E = 0, so there is no flux through that face; the sides have no flux. Flux = EA through the outer face only; enclosed charge σA. So $E A = \frac{σ A}{ε _{0}}$ and $E = \frac{σ}{ε _{0}}$ . For an isolated thin sheet, flux leaves through both faces ( $2 E A$ ), which gives $\frac{σ}{2 ε _{0}}$ .

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 law in symbols.Show answer
$\oint E \cdot d S = \frac{q _{e n c}}{ε _{0}}$ .
2.Field due to an infinitely long wire at distance r?Show answer
$\frac{λ}{2 π ε _{0} r}$ , radial.
3.Field due to an infinite plane sheet?Show answer
$\frac{σ}{2 ε _{0}}$ , independent of distance.
4.Field inside a charged thin spherical shell?Show answer
Zero.
5.Flux through one face of a cube with q at its centre?Show answer
$\frac{q}{6 ε _{0}}$ .
6.Does a charge outside a closed surface change the net flux?Show answer
No, but it changes the field at points of the surface.

Gauss's Law and Its Applications: frequently asked questions

Do I need to prove Gauss's law for CBSE Class 12?

No. The 2026-27 syllabus asks you to state Gauss's law and apply it to an infinitely long straight wire, an infinite plane sheet and a thin spherical shell (inside and outside).

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

The field lines from an infinite sheet stay parallel; they do not spread out as you move away. With the pill-box, the area through which flux passes does not change with distance, so E stays the same.

Why is the field inside a charged spherical shell zero?

Any Gaussian sphere drawn inside the shell encloses no charge. By symmetry E would be the same all over that sphere, so the only way the total flux can be zero is for E to be zero at every point inside.

Can Gauss's law be used to find the field of a finite wire?

Gauss's law is always true, but for a finite wire the field is not the same size all over a cylinder, so the flux cannot be written as E times an area. It is only useful as a shortcut when symmetry is high enough.

What is the unit of electric flux?

N m² C⁻¹, which is the same as V m.