1.13 Gauss’s Law

As a simple application of the notion of electric flux, let us consider the total flux through a sphere of radius \(r\), which encloses a point charge \(q\) at its centre. Divide the sphere into small area elements, as shown in Fig. 1.22.
The flux through an area element \(\Delta \mathbf{S}\) is
\(
\Delta \phi=\mathbf{E} \cdot \Delta \mathbf{S}=\frac{q}{4 \pi \varepsilon_0 r^2} \hat{\mathbf{r}} \cdot \Delta \mathbf{S} \dots(1.28)
\)
where we have used Coulomb’s law for the electric field due to a single charge \(q\). The unit vector \(\hat{\mathbf{r}}\) is along the radius vector from the centre to the area element. Now, since the normal to a sphere at every point is along the radius vector at that point, the area element \(\Delta \mathbf{S}\) and \(\hat{\mathbf{r}}\) have the same direction. Therefore,
\(
\Delta \phi=\frac{q}{4 \pi \varepsilon_0 r^2} \Delta S \dots(1.29)
\)
since the magnitude of a unit vector is 1.
The total flux through the sphere is obtained by adding up flux through all the different area elements:
\(
\phi=\sum_{\text {all } \Delta S} \frac{q}{4 \pi \varepsilon_0 r^2} \Delta S
\)
Since each area element of the sphere is at the same distance \(r\) from the charge,
\(
\phi=\frac{q}{4 \pi \varepsilon_o r^2} \sum_{\text {all } \Delta S} \Delta S=\frac{q}{4 \pi \varepsilon_0 r^2} S
\)
Now \(S\), the total area of the sphere, equals \(4 \pi r^2\). Thus,
\(
\phi=\frac{q}{4 \pi \varepsilon_0 r^2} \times 4 \pi r^2=\frac{q}{\varepsilon_0} \dots(1.30)
\)
Equation (1.30) is a simple illustration of a general result of electrostatics called Gauss’s law.
We state Gauss’s law without proof: Electric flux through a closed surface \(S\)
\(
\begin{aligned}
& =q / \varepsilon_0 \dots(1.31) \\
& q=\text { total charge enclosed by } \mathrm{S} .
\end{aligned}
\)
The law implies that the total electric flux through a closed surface is zero if no charge is enclosed by the surface. We can see that explicitly in the simple situation of Fig. 1.23.

Here the electric field is uniform and we are considering a closed cylindrical surface, with its axis parallel to the uniform field \(\mathbf{E}\). The total flux \(\phi\) through the surface is \(\phi=\phi_1+\phi_2+\phi_3\), where \(\phi_1\) and \(\phi_2\) represent the flux through the surfaces 1 and 2 (of circular cross-section) of the cylinder and \(\phi_3\) is the flux through the curved cylindrical part of the closed surface. Now the normal to the surface 3 at every point is perpendicular to \(\mathbf{E}\), so by definition of flux, \(\phi_3=0\). Further, the outward normal to 2 is along \(\mathbf{E}\) while the outward normal to 1 is opposite to \(\mathbf{E}\). Therefore,
\(
\begin{aligned}
& \phi_1=-E S_1, \quad \phi_2=+E S_2 \\
& S_1=S_2=S
\end{aligned}
\)
where \(S\) is the area of circular cross-section. Thus, the total flux is zero, as expected by Gauss’s law. Thus, whenever you find that the net electric flux through a closed surface is zero, we conclude that the total charge contained in the closed surface is zero.

The great significance of Gauss’s law Eq. (1.31), is that it is true in general, and not only for the simple cases we have considered above. Let us note some important points regarding this law:

• (i) Gauss’s law is true for any closed surface, no matter what its shape or size.
• (ii) The term \(q\) on the right side of Gauss’s law, Eq. (1.31), includes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface.
• (iii) In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field [whose flux appears on the left side of Eq. (1.31)] is due to all the charges, both inside and outside \(S\). The term \(q\) on the right side of Gauss’s law, however, represents only the total charge inside \(S\).
• (iv) The surface that we choose for the application of Gauss’s law is called the Gaussian surface. You may choose any Gaussian surface and apply Gauss’s law. However, take care not to let the Gaussian surface pass through any discrete charge. This is because electric field due to a system of discrete charges is not well defined at the location of any charge. (As you go close to the charge, the field grows without any bound.) However, the Gaussian surface can pass through a continuous charge distribution.
• (v) Gauss’s law is often useful towards a much easier calculation of the electrostatic field when the system has some symmetry. This is facilitated by the choice of a suitable Gaussian surface.
• (vi) Finally, Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. Any violation of Gauss’s law will indicate departure from the inverse square law.

