GAUSS’S LAW

Gauss’s law is an alternative to Coulomb’s law. While completely equivalent to Coulomb’s law, Gauss’s law provides a different way to express the relationship between electric charge and electric field. It was formulated by Carl Friedrich Gauss (1777–1855), one of the greatest mathematicians of all time Fig.1.

Fig.1: Carl Friedrich Gauss helped develop several branches of mathematics, including differential geometry, real analysis, and number theory. The “bell curve” of statistics is one of his inventions. Gauss also made state-of-the-art investigations of the earth’s magnetism and calculated the orbit of the first asteroid to be discovered.

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.

Fig.1: Flux through a sphere enclosing a point charge q at its centre.

The flux through an area element ∆S is

$\Delta \phi=\mathbf{E}.\Delta \mathbf{S}=\frac{q}{4\pi \epsilon_0r^2}\hat{\mathbf{r}}.\Delta \mathbf{S}$ (1)

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 ∆S and $\hat{\mathbf{r}}$ have the same direction. Therefore,

$\Delta \phi=\frac{q}{4\pi \epsilon_0r^2}\Delta S$ (2)

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:

$\phi=\Sigma_{all \ \Delta S}\frac{q}{4\pi \epsilon_0r^2}\Delta S$

Since each area element of the sphere is at the same distance r from the charge,

$\phi=\frac{q}{4\pi \epsilon_0r^2}\Sigma_{all \ \Delta S}\Delta S=\frac{q}{4\pi \epsilon_0r^2}S$

Now S, the total area of the sphere, equals $4\pi r^2$. Thus,

$\phi=\frac{q}{4\pi \epsilon_0r^2}\times 4\pi r^2$ (3)

Equation (3) 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, $\Phi_E=\frac{q}{\epsilon_0}$ (4)

q = total charge enclosed by S.

General Form of Gauss’s Law

The total electric flux through a closed surface is equal to the total (net) electric charge inside the surface, divided by $\epsilon_0$.

$\Phi_E=\oint \mathbf{E.S}=\frac{q_{encl}}{\epsilon_0}$ (Gauss’s law) (5)

Fig 2: Calculation of the flux of uniform electric field through the surface of a cylinder.

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. 2.

Here the electric field is uniform and we are considering a closed cylindrical surface, with its axis parallel to the uniform field E. The total flux φ 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 E, so by definition of flux, $\phi_3=0$. Further, the outward normal to 2 is along E while the outward normal to 1 is opposite to E. Therefore,

$\phi=-ES_1$, $\phi=+ES_2$

$S_1=S_2=S$

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. (4), 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. (4), 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. (4)] 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

The cube in Fig. 3 has edge length 1.40 m and is oriented as shown in a region of uniform electric field. Find the electric flux through the right face if the electric field, in newtons per coulomb, is given by (a) 6.00$\hat{\mathbf{i}}$ , (b) $-2.00\hat{\mathbf{j}}$ , and (c) $-3\hat{\mathbf{i}}+4.00\hat{\mathbf{k}}$ (d) What is the total flux through the cube for each field?

Fig.3

Solution:

We use $\Phi=\mathbf{E.A}$, where $A=A\hat{j}$.

(a) $\Phi=(6.00 / N/C)\mathbf{i}.(1.40 \ m)^2 \hat{j}=0$

(b) $\Phi=(-2.00 / N/C)\mathbf{i}.(1.40 \ m)^2 \hat{j}=-3.92 \ N.m^2/C$

(d) The total flux of a uniform field through a closed surface is always zero.

Example 2

At each point on the surface of the cube shown in Fig. 3, the electric field is parallel to the z axis. The length of each edge of the cube is 3.0 m. On the top face of the cube the field is $E=-34 \hat{\mathbf{k}}$, and on the bottom face it is $E=+20 \hat{\mathbf{k}}$. Determine the net charge contained within the cube.

Solution:

There is no flux through the sides, so we have two “inward” contributions to the flux, one from the top (of magnitude (34)(3.0)²) and one from the bottom (of magnitude (20)(3.0)²).

With “inward” flux being negative, the result is $\Phi=-486 \ N.m^2/C$. Gauss’s law then leads to

$q_{enc}=\epsilon_0 \Phi=(8.85 \times 10^{-12} \ C^/N.m^2)(-486 \ N.m^2/C)=-4.3 \times 10^{-9} \ C$

Example 3

The three small spheres shown in Fig.3 carry charges $q_1$ = 4.00 nC, $q_2$ = -7.80 nC and $q_3$ = 2.40 nC. Find the net electric flux through each of the following closed surfaces shown in cross section in the figure: (a) $S_1$; (b) $S_2$; (c) $S_3$; (d) $S_4$; (e) $S_5$. (f) Do your answers to parts (a)–(e) depend on how the charge is distributed over each small sphere? Why or why not?

