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GAUSS’S LAW for electricity and magnetism derivation examples and applications

What is Gauss’s law?

                                       Gauss’s law states that: “The total electric flux through any closed surface  is equal to 1/ε0 times the total charge enclosed by the surface.”

It is given by Karl Friedrich  Gauss ,named after him gave a relationship between electric flux through a closed surface and the net charge enclosed by the surface.It is applied to calculate the electric intensity due to different charge configuration.In all such cases,an imaginary closed surface is considered which the electric intensity is to be evaluated.This closed surface is known as Gaussian surface.Its choice is such that the flux through it can be easily evaluated.It is given by the formula

Φ=q/∈0

Where ∈0 is the relative permittivity of free space vacuum.

Derivation of Gauss’s law:

gauses law 2

Suppose point charges q1,q2,q3,…….,qare arbitrarily distributed in an arbitrary shaped closed surface shown in figure.Using idea given that the electric flux passing through the closed surface is:

GAUSS'S LAW DERIVATION

This is the mathematical expression of gauss’s law which can be stated as:”The flux through any closed surface is 1/∈0 times the total charge enclosed in it.”

Where Q =q1+q2+q3+…….+qn,is the total charge enclosed by closed surface.

Integral form of gauss’s law:

Since the volume charge density is defined as:

integral form of gauss's law

Equation (4) is the integral form of gauss’s law.

Differential form of gauss’s law:

If the charge is distributed into a volume having uniform volume charge density ‘ρ’.then according to differential form of gauss’s law:

i form of gauss's law

We know by Divergence theorem:

DIFFERENTIAL FORM OF GAUSS'S LAW

This is the differential form of Gauss’s law.

Applications of Gauss’s law:

Gauss’s law is applied to calculate the electric intensity due to different charge configurations.In all such cases,an imaginary closed surface is considered which passes through the point at which the electric intensity is to be evaluated.This closed surface is known as Gaussian surface.Its choice is such that the flux through it can be easily evaluated.Next the charge enclosed by Gaussian surface is calculated and finally the electric intensity is computed by applying Gauss’s law.

(a):Intensity of field inside a hollow charged sphere:

gausian surface

Suppose that a hollow conducting sphere of radius R is given a positive charge +Q.We wish to calculate the field intensity first at a point inside the sphere.

Now imagine a sphere of radius R′ < R to be inscribed within the hollow charged sphere as shown in above figure.The surface of this sphere is the Gaussian surface.Let Φ be the flux through this closed surface.It can be seen in the figure that the charge enclosed by the Gaussian surface is zero.Applying Gaussian law we have:

gaussian law

Since  Φe=E.A=0

as  A≠0

Therefore:

E=0

Thus the interior of a hollow charged metal sphere is a field free region.As a consequence,any apparatus placed within a metal enclosure “from electric fields.

(b):Electric intensity due to an infinite sheet of charge:

infinite sheet of charges

Suppose we have a plane sheet of infinite extent on which positive charges are uniformly distributed.The uniform surface charge density is ,say,σ.A finite part of this sheet is shown in above figure.To calculate the electric intensity E at a point P,close to the sheet,imagine a closed Gaussian surface in the form of a cylinder passing through the sheet,whose one flat face contains point P.From symmetry we can conclude that points at right angle to the end faces and away from the plane.Since parallel to the curved surface of the cylinder,so there is no contribution to flux from the curved wall of the cylinder.While it will be,EA +EA=2 EA,Through the two flat end faces of the closed cylindrical surface,where A is the surface area of the flat faces.As the charge enclosed by the closed surface is σA,therefore,according to Gauss’s law:

Φ=1/∈0  ×charged enclosed by closed surface

Φ=1/∈×σA

Therefore:

2EA =1/∈×σA

OR

E=σ/2∈0

In vector form:

E=σ/2∈

Where rˆ is a unit vector normal to the sheet directed away from it.

you can clear more concepts from given video:

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