JEE Mains Physics · Electrostatics

Electric Flux and Gauss's Law

Flux counts the field through a surface, E·A; through any closed surface it equals the charge inside divided by ε₀, which gives flux by symmetry and, for symmetric charge, the field itself.

All Electrostatics notes JEE Mains Physics strategy

Why this matters

Thirty-five PYQs, twenty-five of them multiple choice, and four from 2026. Eight ask for the flux through a flat surface, or through a cube in a field that changes with position. Sixteen find the enclosed charge, or share a charge's flux among the faces of a cube by symmetry. Eleven use Gauss's law for the field of a sphere, a shell or a cylinder, sometimes starting from a given potential.

Concept 1 of 3: Flux through a surface and through a cube

Flux measures how much field passes through a surface. Only the part of the field along the surface's normal counts, so a surface edge-on to the field carries none. For a cube in a field that points along x and changes with x, only the two faces facing the field carry flux. The net flux is what leaves through the far face minus what enters through the near one.

Definition

$\phi = \vec E \cdot \vec A = EA\cos\theta$ , with $\vec A$ along the normal. Unit: N m² C⁻¹ (= V m).
A surface in the yz-plane has $\vec A$ along $\hat i$ ; one parallel to the xz-plane, along $\hat j$ ; one in the xy-plane, along $\hat k$ .
Field $\vec E = f(x)\,\hat i$ through a cube of face area A between $x_1$ and $x_2$ : $\phi_{\text{net}} = [f(x_2) - f(x_1)]A$ . The four faces parallel to the field carry nothing.
A uniform field gives zero net flux through any closed surface.
Enclosed charge from the net flux: $q = \varepsilon_0\phi_{\text{net}}$ .

Flux and Gauss's law

\phi = \vec E \cdot \vec A, \qquad \phi_{\text{net}} = \oint \vec E \cdot d\vec A = \frac{q_{\text{enc}}}{\varepsilon_0}

Worked example

The field in a region is

\vec E = (5 + 2x)\,\hat i

N/C, with x in metres. A cube of side 0.5 m has its faces at

x = 1

m and

x = 1.5

m. Find the net flux and the charge inside. (

\varepsilon_0 = 8.85 \times 10^{-12}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q116Moderate

Example 1 · Electrostatics · Electric Flux and Gauss's Law

A cubical volume is bounded by the surfaces

x = 0,x = a,y = 0,y = a,z = 0,z = a

. The electric field in the region is given by

\overrightarrow{E}=E_{0}x\widehat{\imath}

. Where

E_{0}= 4 \times10^{4}NC^{- 1}{\text{ }m}^{- 1}

. If

a = 2\text{ }cm

, the charge contained in the cubical volume is

Q \times10^{- 14}C

. The value of

Q

is .___ (Take

\left. \ \epsilon_{0}= 9 \times\frac{10^{- 12}C^{2}}{Nm^{2}} \right)

Name the normal, not the plane

A surface 'in the yz-plane' or 'parallel to the yz-plane' has its normal along x. Taking the field's y-part for it gives the wrong flux.

Only faces facing the field count

When E points along x, the four faces parallel to x carry no flux. Do not multiply the field by the total area of the cube.

Inward flux is negative

Net flux is flux out minus flux in. Adding the two sizes gives a charge that is far too large.

Drill 7 more on flux through a surface and through a cube

Concept 2 of 3: Enclosed charge and sharing flux by symmetry

The net flux through a closed surface depends only on the charge inside, never on the surface's size or shape or on charges outside. When a charge sits on the surface, imagine copies of the box stacked round it until the charge is at the centre of a larger symmetric shape. Its flux then shares equally among the identical faces, and your box gets its fraction.

Definition

$\phi = \dfrac{q_{\text{enc}}}{\varepsilon_0}$ , whatever the shape or size of the surface.
Charges outside change the field on the surface but add nothing to the net flux.
A dipole, or any group with zero net charge, inside a surface gives zero net flux.
A charge on the surface counts by the fraction of space the surface wraps round it: half on a flat face, a quarter on an edge, an eighth at a corner of a cube.
A face that contains the charge carries no flux: the charge's field runs along that face.
A line charge partly inside: only the length inside counts.

Charge q placed at	Flux through the whole cube	Flux through single faces	How to see it
Centre of the cube	$q/\varepsilon_0$	$q/6\varepsilon_0$ through each face	Six identical faces share it equally
Centre of one face	$q/2\varepsilon_0$	Zero through the face it sits on	A second cube on the other side takes the other half
Middle of an edge	$q/4\varepsilon_0$	Zero through the two faces meeting at that edge	Four cubes share the edge
A corner	$q/8\varepsilon_0$	$q/24\varepsilon_0$ through each of the three far faces; zero through the three faces at the corner	Eight cubes meet at the corner The far faces get q/24ε₀ each, not q/8ε₀: three faces share the cube's eighth.
On the axis of a square of side a, at distance a/2	$q/\varepsilon_0$ through the imagined cube of side a	$q/6\varepsilon_0$ through the square	The square is one face of a cube centred on q
Outside the cube	Zero	Inward on some faces, outward on others	What enters also leaves

Stack copies of the box round the charge until it sits at a centre of symmetry, then divide.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q4Moderate

Example 2 · Electrostatics · Electric Flux and Gauss's Law

In a cuboid of dimension

2\text{ }L \times 2\text{ }L \times L

, a charge

q

is placed at the center of the surface '

S

' having area of

4{\text{ }L}^{2}

. The flux through the opposite surface to '

S

' is given by

A charge on the surface counts in part

A charge on a flat face is half inside the box. Counting all of it doubles the flux.

Outside charges do not change the net flux

They change the field at every point of the surface, so the flux through one face can change. The net flux through the closed surface stays q_enc/ε₀.