Fig.3

Solution:

To solve the problem, we will apply Gauss's Law, which states:

$\Phi_E=\oint \mathbf{E.S}=\frac{q_{encl}}{\epsilon_0}$

where:

$\Phi_E$ is the net electric flux through a surface.
$q_{enclosed}$ is the total charge enclosed within the surface.
$\epsilon_0=8.85 \times 10^{-12} \ C^2/N.m^2$ is the permittivity of free space.

(a) Flux through $S_1$:

Surface $S_1$ encloses only $q_1=4.00 \ nC=4.00 \times 10^{-9} \ C$.

$\Phi_E=\frac{q_1}{\epsilon_0}=\frac{(4.00 \times 10^{-9} \ C)}{8.85 \times 10^{-12} \ C^2/N.m^2}$

$\Phi_E= 4.52 \times 10^{2} \ N.m^2/C$

(b) Flux through $S_2$:

Surface $S_2$ encloses only $q_1=-7.80 \ nC=-7.80 \times 10^{-9} \ C$.

$\Phi_E=\frac{q_2}{\epsilon_0}=\frac{(-7.80 \times 10^{-9} \ C)}{8.85 \times 10^{-12} \ C^2/N.m^2}$

$\Phi_E= -8.81\times 10^{2} \ N.m^2/C$

(c) Flux through $S_3$:

Surface $S_3$ encloses only $q_1+q_2=4.00 \ nC-7.80 \ nC=-3.80 \times 10^{-9} \ C$.

$\Phi_E=\frac{q_1+q_2}{\epsilon_0}=\frac{(-3.80 \times 10^{-9} \ C)}{8.85 \times 10^{-12} \ C^2/N.m^2}$

$\Phi_E= -4.29 \times 10^{2} \ N.m^2/C$

(d) Flux through $S_4$:

Surface $S_4$ encloses only $q_1+q_3=4.00 \ nC+2.40 \ nC=6.40 \times 10^{-9} \ C$.

$\Phi_E=\frac{q_1+q_3}{\epsilon_0}=\frac{(6.40 \times 10^{-9} \ C)}{8.85 \times 10^{-12} \ C^2/N.m^2}$

$\Phi_E= 7.23 \times 10^{2} \ N.m^2/C$

(e) Flux through $S_5$:

Surface $S_5$ encloses only $q_1+q_2+q_3=4.00 \ nC-7.80 \ nC+2.40 \ nC=-1.40 \times 10^{-9} \ C$.

$\Phi_E=\frac{q_1+q_2+q_3}{\epsilon_0}=\frac{(-1.40 \times 10^{-9} \ C)}{8.85 \times 10^{-12} \ C^2/N.m^2}$

$\Phi_E= -1.58 \times 10^{2} \ N.m^2/C$

(f) Dependence on charge distribution:

The net flux through a closed surface depends only on the total charge enclosed within the surface, not on how the charge is distributed. This is because Gauss's Law accounts for the total enclosed charge, irrespective of its configuration or placement inside the surface.

Example 4

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 E = 200 $\hat{\mathbf{i}}$ N/C for x > 0 and E = –200 $\hat{\mathbf{i}}$ N/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 cm and the other is at x = –10 cm (Fig. 4). (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 E and ∆S are parallel. Therefore, the outward flux is

$\phi_L=\mathbf{E}.\Delta \mathbf{S}=-200 \hat{\mathbf{i}}\Delta \mathbf{S}$

= +200 ∆S, since $\hat{\mathbf{i}}\Delta \mathbf{S} = – \Delta S$

= +200 × π$(0.05)^2$

= +1.57 $N.m^2.C^{–1}$

On the right face, E and ∆S are parallel and therefore

$\phi_R=\mathbf{E}.\Delta \mathbf{S}= + 1.57 N.m^2.C^{–1}$

(b) For any point on the side of the cylinder E is perpendicular to ∆S and hence E.∆S = 0. Therefore, the flux out of the side of the cylinder is zero.

$\phi$ = 1.57 + 1.57 + 0 = 3.14 $N.m^2.C^{–1}$