MHT-CET Physics · Electrostatics

Gauss's Law and Electric Flux

The total electric flux out of any closed surface equals the charge inside divided by ε₀ — whatever the surface's size or shape — and for symmetric charge the same law gives the field in one line.

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Why this matters

20 PYQs, none HARD: this is the chapter's most reliable marks. Two in three ask only 'what is the flux?', and the answer never depends on the surface's size. The rest ask for the field of a symmetric body — sphere, sheet, cylinder, or the surface of a conductor.

Concept 1 of 2: Electric Flux and Gauss's Law

Flux counts field lines crossing a surface. Every line leaving a charge must cross any closed surface around it once, however big or odd the surface — so the total flux out depends only on the charge inside. Charges outside send lines in and out again and add nothing.

Definition

$\phi = \oint \vec E\cdot d\vec A = \dfrac{q_{\text{enclosed}}}{\varepsilon_0}$ ; unit V m (= N m² C⁻¹).
Net flux = flux leaving − flux entering: $q = \varepsilon_0(\phi_2 - \phi_1)$ .
Growing the surface (a bigger sphere, an inflating balloon) with the same charge inside leaves $\phi$ unchanged.
Charge at the centre of a cube: each face carries $\dfrac{q}{6\varepsilon_0}$ , two opposite faces $\dfrac{q}{3\varepsilon_0}$ .
Closed cylinder with charge inside and flux $\phi$ through the curved face: each flat end carries $\dfrac{1}{2}\left(\dfrac{q}{\varepsilon_0} - \phi\right)$ .
Charged sphere of surface density $\sigma$ , radius $r$ : $q = 4\pi r^2\sigma$ , $\phi = \dfrac{4\pi r^2 \sigma}{\varepsilon_0}$ . Watch diameter versus radius.

Gauss's law

\oint \vec E\cdot d\vec A = \frac{q_{\text{enc}}}{\varepsilon_0}

Worked example

A conducting sphere of diameter 10 cm carries

8.85\,\mu\text{C m}^{-2}

on its surface. Total flux leaving it? (

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

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 2 · Q30Easy

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

If the electric flux entering and leaving an enclosed surface are

\phi_1

and

\phi_2

respectively, the electric charge inside the surface will be

Subtracting the wrong way round

Gauss's law counts OUTWARD flux as positive, so the net flux is leaving minus entering:

q = \varepsilon_0(\phi_2 - \phi_1)

. The options pair both orders with both

\varepsilon_0

placements — fix the order first, then multiply.

Drill 12 more on electric flux and gauss's law

Concept 2 of 2: Fields of Spheres, Sheets, Cylinders and Conductors

Pick a surface that matches the symmetry, so the field has one value over it; then Gauss's law gives the field at once. Outside any spherical charge the field is as if the whole charge sat at the centre. Inside a conductor it is zero, and just outside it points straight out.

Definition

Sphere or shell of total charge $Q$ , outside ( $r \ge R$ ): $E = \dfrac{kQ}{r^2}$ . With surface density $\sigma$ : $E = \dfrac{\sigma R^2}{\varepsilon_0 r^2}$ , $r$ measured from the CENTRE.
Inside a conductor or a hollow shell: $E = 0$ .
Uniform solid sphere of density $\rho$ : $E = \dfrac{\rho r}{3\varepsilon_0}$ inside, $\dfrac{\rho R}{3\varepsilon_0}$ on the surface.
Infinite sheet: $E = \dfrac{\sigma}{2\varepsilon_0}$ , independent of distance. Just outside a conductor: $E = \dfrac{\sigma}{\varepsilon_0}$ , normal to the surface.
Long charged cylinder: outside $E \propto \dfrac{1}{r}$ ( $\dfrac{\lambda}{2\pi\varepsilon_0 r}$ ); inside a uniformly charged one $E \propto r$ .
Sphere $+3Q$ inside a shell $-Q$ : between them only the inner charge counts: $E = \dfrac{3Q}{4\pi\varepsilon_0 r^2}$ .

Three results to recall

E_{\text{sheet}} = \frac{\sigma}{2\varepsilon_0}, \qquad E_{\text{conductor}} = \frac{\sigma}{\varepsilon_0}, \qquad E_{\text{solid sphere}}(r \le R) = \frac{\rho r}{3\varepsilon_0}

Worked example

A charge of

8.85 \times 10^{-7}

C is spread over a large sheet of area 2 m². Field near it?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q20Moderate

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

The electric field intensity on the surface of a solid charged sphere of radius '

r

' and volume charge density '

\rho

' is (

\epsilon_0=

permittivity of free space)

Distance from the surface is not distance from the centre

'0.2 m from a point on the surface' of a 0.1 m sphere puts the point 0.3 m from the centre. The field formula uses the distance from the centre; the other reading lands on an option too.

Sheet versus conductor

A thin sheet of charge sends half its field each way:

\frac{\sigma}{2\varepsilon_0}

. A conductor has field only on its outside:

\frac{\sigma}{\varepsilon_0}

. The paper offers both with both directions.

Drill 6 more on fields of spheres, sheets, cylinders and conductors

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)

Electric Flux and Gauss's Law
Gauss's law
$\oint \vec E\cdot d\vec A = \frac{q_{\text{enc}}}{\varepsilon_0}$
Fields of Spheres, Sheets, Cylinders and Conductors
Three results to recall
$E_{\text{sheet}} = \frac{\sigma}{2\varepsilon_0}, \qquad E_{\text{conductor}} = \frac{\sigma}{\varepsilon_0}, \qquad E_{\text{solid sphere}}(r \le R) = \frac{\rho r}{3\varepsilon_0}$

Watch out for (3)

Subtracting the wrong way round
→ Electric Flux and Gauss's Law
Distance from the surface is not distance from the centre
→ Fields of Spheres, Sheets, Cylinders and Conductors
Sheet versus conductor
→ Fields of Spheres, Sheets, Cylinders and Conductors

Test yourself on Electrostatics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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Related notes

Electric Potential — the same bodies, as potentials →