Gauss’s Law

You have already seen that the electric flux through a surface tells you how many field lines pass through it. Now comes a remarkable result: if you wrap a closed surface around a charge and add up the flux through every tiny patch of that surface, the answer depends on nothing but the amount of charge sitting inside. It does not matter how big the surface is, what shape it takes, or where exactly the charge sits within it. This is Gauss’s law, and it is one of the most powerful tools in all of electrostatics.

Building Up to the Law: Flux Through a Sphere

The cleanest way to see why Gauss’s law works is to start with the simplest possible case. Place a single point charge $q$ at the centre of a sphere of radius $r$, and calculate the total flux through the sphere step by step.

Fig 1.22: Flux through a sphere enclosing a point charge $q$ at its centre

Step 1: Write the flux through one small patch.

Pick any small area element $\Delta\mathbf{S}$ on the surface of the sphere. The flux through this patch is the dot product of the electric field with the area vector:

$\Delta\phi = \mathbf{E} \cdot \Delta\mathbf{S} = \frac{q}{4\pi\varepsilon_0 r^2} \hat{\mathbf{r}} \cdot \Delta\mathbf{S} \qquad \text{(1.28)}$

Here the field $\mathbf{E}$ comes straight from Coulomb’s law for a point charge. The unit vector $\hat{\mathbf{r}}$ points radially outward from the charge at the centre to the area element on the surface.

Step 2: Recognise that $\hat{\mathbf{r}}$ and $\Delta\mathbf{S}$ are parallel.

On a sphere, the outward normal at every point lies along the radius at that point. This means $\hat{\mathbf{r}}$ and $\Delta\mathbf{S}$ point in the same direction everywhere on the surface. Their dot product simply gives the magnitude of $\Delta\mathbf{S}$:

$\hat{\mathbf{r}} \cdot \Delta\mathbf{S} = \Delta S$

since the magnitude of a unit vector is 1. So the flux through one patch simplifies to:

$\Delta\phi = \frac{q}{4\pi\varepsilon_0 r^2} \Delta S \qquad \text{(1.29)}$

Step 3: Add up the flux from every patch.

To get the total flux, sum $\Delta\phi$ over all area elements covering the entire sphere:

$\phi = \sum_{\text{all } \Delta S} \frac{q}{4\pi\varepsilon_0 r^2} \Delta S$

Every element sits at the same distance $r$ from the charge (they all lie on a sphere of radius $r$), so the factor $\frac{q}{4\pi\varepsilon_0 r^2}$ is the same for every element and can be pulled out of the sum:

$\phi = \frac{q}{4\pi\varepsilon_0 r^2} \sum_{\text{all } \Delta S} \Delta S = \frac{q}{4\pi\varepsilon_0 r^2} \times S$

where $S$ is the total surface area of the sphere.

Step 4: Substitute the surface area and simplify.

The surface area of a sphere is $S = 4\pi r^2$. Plugging this in:

$\phi = \frac{q}{4\pi\varepsilon_0 r^2} \times 4\pi r^2 = \frac{q}{\varepsilon_0} \qquad \text{(1.30)}$

The $r^2$ in the denominator (from Coulomb’s law) cancels perfectly with the $r^2$ in the surface area. This cancellation is not a coincidence; it happens precisely because the electric field falls off as the inverse square of distance. The final result is beautifully simple: the total flux depends only on the charge and the permittivity of free space, with no dependence on the radius of the sphere at all.

The General Statement of Gauss’s Law

Equation (1.30) was derived for a sphere, but the result holds far more generally. Gauss’s law states:

The total electric flux through any closed surface $S$ equals the total charge enclosed by that surface divided by $\varepsilon_0$:

$\phi = \frac{q}{\varepsilon_0} \qquad \text{(1.31)}$

where $q$ is the algebraic sum of all charges inside $S$.

The immediate consequence is this: if no charge sits inside a closed surface, the net flux through it is zero. Whatever field lines enter the surface from one side must leave from another, and the inward and outward contributions cancel exactly.

Verifying Zero Flux with a Cylinder

To see this cancellation in action, consider a closed cylinder placed with its axis parallel to a uniform electric field $\mathbf{E}$. The cylinder has two flat circular end faces (call them face 1 on the left and face 2 on the right) and a curved side surface (face 3).

Fig 1.23: Flux of a uniform electric field through a closed cylindrical surface

The total flux is $\phi = \phi_1 + \phi_2 + \phi_3$.

