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.