This is Part one of my eight part series on Maxwell Equations . We would be discussing – Integral form of Gauss’s Law For Electric Field

Although student of any level can read and understand them as I try to keep things to basic level but these are meant for Undergraduate level in general.

So we I assume that you are a grade 12 student or already in your college and you have basic introduction of electricity and magnetism as taught in class 12. So you probably knew about Maxwell’s equations. While studying Maxwell’s Equations you encounter two kinds of electric field:

1. the electrostatic field produced by electric charge and

2. the induced electric field produced by a changing magnetic field.

In this part one I’ll cover ” Gauss’s law for electric fields ” which is among one of the Maxwell’s equations.In this article I’ll try to cover Integral form of Gauss’s Law.

Gauss’s law for electric fields as we all know deals with the electrostatic field, this law to be a powerful tool because it relates the spatial behavior of the electrostatic field to the charge distribution that produces it.

**Integral Form of Gauss’s Law**

In integral form we write Gauss’s Law as

$ \oint\limits_s {\overrightarrow E \cdot \widehat nda = \frac{{{q_{enc}}}}{{{\varepsilon _0}}}} $

So main idea of Gauss’s law is

Electric charge produces an electric field, and the flux of that field passing through any closed surface is proportional to the total charge contained within that surface.

This is nothing but the statement of Gauss’s Law. So throughout this post you will hang on to this main idea while I explain the meaning of each and every symbol used in the integral form of Gauss’s Law.

- Integral sign : Tells you to sum up the contributions from each portion of the surface
- circle in integral sign : tells you that integral is taken over the closed surface
- S in integral subscript : It mentions that integral is to be taken over a surface so the integral is not a volume or line integral.
- $E$ : It is the electric field measured in N/C. It represents the total electric field at each point on the surface under consideration. The surface may be real or imaginary. For more follow this link
- $\rightarrow $ : above $E$ shows that electric field is a vector quantity.
- $\cdot$ : Dot product tells you to find the part of electric field that is perpendicular to the surface. It stands for vector scalar product. In Gauss’s law, the circle between $\vec{E}$ and $\hat{n}$ represents the dot product (or ‘‘scalar product’’) between the electric field vector $\vec{E}$ and the unit normal vector $\hat{n}$ and the dot product between two vectors gives a scalar result.
- $\hat{n}$ : It is the unit vector normal to the surface. The concept of the unit normal vector that at any point on a surface is simple, for this imagine a vector with length of one pointing in the direction per pendicular to the surface. It is called a ‘‘unit’’ vector because its length is unity and ‘‘normal’’ because it is perpendicular to the surface.
- $\vec{E}\cdot\hat{n}$ : this expression represents the component of the electric field vector that is perpendicular to the surface under consideration. It is the projection of first vector on second vector multiplied by the length of the second vector. that is

$\vec{E}\cdot \hat{n}=|\vec{E}||\hat{n}|cos\theta =|\vec{E}|cos\theta $

where $\theta$ is the angle between unit normal $\hat{n}$ and $\vec{E}$

for $\theta=90^{0}$ , $|\vec{E}|cos\theta=0$ that is component of electric field perpendicular to the surface is zero.

For $\theta=0^{0}$ , $|\vec{E}|cos\theta=|\vec{E}|$ that is component of electric field perpendicular to the surface is the entire length of vector $\vec{E}$ . - $da$ : is the surface area in $m^{2}$
- $q$ : is amount of charge in Coulomb.
- subscript $enc$ as in $q_{enc}$ : Reminder that only the charge enclosed in the surface contributes to the integral.
- $\varepsilon_{0}$ : is the electric permitivity of free space

Now there are two basic types of problems that you can solve using this equation

- First case is where you have information about distribution of electric charge and you want to find out the electric flux through the surface that encloses the charge
- In second case you have information about the electric field through a closed surface and you want to find out the total charge enclosed by that surface.

Also I am sure you know that the Gauss’s Law is highly symmetric and a powerful tool and you can use the law to find electric field itself rather than just finding the electric flux over a surface.

In my next post I’ll write about “Differential form of the Gauss’s Law”

Maxwell’s Equations – Differential form of Gauss’s Law For Electric Field(Part 2)

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