Curved surface (face 3): At every point on the curved side, the outward normal is perpendicular to the cylinder’s axis, which means it is perpendicular to $\mathbf{E}$. Since $\mathbf{E} \cdot \Delta\mathbf{S} = E\,\Delta S \cos 90° = 0$, the flux through the curved surface is:

$\phi_3 = 0$

Left face (face 1): The outward normal points to the left, opposite to $\mathbf{E}$ (which points right). The angle between them is $180°$, so:

$\phi_1 = -E\,S_1$

Right face (face 2): The outward normal points to the right, in the same direction as $\mathbf{E}$. The angle is $0°$, so:

$\phi_2 = +E\,S_2$

Since the two faces have equal area ($S_1 = S_2 = S$), the total flux is:

$\phi = -ES + ES + 0 = 0$

The net flux is zero, exactly as Gauss’s law predicts for a region containing no charge.

This example also works in reverse: whenever you find that the net flux through a closed surface is zero, you can conclude that the total charge inside is zero.

Six Key Properties of Gauss’s Law

Gauss’s law looks simple, but several subtleties deserve careful attention.

1. Valid for any closed surface. The closed surface you choose for applying Gauss’s law is called the Gaussian surface (the surface through which you evaluate the total flux). Its shape and size do not matter. A sphere, a cube, an irregular potato-shaped blob: Gauss’s law works for all of them. You choose whichever shape makes the calculation easiest.

2. $q$ includes all enclosed charges. The right-hand side of the law uses the algebraic sum of every charge sitting anywhere inside the surface. Positive and negative charges within the surface can partially or fully cancel each other.

3. The field on the surface is due to all charges, but the flux depends only on internal ones. This is a point that often causes confusion. The electric field $\mathbf{E}$ at any point on the Gaussian surface is produced by every charge in the universe, both inside and outside the surface. However, when you integrate this field over the entire closed surface, the contributions from outside charges produce equal amounts of inward and outward flux that cancel perfectly. Only the charges inside survive in the net flux.

4. The Gaussian surface must not pass through a discrete point charge. The electric field from a point charge blows up (grows without bound) as you approach the charge, so the field is not well defined at the location of a discrete charge. Your chosen surface must avoid such points. However, the surface can pass through a continuous charge distribution, where the field remains finite everywhere.

5. Symmetry makes Gauss’s law a powerful calculation tool. Although Gauss’s law is always true, it becomes especially useful for computing electric fields when the charge distribution has spherical, cylindrical, or planar symmetry. In such cases, you can pick a Gaussian surface that matches the symmetry, making the flux integral straightforward to evaluate.

6. Rooted in the inverse-square law. The entire derivation relied on the $1/r^2$ dependence of Coulomb’s law, which allowed the $r^2$ terms to cancel. If the force law had any other power of $r$, the cancellation would fail and Gauss’s law would break down. This means any experimental violation of Gauss’s law would be direct evidence that the electrostatic force does not follow an exact inverse-square dependence.

Solved Example 1.10: Flux and Charge in a Cube

Problem: The electric field components in Fig. 1.24 are $E_x = \alpha x^{1/2}$, $E_y = E_z = 0$, where $\alpha = 800$ N/C m$^{1/2}$. The cube has side $a = 0.1$ m, with its left face at $x = a$ and right face at $x = 2a$. Calculate (a) the flux through the cube, and (b) the charge within the cube.

Fig 1.24: Cube of side $a$ in a non-uniform electric field along the $x$-axis

Solution

(a) Finding the flux

Since $\mathbf{E}$ has only an $x$-component, the four faces whose outward normals point in the $\pm y$ or $\pm z$ directions are perpendicular to $\mathbf{E}$. The dot product $\mathbf{E} \cdot \Delta\mathbf{S}$ is zero for each of these four faces, so they contribute no flux. Only the left face (at $x = a$) and the right face (at $x = 2a$) matter.

