Gauss Law And It's Application

Gauss's Law is a fundamental principle in electromagnetism that elegantly connects electric fields to the distribution of electric charges. Formulated by Carl Friedrich Gauss, this law states that the electric flux through a closed surface is directly proportional to the total charge enclosed within that surface. Gauss's Law simplifies complex problems in electrostatics, making it easier to calculate electric fields, especially for symmetrical charge distributions.

In real life, Gauss's Law is crucial in designing devices like capacitors, where it helps in understanding the behaviour of electric fields and charge distribution. For instance, in a coaxial cable, which is widely used in telecommunications, Gauss's Law aids in determining the electric field around the inner conductor, ensuring efficient signal transmission. This law also plays a significant role in shielding sensitive electronic equipment from external electric fields, ensuring their proper functioning in diverse environments.

What is Gauss's Law?

Gauss's law is one of the fundamental laws of physics which states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge. It is one of the four equations of Maxwell’s laws of electromagnetism.

The diagram below shows a locally uniform electric field E. The lines are parallel and have constant density. The same surface is inserted in three different orientations. The maximum number of field lines is intercepted when the unit vector normal to the surface, n, is parallel to the field E, while no field lines pass through the surface when n is perpendicular to the field. In general, the number of field lines passing through an area A is directly proportional to A*cosθ, where θ is the angle between the field direction and the unit vector n normal to the surface. This leads to the definition of the electric flux.

The surface on which Gauss's law is applied is called Gaussian Surface (as shown in the below figure).

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Note- Remember that the closed surface in Gauss's law is imaginary. There need not be any material object at the position of the surface.

Gauss's law for a closed surface states that $\phi_{\text {closed }}=\frac{q_{\text {net }}}{\epsilon_0} q_{\text {net }}$ is the total charge inside the closed surface. The closed surface on which we apply Gauss law is called the Gaussian surface.

Also, the flux can be written in the integral form as:

$\phi=\int \bar{E} d \bar{S}$ S is the area enclosed and E is the electric field intensity passing through it.

The Usefulness of Gauss's Law :

1. Gauss's law is useful when the Gaussian surface has symmetry about the charge.

2. We can take any Gaussian surface but the Gaussian surface should not pass through the charge. iI can pass through the charge but it can pass through continuous charge distribution.

The limitation of Gauss law

Gauss law is applicable to certain symmetrical shapes it cannot be used for disk and ring and an electric dipole etc.

Applications of Gauss Law

Electric Field Due to Infinite Linear Charge

The electric field of an infinite line charge with a uniform linear charge density can be obtained by using Gauss's law. Considering a Gaussian surface in the form of a cylinder at radius r, the electric field has the same magnitude at every point of the cylinder and is directed outward. The electric flux is then just the electric field times the area of the cylinder.

Earlier we derived the electric field due to the infinite line charge as :

$E=\frac{\lambda}{2 \pi \epsilon_0 r}$ where $\lambda$ is linear charge density. Here the Gaussian surface will be the cylinder around the linear charge of length l.

Gauss's law :

$\phi=\oint E \cdot d A=\frac{Q_{\text {inside }}}{\epsilon_0}$

Here, the cylinder has 3 surfaces: 1. Upper 2. lower 3. Curved

$\begin{aligned} \phi_{\text {cylinder }} & =\phi_1+\phi_2+\phi_3 \\ & =0+0+\int E d A \cos 0 \\ \Longrightarrow \phi & =\int E d A=\frac{Q_{\text {inside }}}{\epsilon_0} \\ \Longrightarrow \phi & =E \int d A=\frac{Q_{\text {inside }}}{\epsilon_0} \\ \Longrightarrow \phi & =E(2 \pi r l)=\frac{Q_{\text {inside }}}{\epsilon_0} \\ \Longrightarrow E & =(2 \pi r l)=\frac{\lambda l}{2 \pi a \times l \times \epsilon_0} \text { since, } \lambda=\frac{Q_{\text {inside }}}{l} \\ \Longrightarrow E & =\frac{\lambda}{2 \pi \epsilon r}\end{aligned}$

Therefore E is inversely proportional to r.

The Electric Field in a Conductor

We have studied that conductors have a large number of free electrons that are free to move inside the conductors but metal ions are fixed. Now if we place this conductor in an electric field, due to the electric field, electrons will experience a force.

