Electric Flux & Gauss's Law

PHYGauss's Law & Flux

Electric flux measures the number of field lines crossing a surface, and Gauss's law states that the net flux through any closed surface equals the enclosed charge divided by $\varepsilon_0$ . Its power is in computing fields of symmetric charge distributions — a line, a sheet, a sphere — without integration.

ISC frequently tests the standard results for an infinite wire, an infinite sheet and a charged conductor.

Electric flux

\phi_E = \vec{E}\cdot\vec{A} = EA\cos\theta

\theta

is the angle between

\vec{E}

and the area normal

\vec{A}

; units

\text{N m}^2\,\text{C}^{-1}

Gauss's law

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

q_{enc}

is the total charge enclosed by the closed (Gaussian) surface; only enclosed charge counts

Derivation

Enclose a point charge $q$ in a sphere of radius $r$ . By symmetry $\vec{E}$ is radial and uniform over the surface, so the flux is field times area:
$\phi_E = \oint \vec{E}\cdot d\vec{A} = E\,(4\pi r^2)$
Substitute the point-charge field $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ :
$\phi_E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}\,(4\pi r^2) = \dfrac{q}{\varepsilon_0}$
The $r^2$ cancels, so the flux is independent of the surface's size or shape and counts only the enclosed charge:
$\oint \vec{E}\cdot d\vec{A} = \dfrac{q_{enc}}{\varepsilon_0}$

Field of an infinite line charge

E = \dfrac{\lambda}{2\pi\varepsilon_0 r}

\lambda

linear charge density (

\text{C m}^{-1}

r

perpendicular distance;

E\propto 1/r

Derivation

Use a coaxial cylinder (radius $r$ , length $l$ ) as the Gaussian surface. By symmetry $\vec{E}$ is radial, so flux crosses only the curved surface:
$\phi_E = E\,(2\pi r l)$
The enclosed charge is $q_{enc} = \lambda l$ , so Gauss's law reads:
$E\,(2\pi r l) = \dfrac{\lambda l}{\varepsilon_0}$
Cancel $l$ and solve for $E$ :
$E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$

Field of an infinite charged sheet

E = \dfrac{\sigma}{2\varepsilon_0}

\sigma

surface charge density (

\text{C m}^{-2}

); uniform and independent of distance

Derivation

Take a Gaussian pillbox of cross-section $A$ piercing the sheet. By symmetry $\vec{E}$ is perpendicular and emerges from both flat faces:
$\phi_E = EA + EA = 2EA$
The enclosed charge is $q_{enc} = \sigma A$ , so Gauss's law gives:
$2EA = \dfrac{\sigma A}{\varepsilon_0}$
Cancel $A$ :
$E = \dfrac{\sigma}{2\varepsilon_0}$

Field of a charged conducting sphere

E_{out} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}, \qquad E_{in}=0

outside (

r\ge R

) it behaves like a point charge at the centre; inside the conductor the field is zero

Derivation

Outside ( $r \ge R$ ): a concentric Gaussian sphere encloses the entire charge $q$ , so by symmetry:
$E\,(4\pi r^2) = \dfrac{q}{\varepsilon_0} \implies E_{out} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$
Inside ( $r < R$ ): all charge sits on the surface, so a Gaussian sphere within the conductor encloses none:
$E\,(4\pi r^2) = \dfrac{0}{\varepsilon_0} \implies E_{in} = 0$
So outside it behaves like a point charge at the centre, while the field jumps to zero on entering the conductor.

Net flux through a closed surface depends only on the enclosed charge — not on its position inside, the surface shape, or external charges.
An external charge contributes zero net flux (lines that enter also leave), though it does affect the field at individual points.
For an infinite sheet $E=\sigma/2\varepsilon_0$ is independent of distance; just outside a charged conductor the field is $E=\sigma/\varepsilon_0$ , where $\sigma$ is the local surface charge density.
The line-charge field falls as $1/r$ , the sheet field is constant, and the point/sphere field falls as $1/r^2$ — know which symmetry gives which dependence.
Field is zero everywhere inside a conductor in electrostatic equilibrium, and all excess charge resides on the outer surface.
At the surface of a conducting sphere of radius $R$ , $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R^2}$ ; just inside the metal $E=0$ .
Choose a Gaussian surface matching the symmetry (cylinder for line/sheet, sphere for point/sphere) so $E$ is constant over it.

Where the marks go

Confusing the infinite-sheet field $\sigma/2\varepsilon_0$ with the conductor-surface field $\sigma/\varepsilon_0$ (factor of 2).
Thinking flux changes when the charge moves around inside the surface or when the surface is reshaped — it does not.
Using the point-charge $1/r^2$ formula for a line charge; the infinite line gives $1/r$ with the constant $1/2\pi\varepsilon_0$ .
Computing field inside a conductor as non-zero, or putting charge throughout its volume rather than on the outer surface.

How the board asks it

Define / stateelectric flux and gauss's law statement
Define electric flux and state its SI unit. State Gauss's law in electrostatics.
Derive / provefield of an infinite charged sheet using a cylindrical gaussian surface
Using Gauss's law, derive an expression for the electric field intensity at a point near an infinite plane sheet of charge having uniform surface charge density $\sigma$ . Hence show that $E$ is independent of the distance from the sheet.
Numericalfield of an infinite line charge, $E=\lambda/2\pi\varepsilon_0 r$
An infinitely long straight wire carries a uniform linear charge density $\lambda = 4\times10^{-6}\ \text{C m}^{-1}$ . Calculate the electric field intensity at a point $0.2\ \text{m}$ from the wire. $\left(\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \text{N m}^2\text{C}^{-2}\right)$
Give reasonsnet flux depends only on the enclosed charge
A point charge is placed at the centre of a cube. State, giving a reason, how the total electric flux through the cube changes if (i) the charge is moved to a corner of the cube and (ii) the side of the cube is doubled.
Applicationfield at the surface of a charged conductor; $E=0$ inside
Show that the electric field just outside a charged conductor is $E=\sigma/\varepsilon_0$ , where $\sigma$ is the local surface charge density, and state the value of the field inside the conductor.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.