Application of Gauss's Law: Definition, Application & Notes

As per the textbook definition," The Gauss law states that net electric flux through a surface will be 1/∈ times the total enclosed charge."

In other words, you can say that the electric flux is directly proportional to the enclosed charge. The more charge enclosed within the surface, the more electric flux passes through the surface.

Mathematically, Gauss's Law is expressed in terms of charge enclosed:

${\mathrm{\Phi }}_E=\text{∮}\mathrm{}\mathrm{}\vec{E}\cdot d\vec{A}=\frac{q_{\mathrm{e}\mathrm{n}\mathrm{c}}}{{ϵ}_0}$

where:

${\mathrm{\Phi }}_E$ : electric flux 

$\vec{E}$ : electric field 

$d\vec{A}:$ area element vector

$q_{\text{enc}}$ : enclosed charge

${ϵ}_0\approx 8.854\times {10}^{-12}{\mathrm{C}}^2{\mathrm{N}}^{-1}{\mathrm{m}}^2$

Gauss's Law is a key concept in electrostatics that helps in calculating the electric field. You can apply gauss law to find the electric field intensity of various charge distributions. It is more useful for highly symmetrical charge distributions. Check the applications of Gauss law below.

Electric Field due to an Infinitely Long Straight Uniformly Charged Wire

Let's assume there is a long, straight, uniformly charged wire. The wire has a Gaussian surface of cross-sectional radius $r$, and length $L$. All the electric flux passes through the Gaussian surface (an imaginary closed cylindrical surface) around the wire. If the charge density of wire is  $\lambda$

$$E = \frac{\lambda}{2 \pi \varepsilon_0 r}$$This formula defines that the net electric field around the wire is directly proportional to the linear charge density ( $\lambda$) and inversely proportional to the distance (r) between the wire and the point.

Electric Field due to an Infinite Plane Sheet of Charge

Let's assume there is an infinitely plane, uniformly charged sheet. The imaginary Gaussian surface around the charge sheet is a cylindrical box that has flat faces parallel to the sheet. All the electric flux passes through the parallel faces of the imaginary Gaussian cylindrical surface around the sheet. If the surface charge density of the charged sheet is $\sigma$ $,\; then\; the\; elctric\; field\; due\; to\; straight\; uniformely\; charged\; sheet\; as\; per\; the\; gauss\; law:$

$$E = \frac{\sigma}{2 \varepsilon_0}$$.According to the formula, the net electric field due to the infinite charged sheet is directly proportional to the surface charge density.

Electric Field due to a Uniformly Charged Thin Spherical Shell

If there is a spherical shell, it has three specific points to find the net electric field: one outside, other on the surface, and one inside the shell. Let's take the spherical shell of radius is R, and the Gaussian surface has a similar symmetrical surface as the spherical shell. Let's take all three cases separately;

 

• Outside the Shell (r > R)

The net distance between the center and the point where the electric field is to be found, is larger than the radius R, The electric field using gauss law.

$$E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}$$

• On the Surface (r = R):

When the point is on the surface.

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

• Inside the Shell (r < R):

When the point is inside the spherical shell.

$$E = 0$$

Practice the questions given below;

Question: A thin spherical shell of radius 0.2 m has charge $5\mu \mathrm{C}$ . Find the field at 0.3 m from the center and inside the shell.

$\begin{array}{rr}{q}_{\mathrm{e}\mathrm{n}\mathrm{c}}=5\times {10}^{-6}\mathrm{C},\mathrm{}{\mathrm{\Phi }}_E& =E\cdot 4\pi (0.3{)}^2=\frac{5\times {10}^{-6}}{8.854\times {10}^{-12}{\mathrm{C}}^2{\mathrm{N}}^{-1}{\mathrm{m}}^2}\\ E& \approx 4.98\times {10}^5\mathrm{N}{\mathrm{C}}^{-1}\end{array}$

• Inside E=0, according to the gauss law.

Question: An infinite sheet has surface charge density $\sigma =2\mu {\mathrm{C}\mathrm{m}}^{-2}$ . What is the net electric field due to this charged sheet.

Using gauss law, $E=\frac{\sigma }{2{ϵ}_0}\approx 1.13\times {10}^5\mathrm{N}{\mathrm{C}}^{-1}$ .

 

You can access complete study material for Class 12 CBSE exams based on NCERT Textbooks. Check below;

Complete CBSE Class 12 Study Material
NCERT Class 12 Maths Solutions	NCERT Physics Class 12 Solutions	Class 12 Chemistry NCERT Solutions
Class 12 Maths NCERT Notes	Physics Class 12 NCERT Notes	Class 12 Chemistry NCERT Notes
CBSE Sample papers for class 12 Maths	CBSE Class 12 Physics Sample Papers	CBSE Class 12 Sample Paper for Chemistry
CBSE Class 12 Maths Previous Year Question Papers	CBSE Class 12 Physics Previous Year Question Papers	CBSE Chemistry Class 12 Question Papers