The total electric field at any point in the conductor is the vector sum of the original electric field and the electric field due to the redistributed charged particles. Since they are oppositely directed, the two contributions to the electric field inside the conductor tend to cancel each other. Now comes the profound part of the argument: the two contributions to the electric field at any point in the conductor exactly cancel. We know they have to completely cancel because, if they didn’t, the free-to-move-charge in the conductor would move as a result of the force exerted on it by the electric field. The force on the charge is always in a direction that causes the charge to be redistributed to positions in which it will create its own electric field that tends to cancel the electric field that caused the charge to move. The point is that the charge will not stop responding to the electric field until the net electric field at every point in the conductor is zero.

The electric field is zero at all points inside the conductor, and, while the total charge is still zero, the charge has been redistributed as in the following diagram:

These charges which are appearing on the surface are called induced charges. Due to these induced charges, the electric field will be produced and that is the induced electric field.

The direction of this electric field will be from positive to negative.
$
E_{\text {net }}=E_{\text {in }}+E_{\text {ext }}
$

and $E_{\text {net }}=0$ inside the conductor.

now if we make a Gaussian surface inside the conductor

We know that E_inside =0

Therefore, $\phi=\oint E \cdot d s=\frac{q_{\text {inside }}}{\epsilon_0}=0$

Hence, the charge inside the conductor is zero.

Electric Field Due to Cylinders

1. Solid Conducting/ Hollow Conducting

The electric field of an infinite cylindrical conductor with a uniform linear charge density can be obtained by using Gauss' law. Considering a Gaussian surface in the form of a cylinder at radius r > R, the electric field has the same magnitude at every point of the cylinder and is directed outward. The electric flux is then just the electric field times the area of the cylinder.

when the charge is on the surface we need to take account of the surface charge density.

$\sigma=\frac{\text { charge }}{\text { area }}$

For a uniformly charged cylinder of radius, R. We have to consider these three areas where

a) Inside the cylinder (r<R)

b) At the surface (r=R)

c) outside the cylinder (r>R).

(a). Inside (r<R)

E =0

(b). Outside (r>R)
$
E=\frac{\sigma R}{\epsilon_0 r} \text { or } E=\frac{\lambda}{2 \pi \varepsilon_0 r}
$

(c). At the surface $(\mathrm{r}=\mathrm{R})$

$
E=\frac{\sigma}{\epsilon_0} \text { or } E=\frac{\lambda}{2 \pi \varepsilon_0 R}
$

2. Solid Non-Conducting

In the case of a solid non-conducting cylinder, the charge is not only on the surface but also distributed through the whole volume. Therefore, volume charge density, $\rho=\frac{\text { charge }}{\text { volume }}$

(a). Inside (r<R)
$
E=\frac{\rho r}{2 \epsilon_0} \text { or } E=\frac{\lambda r}{2 \pi \varepsilon_0 R^2}
$

(b). Outside ( $r>R$ )

$
E=\frac{\rho R^2}{2 \epsilon_0 r}
$

(c). At the surface ( $\mathrm{r}=\mathrm{R}$ )

$
E=\frac{\rho R}{2 \epsilon_0}(\max .)
$

Field of an Infinite Plane Sheet

It is one of the very important cases and we are going to find the electric field of an infinite plane sheet with the help of Gauss's law -

Let us consider a thin, flat, infinite sheet which consists of uniform positive charge per unit area $\sigma$. We can see that there is symmetry in this lamina. So, to take advantage of these symmetry properties, we use a cylinder as our Gaussian surface whose axis is perpendicular to the sheet of charge, with ends of area A.

We can also observe that the charged sheet passes through the middle of the cylinder's length and because of this flux through each end is EA. This is because $\vec{E}$ is perpendicular to the charged sheet and parallel to the area vector of the flat face. The $\vec{E}$ is along the curved surface i.e. perpendicular to the area vector so, the flux will be zero through this. Then the total flux will be 2EA. Now the net charge within the Gaussian surface can be calculated as

$Q_{\text {enclosed }}=\sigma A$

So we can write that by Gauss's law

$\begin{aligned} 2 E A & =\frac{\sigma A}{\varepsilon_o} \\ E & =\frac{\sigma}{2 \varepsilon_o}\end{aligned}$

If the charge is negative, $\vec{E}$ will be toward the sheet.

Since in this nature nothing is infinitely large, this assumption is valid if we place a unit point charge in front of a large sheet.