Field at the left face ($x = a$):

$E_L = \alpha\, a^{1/2}$

The outward normal on the left face points in the $-x$ direction, which is opposite to $\mathbf{E}$. The angle between them is $180°$, so:

$\phi_L = E_L \times a^2 \times \cos 180° = -E_L\, a^2 = -\alpha\, a^{1/2} \times a^2 = -\alpha\, a^{5/2}$

Field at the right face ($x = 2a$):

$E_R = \alpha\,(2a)^{1/2}$

The outward normal on the right face points in the $+x$ direction, same as $\mathbf{E}$. The angle is $0°$, so:

$\phi_R = E_R \times a^2 \times \cos 0° = +E_R\, a^2 = \alpha\,(2a)^{1/2} \times a^2$

Net flux through the cube:

$\phi = \phi_R + \phi_L = \alpha\,(2a)^{1/2}\, a^2 - \alpha\, a^{1/2}\, a^2$

Factor out $\alpha\, a^{1/2}\, a^2 = \alpha\, a^{5/2}$:

$\phi = \alpha\, a^{5/2}\left(\sqrt{2} - 1\right)$

Substitute the numbers ($\alpha = 800$ N/C m$^{1/2}$, $a = 0.1$ m):

$\phi = 800 \times (0.1)^{5/2} \times (\sqrt{2} - 1)$

$(0.1)^{5/2} = (0.1)^2 \times (0.1)^{1/2} = 0.01 \times 0.3162 = 3.162 \times 10^{-3}$

$\phi = 800 \times 3.162 \times 10^{-3} \times 0.4142 = 1.05 \text{ N m}^2 \text{ C}^{-1}$

(b) Finding the enclosed charge

Apply Gauss’s law: $\phi = q/\varepsilon_0$, which gives $q = \phi\,\varepsilon_0$.

$q = 1.05 \times 8.854 \times 10^{-12} = 9.27 \times 10^{-12} \text{ C}$

The cube encloses approximately $9.27 \times 10^{-12}$ C of charge.

Solved Example 1.11: Flux Through a Cylinder in a Symmetric Field

Problem: An electric field is uniform and points in the $+x$ direction for positive $x$, and uniform with the same magnitude but in the $-x$ direction for negative $x$. Specifically, $\mathbf{E} = 200\,\hat{\mathbf{i}}$ N/C for $x > 0$ and $\mathbf{E} = -200\,\hat{\mathbf{i}}$ N/C for $x < 0$. A right circular cylinder of length 20 cm and radius 5 cm is centred at the origin with its axis along the $x$-axis, so one flat face is at $x = +10$ cm and the other at $x = -10$ cm. Find (a) the net outward flux through each flat face, (b) the flux through the curved side, (c) the net outward flux through the cylinder, and (d) the net charge inside the cylinder.

Fig 1.25: Cylinder centred at the origin in a symmetric field that reverses direction at $x = 0$

Solution

(a) Flux through each flat face

Left face (at $x = -10$ cm): Here the field is $\mathbf{E} = -200\,\hat{\mathbf{i}}$ N/C (pointing in the $-x$ direction). The outward normal to the left face also points in the $-x$ direction. Since $\mathbf{E}$ and the outward normal are parallel:

$\phi_L = E \times \Delta S = 200 \times \pi(0.05)^2 = 200 \times \pi \times 0.0025 = +1.57 \text{ N m}^2 \text{ C}^{-1}$

(The flux is positive because the field is leaving the cylinder through this face.)

Right face (at $x = +10$ cm): Here the field is $\mathbf{E} = 200\,\hat{\mathbf{i}}$ N/C (pointing in the $+x$ direction). The outward normal to the right face also points in the $+x$ direction. Again, $\mathbf{E}$ and the outward normal are parallel:

$\phi_R = E \times \Delta S = 200 \times \pi(0.05)^2 = +1.57 \text{ N m}^2 \text{ C}^{-1}$

Both flat faces give the same outward flux of $+1.57$ N m$^2$ C$^{-1}$.

(b) Flux through the curved side

At every point on the curved surface, $\mathbf{E}$ (pointing along $\pm x$) is perpendicular to the outward normal (which points radially away from the axis). Since $\mathbf{E} \cdot \Delta\mathbf{S} = 0$ everywhere on the curved surface:

$\phi_{\text{side}} = 0$

(c) Net outward flux through the cylinder

$\phi = \phi_L + \phi_R + \phi_{\text{side}} = 1.57 + 1.57 + 0 = 3.14 \text{ N m}^2 \text{ C}^{-1}$

(d) Net charge inside the cylinder

Using Gauss’s law, $q = \varepsilon_0\,\phi$:

$q = 8.854 \times 10^{-12} \times 3.14 = 2.78 \times 10^{-11} \text{ C}$

The cylinder encloses a net charge of $2.78 \times 10^{-11}$ C. The non-zero flux (and hence the non-zero enclosed charge) arises because the field points outward on both flat faces, so flux leaves the cylinder from both ends without any compensating inward flux.