Example 1.10: The electric field components in Fig. 1.24 are \(E_x=\alpha x^{1 / 2}, E_y=E_z=0\), in which \(\alpha=800 \mathrm{~N} / \mathrm{C} \mathrm{m}^{1 / 2}\). Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that \(a=0.1 \mathrm{~m}\).

(a) Since the electric field has only an \(x\) component, for faces perpendicular to \(x\) direction, the angle between \(\mathbf{E}\) and \(\Delta \mathbf{S}\) is \(\pm \pi / 2\). Therefore, the flux \(\phi=\mathbf{E} . \Delta \mathbf{S}\) is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is
\(
E_L=\alpha x^{1 / 2}=\alpha a^{1 / 2}
\)
( \(x=a\) at the left face).
The magnitude of electric field at the right face is
\(
E_R=\alpha x^{1 / 2}=\alpha(2 a)^{1 / 2}
\)
( \(x=2 a\) at the right face).
The corresponding fluxes are
\(
\begin{aligned}
\phi_L & =\mathbf{E}_L \cdot \Delta \mathbf{S}=\Delta \mathbf{S} \mathbf{E}_L \cdot \hat{\mathbf{n}}_L=E_L \Delta \mathbf{S} \cos \theta=-E_L \Delta \mathbf{S}, \text { since } \theta=180^{\circ} \\
& =-E_L a^2 \\
\phi_R & =\mathbf{E}_R \cdot \Delta \mathbf{S}=E_R \Delta \mathbf{S} \cos \theta=E_R \Delta \mathbf{S}, \quad \text { since } \theta=0^{\circ} \\
& =E_R a^2
\end{aligned}
\)
Net flux through the cube
\(
\begin{aligned}
& =\phi_R+\phi_L=E_R a^2-E_L a^2=a^2\left(E_R-E_L\right)=\alpha a^2\left[(2 a)^{1 / 2}-a^{1 / 2}\right] \\
& =\alpha a^{5 / 2}(\sqrt{2}-1) \\
& =800(0.1)^{5 / 2}(\sqrt{2}-1) \\
& =1.05 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}
\end{aligned}
\)

(b) We can use Gauss’s law to find the total charge \(q\) inside the cube. We have \(\phi=q / \varepsilon_0\) or \(q=\phi \varepsilon_0\). Therefore,
\(
q=1.05 \times 8.854 \times 10^{-12} \mathrm{C}=9.27 \times 10^{-12} \mathrm{C}
\)

Example 1.11: An electric field is uniform and in the positive \(x\) direction for positive \(x\), and uniform with the same magnitude but in the negative \(x\) direction for negative \(x\). It is given that \(\mathbf{E}=200 \hat{\mathbf{i}} \mathrm{~N} / \mathrm{C}\) for \(x>0\) and \(\mathbf{E}=-200 \hat{\mathbf{i}} \mathrm{~N} / \mathrm{C}\) for \(x<0\). A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the \(x\)-axis so that one face is at \(x=+10 \mathrm{~cm}\) and the other is at \(x=-10 \mathrm{~cm}\) (Fig. 1.25). (a) What is the net outward flux through each flat face? (b) What is the flux through the side of the cylinder? (c) What is the net outward flux through the cylinder? (d) What is the net charge inside the cylinder?

Solution: (a) We can see from the figure that on the left face \(\mathbf{E}\) and \(\Delta \mathbf{S}\) are parallel. Therefore, the outward flux is
\(
\begin{aligned}
& \phi_L=\mathbf{E} \cdot \Delta \mathbf{S}=-200 \hat{\mathbf{i}} \cdot \Delta \mathbf{S} \\
& =+200 \Delta \mathbf{S}, \text { since } \hat{\mathbf{i}} \cdot \Delta \mathbf{S}=-\Delta \mathrm{S} \\
& =+200 \times \pi(0.05)^2=+1.57 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}
\end{aligned}
\)
On the right face, \(\mathbf{E}\) and \(\Delta \mathbf{S}\) are parallel and therefore
\(
\phi_R=\mathbf{E} \cdot \Delta \mathbf{S}=+1.57 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}
\)
(b) For any point on the side of the cylinder \(\mathbf{E}\) is perpendicular to \(\Delta \mathbf{S}\) and hence \(\mathbf{E} \cdot \Delta \mathbf{S}=0\). Therefore, the flux out of the side of the cylinder is zero.
(c) Net outward flux through the cylinder
\(
\phi=1.57+1.57+0=3.14 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}
\)

(d) The net charge within the cylinder can be found by using Gauss’s law which gives
\(
\begin{aligned}
q & =\varepsilon_0 \phi \\
& =3.14 \times 8.854 \times 10^{-12} \mathrm{C} \\
& =2.78 \times 10^{-11} \mathrm{C}
\end{aligned}
\)