Gauss's Law

Continuous charge distribution (NCERT 1.12, folded in here)

Before Gauss's law can be applied, the charge on an extended body has to be described by a density rather than by listing individual charges.

• Surface charge density $\sigma = \dfrac{\Delta Q}{\Delta S}$, unit $\text{C m}^{-2}$, for charge spread over a surface such as a conductor or a sheet.
• Linear charge density $\lambda = \dfrac{\Delta Q}{\Delta l}$, unit $\text{C m}^{-1}$, for charge along a wire.
• Volume charge density $\rho = \dfrac{\Delta Q}{\Delta V}$, unit $\text{C m}^{-3}$, for charge distributed through a volume.
• In each case the element ($\Delta l$, $\Delta S$, $\Delta V$) is small on the macroscopic scale but contains a very large number of microscopic charges, so the density is a smoothed-out average that ignores both the quantisation of charge and the microscopic discontinuity of the distribution. It is the same idealisation as treating a liquid as continuous rather than molecular.
• The field of a continuous distribution follows from Coulomb's law plus superposition: the element $\rho\,\Delta V$ contributes $\Delta\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\rho\,\Delta V}{r'^2}\hat{r}'$, where $r'$ is the distance from the element to the field point, and the total field is $\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\sum_{\text{all }\Delta V}\dfrac{\rho\,\Delta V}{r'^2}\hat{r}'$, an integral in the limit $\Delta V\to0$.
• Useful conversions: charge on length $l$ of a wire is $\lambda l$; on area $A$ of a sheet is $\sigma A$; on a shell of radius $R$ is $q = 4\pi R^2\sigma$.

Deriving the law from a sphere around a point charge

Put a point charge $q$ at the centre of a sphere of radius $r$ and divide the sphere into small area elements $\Delta\vec{S}$.

$$\Delta\phi = \vec{E}\cdot\Delta\vec{S} = \dfrac{q}{4\pi\varepsilon_0 r^2}\,\hat{r}\cdot\Delta\vec{S}$$

The normal to a sphere at every point is along the radius, so $\hat{r}$ and $\Delta\vec{S}$ are parallel and $\hat{r}\cdot\Delta\vec{S} = \Delta S$. Hence

$$\Delta\phi = \dfrac{q}{4\pi\varepsilon_0 r^2}\,\Delta S$$

Every element is at the same distance $r$, so summing over the whole sphere and using $\sum\Delta S = 4\pi r^2$,

$$\phi = \dfrac{q}{4\pi\varepsilon_0 r^2}\times 4\pi r^2 = \dfrac{q}{\varepsilon_0}$$

The result is independent of $r$: the flux is the same through every sphere around the charge. This is the simplest illustration of the general law.

Statement of Gauss's law

The total electric flux through any closed surface in vacuum is equal to $\dfrac{1}{\varepsilon_0}$ times the total charge enclosed by that surface.

$$\phi = \oint_S \vec{E}\cdot d\vec{S} = \dfrac{q_{\text{enclosed}}}{\varepsilon_0}$$

If the enclosed region is filled with a medium of dielectric constant $\varepsilon_r$, the flux becomes $\dfrac{q_{\text{enclosed}}}{\varepsilon_0\varepsilon_r}$.

The four points NCERT makes about it

• Gauss's law is true for any closed surface, whatever its shape or size.
• The $q$ on the right side is the sum of all charges enclosed by the surface, wherever inside they are placed.
• If some charges are inside and some outside, the $\vec{E}$ appearing in the integral is the field of all the charges, inside and outside; but only the charge inside contributes to $q$ on the right.
• The surface chosen for applying the law is called the Gaussian surface. It may be any imagined closed surface, but it must not be allowed to pass through a discrete point charge, since the field is not defined at the location of a point charge. It may pass through a continuous charge distribution.

Why the shape and size do not matter — the justification CBSE asks for

Place the same charge $Q$ once inside a sphere and once inside a cube. The number of field lines emanating from $Q$ is fixed, and every one of those lines must cross any closed surface enclosing $Q$ exactly once. So the same number of lines crosses the sphere as crosses the cube, and

$$\phi_{\text{net}} = \dfrac{Q}{\varepsilon_0}$$

in both cases. Increasing the radius of a spherical Gaussian surface increases its area but leaves the enclosed charge unchanged, so the flux is unchanged.

Zero flux

• If no charge is enclosed, the net flux is zero. NCERT's illustration: a closed cylinder with its axis parallel to a uniform field $\vec{E}$ has $\phi_3 = 0$ through the curved part (normal perpendicular to $\vec{E}$), $\phi_1 = -ES$ through the face whose outward normal opposes $\vec{E}$ and $\phi_2 = +ES$ through the other face, giving $\phi = 0$ in total.
• Conversely, if the net flux through a closed surface is zero, we may infer that the net charge enclosed is zero. We may not infer that the field inside is zero, or uniform, or that the potential is constant.
• A cube of side $1\,\text{cm}$ enclosing an electric dipole encloses net charge $+q-q = 0$, so the flux through it is zero, whatever the size of the cube.

Signed sums and outside charges

For a closed surface enclosing $-q_1$ and $+q_2$ while $+q_3$ lies outside, the flux is

$$\phi = \dfrac{q_2-q_1}{\varepsilon_0}$$

The outside charge $q_3$ contributes nothing to the flux, although it does contribute to $\vec{E}$ at every point of the surface.

Concentric surfaces

Two concentric closed surfaces $S_1$ and $S_2$, with $S_1$ enclosing $2Q$ and the shell between them carrying a further $4Q$:

$$\phi_1 = \dfrac{2Q}{\varepsilon_0},\qquad \phi_2 = \dfrac{2Q+4Q}{\varepsilon_0} = \dfrac{6Q}{\varepsilon_0},\qquad \phi_1:\phi_2 = 1:3$$

If a medium of dielectric constant $\varepsilon_r$ fills the space inside $S_1$, then $\phi_1' = \dfrac{2Q}{\varepsilon_0\varepsilon_r}$.

Charge at the centre of a cube, and the square trick

A point charge $q$ at the centre of a cube sends flux $\dfrac{q}{\varepsilon_0}$ through the whole cube, and by symmetry each of the six faces takes an equal share:

$$\phi_{\text{face}} = \dfrac{q}{6\varepsilon_0}$$

The board's favourite dressing of this: a charge $q$ sits at height $\dfrac{d}{2}$ directly above the centre of a square of side $d$. Imagine the square as one face of a cube of edge $d$; the charge is then at the centre of that cube, so the flux through the square is $\dfrac{q}{6\varepsilon_0}$. If the height is doubled to $d$ and the side doubled to $2d$, the charge again sits at the centre of an imaginary cube, this time of edge $2d$, so the flux is unchanged at $\dfrac{q}{6\varepsilon_0}$.

Flux from a position-dependent field

Where the field varies with position, evaluate the flux face by face. For $\vec{E} = 2x\hat{i}$ and a cube of side $a$ with one face at $x=0$ and the opposite face at $x=a$: only the two faces perpendicular to the $x$-axis matter. At the left face $E_L = 0$; at the right face $E_R = 2a$ with the outward normal along $+\hat{i}$. Hence

$$\phi = 0+2a\times a^2 = 2a^3,\qquad q = \varepsilon_0\phi = 2a^3\varepsilon_0$$

True or false, quickly