Electric Field Due to Uniform Charged Sphere

The sphere may be hollow or solid and both hollow and solid spheres may be conducting or non-conducting. So let us know the electric field due to all these cases

In the case of a conducting sphere, the whole charge will come on the surface of the sphere but when the sphere is non-conducting then the whole charge is distributed all over the sphere.

Electric field Due to Hollow Conducting/ Non-Conducting and Solid Conducting Sphere

1. For a point outside the sphere

We first consider the field outside the conductor, so we choose $r>R$. The entire conductor is within the Gaussian surface, so the enclosed charge is $q$. The area of the Gaussian surface is $4 \pi r^2$; $\vec{E}$ is uniform over the surface and perpendicular to it at each point. The flux integral $\oint E_{\perp} d A$ in Gauss's law is therefore $E\left(4 \pi r^2\right)$ which gives

$
E\left(4 \pi r^2\right)=\frac{q}{\varepsilon_0}
$

or, $\quad E=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} \quad$ (outside a charged conducting sphere)

For all points outside the shell, the field due to the uniformly charged shell is such that the entire charge is concentrated at the centre of the shell.

2. For a point inside the sphere (r<R)

Since the charge will be zero in this case for the hollow sphere and conducting solid sphere. So, E = 0 inside the sphere.

3. At surface (r = R)

$E=\frac{1}{4 \pi \varepsilon_0} \frac{q}{R^2}$

Electric field Due to Solid Non-Conducting

1. Outside ( $r>R$ )

$
E=\frac{\rho R^3}{3 \varepsilon_o r^2}
$

2. Inisde ( $r<R$ )

$
E=\frac{\rho r}{3 \varepsilon_o}
$

Solved Examples Based on Gauss law

Example 1: A charged particle q is placed at the centre O of the cube of side length L (ABCDEFGH) Another same charge q is placed at a distance L from O . Then the electric flux through ABCD is

1) $\frac{q}{4 \epsilon_0}$
2) Zero
3) $\frac{q}{2 \epsilon_0}$
4) $\frac{q}{6 \varepsilon_0}$

Solution:

According to Gauss's Law

The total flux is linked with a closed surface called the Gaussian surface.

Formula:

$\phi=\oint \vec{E} \cdot d \vec{s}=\frac{Q_{e n c}}{\epsilon_0}$

For the charge placed outside the flux through the surface, ABCD will be zero. But for the charge inside the cube, it is $\frac{q}{\varepsilon_0}$ through all the surfaces. For one surface, it is $\frac{q}{6\varepsilon_0}$ .

Net flux = $0+\frac{q}{6 \epsilon_0}=\frac{q}{6 \epsilon_0}$

Hence, the answer is the option (4).

Example 2: If the electric flux entering and leaving an enclosed surface respectively is $\phi_1$ and $\phi_2$, the electric charge inside the surface will be:

1) $\left(\phi_2-\phi_1\right) \varepsilon_0$
2) $\left(\phi_1+\phi_2\right) / \varepsilon_0$
3) $2\left(\phi_2-\phi_1\right) / \varepsilon_0$
4) $2\left(\phi_1+\phi_2\right) / \varepsilon_0$

Solution:

According to Gauss theorem,

$\left(\phi_2-\phi_1\right)=\frac{Q}{\varepsilon_0} \Rightarrow Q=\left(\phi_2-\phi_1\right) \varepsilon_0$

The flux enters the enclosure if one has a negative charge $\left(-q_2\right)$ and flux goes out if one has a +ve charge $\left(+q_1\right)$.

As one does not know whether,

$\phi_1>\phi_2$ or $\phi_2>\phi_1, Q=\left(\phi_2-\phi_1\right) \varepsilon_0$

Hence, the answer is the option (1).

