Gauss's law and applications

Gauss’s law and applications

            Suppose a aphere enclosing a charge q of radius R

Electric field at the surface of the sphere at all points is:

                        E=q/R²

            Because of the spherical symmetry of the electric field of the charge.

Flux through sphere is:

                        Φ = ∫E.da = (q/R²)(4пR²)

                                      

                                         = 4пq

Hence

                        Φ = 4пq

Therefore we have a general law known as the gauss’s law which states that:

“The amount of electric flux through a closed surface is equal to 4п times the total algebraic sum of the charges enclosed by that surface in vacuum.”

Gauss’s law is a very effective way to calculate symmetrical fields. However for unsymmetrical fields the law becomes exceptionally difficult to solve although it still holds for them.

Applications:

            Field of a spherical charge distribution          

                       

                        Assume a sphere of radius R and volume charge density ρ

                        Take a Gaussian surface in the form of a sphere of radius r as shown.

Two cases will arise here:

                        A) r<R and

                        B) r>R

                       

                        Case A:

                        Flux through the surface is

                                    Φ = ∫E.da

                                    Φ = E(4пr²)      …..1

                        According to gauss’s law

                                    Φ = 4пq

                       

                                    q = (ρ) x (Volume)

                                       = ρ x (4/3)пr³

                                   

                                    Φ = 4пρ((4/3)пr³)   …..2

                                   

                        Equating 1 and 2

                       

                                    E(4пr²) = 4пρ((4/3)пr³)

                                   

                                    E = (4/3)пρr

                                    ρ = q/((4/3)пR³)

                                    E = (qr/R³)ř

                        Case B:

                                    Φ = ∫E.da = E(4пr²)    ……..1

                        According to gauss’s law

                                    Φ = 4пq = 4пρ((4/3)пR³)    ……….2

                                    Equating 1 and 2 we get

                                   

                                                E = (q/r²)ř

RESULT:  This is as if the entire sphere behaves like a point charge of magnitude q placed at its center.

Similarly we can say that the field of a hollow sphere with a surface charge q the electric field is zero inside it, which is in accordance with the result we have already proven.       

Introduction to E.M.

Quantization of charge

Electric charges always come in integral multiples of charge on an electron/proton. However quarks(building blocks of electrons) have charges n(e/3) where n=1, 2. because they have not been isolated they are counted as the quantized charge.

Coloumbs law

            Coloumbs law into its simplest form can be written as:

                       

                        F=(Qq)ř/r²

            The force is always along the line of separation between the charges

Energy of a system of charges

The energy of a system of charges is defined as the amount of work done in bringing the charges into the required configuration without acceleration.

We know that work done

            W=F.ds

Let two charges Q and q be brought from infinity to a separation r without acceleration.

Work done is:

            W=∫(Qq)/r²(-dr)

The negative sign indicates that there is a decrement in the distance

Solving we get

            U=W=Qq/r

Generalizing

            U=(1/2)∑∑(Qq)/r

Flux

Flux of an electric field is the amount of electric field lines passing through a

given surface area.

            Φ=∫E.da