The flat face of a hemisphere

With the charge at the centre of the flat face, the field runs along that face and its flux is zero. The q/2ε₀ goes through the curved part.

Drill 15 more on enclosed charge and sharing flux by symmetry

Concept 3 of 3: Gauss's law for spheres, shells and cylinders

When charge is arranged with full symmetry, the field has the same size everywhere on a matching surface: a sphere for a sphere, a cylinder for a long line. Gauss's law then reduces to field times area equals enclosed charge over ε₀. Inside a uniformly charged solid sphere, only the charge closer to the centre counts, so the field grows with r and peaks at the surface.

Definition

Uniform solid sphere (insulating), charge Q, radius R: $E = \dfrac{kQr}{R^{3}} = \dfrac{\rho r}{3\varepsilon_0}$ inside, $\dfrac{kQ}{r^{2}}$ outside, largest at r = R.
Thin shell or conducting sphere: zero inside, $\dfrac{kQ}{r^{2}}$ outside, $\dfrac{\sigma}{\varepsilon_0}$ just outside the surface.
Long cylinder of uniform ρ and radius R: $E = \dfrac{\rho r}{2\varepsilon_0}$ inside, $\dfrac{\rho R^{2}}{2\varepsilon_0 r}$ outside.
Density that varies with r: $q_{\text{enc}} = \displaystyle\int_0^{r} \rho(r')\,4\pi r'^{2}\,dr'$ , then $E\,4\pi r^{2} = q_{\text{enc}}/\varepsilon_0$ .
Given V(r): $E = -\dfrac{dV}{dr}$ , then Gauss's law gives $q_{\text{enc}}$ , or $\rho = \dfrac{\varepsilon_0}{r^{2}}\dfrac{d}{dr}(r^{2}E)$ .
Conductors: zero field in the metal and in an empty cavity; excess charge sits on the surface. A charge q inside a neutral conducting shell induces −q on the inner surface and +q on the outer.

Fields from Gauss's law

E_{\text{sphere, in}} = \frac{\rho r}{3\varepsilon_0}, \qquad E_{\text{cylinder, in}} = \frac{\rho r}{2\varepsilon_0}, \qquad E_{\text{out}} = \frac{kQ}{r^{2}}

Worked example

An insulating sphere of radius 10 cm carries

2\ \mu\text{C}

spread uniformly through its volume. Find the field at 5 cm, at the surface and at 20 cm from the centre.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q119Moderate

Example 3 · Electrostatics · Electric Flux and Gauss's Law

For a charged spherical ball, electrostatic potential inside the ball varies with

r

V = 2ar^{2}+ b

. Here,

a

and

b

are constant and

r

is the distance from the center. The volume charge density inside the ball is

- \lambda a\varepsilon

. The value of

\lambda

is_____

\varepsilon =

permittivity of the medium

Inside a solid sphere the field grows

For a uniformly charged insulating sphere, E rises linearly from zero at the centre to its peak at the surface. Only outside does it fall as 1/r².

'On the surface' of a shell

Just outside a shell the field is σ/ε₀; just inside it is zero. Read which side the question means.

Varying density needs an integral

When ρ changes with r, the enclosed charge is ∫ρ 4πr² dr. Multiplying ρ(r) by the volume gives the wrong power of r.

Drill 10 more on gauss's law for spheres, shells and cylinders

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Flux through a surface and through a cube
Flux and Gauss's law
$\phi = \vec E \cdot \vec A, \qquad \phi_{\text{net}} = \oint \vec E \cdot d\vec A = \frac{q_{\text{enc}}}{\varepsilon_0}$
Gauss's law for spheres, shells and cylinders
Fields from Gauss's law
$E_{\text{sphere, in}} = \frac{\rho r}{3\varepsilon_0}, \qquad E_{\text{cylinder, in}} = \frac{\rho r}{2\varepsilon_0}, \qquad E_{\text{out}} = \frac{kQ}{r^{2}}$

Reference tables (1)

Enclosed charge and sharing flux by symmetry6 rows

Charge q placed at	Flux through the whole cube	Flux through single faces	How to see it
Centre of the cube	$q/\varepsilon_0$	$q/6\varepsilon_0$ through each face	Six identical faces share it equally
Centre of one face	$q/2\varepsilon_0$	Zero through the face it sits on	A second cube on the other side takes the other half
Middle of an edge	$q/4\varepsilon_0$	Zero through the two faces meeting at that edge	Four cubes share the edge
A corner	$q/8\varepsilon_0$	$q/24\varepsilon_0$ through each of the three far faces; zero through the three faces at the corner	Eight cubes meet at the corner The far faces get q/24ε₀ each, not q/8ε₀: three faces share the cube's eighth.
On the axis of a square of side a, at distance a/2	$q/\varepsilon_0$ through the imagined cube of side a	$q/6\varepsilon_0$ through the square	The square is one face of a cube centred on q
Outside the cube	Zero	Inward on some faces, outward on others	What enters also leaves

Stack copies of the box round the charge until it sits at a centre of symmetry, then divide.

Watch out for (9)

Name the normal, not the plane
→ Flux through a surface and through a cube
Only faces facing the field count
→ Flux through a surface and through a cube
Inward flux is negative
→ Flux through a surface and through a cube
A charge on the surface counts in part
→ Enclosed charge and sharing flux by symmetry
Outside charges do not change the net flux
→ Enclosed charge and sharing flux by symmetry
The flat face of a hemisphere
→ Enclosed charge and sharing flux by symmetry
Inside a solid sphere the field grows
→ Gauss's law for spheres, shells and cylinders
'On the surface' of a shell
→ Gauss's law for spheres, shells and cylinders
Varying density needs an integral
→ Gauss's law for spheres, shells and cylinders

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