Example 3: Let $P(r)=\frac{Q}{\pi R^4} r$ be the charge density distribution for a solid sphere of radius R and total charge Q . For a point p inside the sphere at a distance r₁ from the centre of the sphere, the magnitude of the electric field is

1) 0
2) $\frac{Q}{4 \pi \varepsilon_0 r_1^2}$
3) $\frac{Q r_1^2}{4 \pi \varepsilon_0 R^4}$
4) $\frac{Q r_1^2}{3 \pi \varepsilon_0 R^4}$

Solution:

If the charge density
$
\rho=\frac{Q}{\pi R^4} r
$
From Gauss Theorem

$
\begin{aligned}
& E\left(4 \pi r_1^2\right)=\frac{q}{\varepsilon_o} \\
& E\left(4 \pi r_1^2\right)=\frac{1}{\varepsilon_o} \int \rho d V \\
& E 4 \pi r_1^2=\frac{1}{\varepsilon_o} \int_0^{r_1} \frac{Q r}{\pi R^4} 4 \pi r^2 d r \\
& \therefore E=\frac{Q r_1^2}{4 \pi \varepsilon_o R^4}
\end{aligned}
$

Hence, the answer is the option (3).

Example 4: Let there be a spherically symmetric charge distribution with charge density varying as $\rho(r)=\rho_0\left(\frac{5}{4}-\frac{r}{R}\right)$ upto $r=R$, and $\rho(r)=0$ for $r>R$, where r is the distance from the origin. The electric field at a distance $r(r<R)$ from the origin is given by

$\begin{aligned} & \text { 1) } \frac{\rho_0 r}{3 \varepsilon_0}\left(\frac{5}{4}-\frac{r}{R}\right) \\ & \text { 2) } \frac{4 \pi \rho_0 r}{3 \varepsilon_0}\left(\frac{5}{3}-\frac{r}{R}\right) \\ & \text { 3) } \frac{\rho_0 r}{4 \varepsilon_0}\left(\frac{5}{3}-\frac{r}{R}\right) \\ & \text { 4) } \frac{4 \rho_0 r}{3 \varepsilon_0}\left(\frac{5}{4}-\frac{r}{R}\right)\end{aligned}$

Solution:

Gauss's Law

The total flux is linked with a closed surface called the Gaussian surface.

$\begin{aligned} & \text { Volume of shell } d V=4 \pi x^2 d x \\ & Q_{i n}=\int_0^r \rho d v=\int_0^r \rho_o\left(\frac{5}{4}-\frac{x}{R}\right) 4 \pi x^2 d x \\ & =4 \pi \rho_o \int_0^r\left(\frac{5}{4} x^2-\frac{x^3}{R}\right) d x \\ & =4 \pi \rho_o\left[\frac{5}{12} x^3-\frac{x^4}{4 R}\right]_o^r \\ & =4 \pi \rho_o\left[\frac{5}{12} r^3-\frac{r^4}{4 R}\right] \\ & =\frac{4 \pi \rho_o}{4}\left[\frac{5}{3} r^3-\frac{r^4}{R}\right] \\ & =\pi \rho_o\left[\frac{5}{3} r^3-\frac{r^4}{R}\right] \\ & E .4 \pi r^2=\frac{q_{i n}}{\varepsilon_o} \\ & E .4 \pi r^2=\frac{\pi \rho_o}{\varepsilon_o}\left[\frac{5}{3} r^3-\frac{r^4}{R}\right] \\ & E=\frac{\pi \rho_o r^3}{4 \pi r^2 \varepsilon_o}\left[\frac{5}{3}-\frac{r}{R}\right]\end{aligned}$

Hence, the answer is the option (3).

Example 5: Charge Q is placed at a distance a/2 above the centre of the square surface of the edge as shown in the figure

The electric flux through the square surface is :

1) $\frac{Q}{\epsilon_0}$
2) $\frac{Q}{2 \epsilon_0}$
3) $\frac{Q}{3 \epsilon_0}$
4) $\frac{Q}{6 \epsilon_0}$

Solution:

Gauss's Law

The total flux is linked with a closed surface called the Gaussian surface.
$
\phi=\oint \vec{E} \cdot d \vec{s}=\frac{Q_{e n c}}{\epsilon_0}
$
So
We can construct a cube with side a \& charge $Q$ at its centre then total flux $=\frac{Q}{\epsilon_0}$
flux through one side $=\frac{Q}{6 \epsilon_0}$

Hence, the answer is the option (4).

Summary

Gauss's Law, a key concept in electromagnetism, relates the electric flux through a closed surface to the enclosed charge, making it invaluable for solving problems involving electric fields, particularly with symmetrical charge distributions. This law has practical applications in designing capacitors, calculating electric fields in conductors and cylinders, and understanding the behaviour of charges in various geometries. Its utility is seen in real-life scenarios like telecommunications and electronic shielding, although it has limitations when applied to non-symmetrical shapes such as disks or rings.