Application of gauss law

 Application of gauss Law

1) Electric field intensity due to Uniformly charged  spherical shell or hollow sphere.

Consider a sphere of radius R  charged with a uniform charged density  σ in a dielectric medium ε. Consider a gaussian surface as a sphere with radius r surrounding the charged sphere.Consider a point P where electric field intensity is to be measured.

The total charge on a sphere is given by 

            q = σA

                                             q=σ 4πR² ---------------(3)

By gauss law, net electric flux is given by  ∅ = q/ε_{0  ------------from (1)}    

_{Now due to symmetry , the electric field at each and every point is uniform  and directed radially outward and the direction of area is vector is parallel to direction of electric field .ie.} θ= 0 .

We know that , Electric flux is given by,  

  ∅ = ∫ E .ds cos θ------------ from (2)

                            ∅ = ∫ E .ds cos0  ( θ=0)

           ∅ = ∫ E .ds  

                                ∅ =  E .4πr² -------------(4)

                                                                   

          From (1) and (4)

                         q/ε_{0 =} E .4πr²

                                      E =  q/ε₀4πr²

                                     E =σ 4πR² /ε₀4πr^{2   ------------------from(3)}

                                          E= σ R² /ε₀r^{2 -----------------(A)}

                                      Equation (A)  gives Electric field intensity at point P 

                  Case 1: 

 if point P lies on the sphere then  R= r  , then equation (A) changes to 

                  E= σ  /ε₀

                Case 2: 

if point P lies inside the sphere , then E= 0  ( electric field inside a sphere is always zero)

2) Electric Field Intensity Due to an Infinitely Long Straight Charged Conductor

Consider a Long charged cylinder of radius R  charged with a uniform charged density  l in a dielectric medium ε. Consider a gaussian surface as a co-axial cylinder with radius r  and length l,surrounding the charged cylinder. Consider a point P where electric field intensity is to be measured.

The total charge on a cylinder  is given by 

            q = ll   ---------------(1)

    

By gauss law, net electric flux is given by  ∅ = q/ε_{0  -----(2)}   

_{Now due to symmetry , the electric field at each and every point is uniform  and directed radially outward and the direction of area is vector is parallel to direction of electric field .ie.} θ= 0 .

          We know that , Electric flux is given by,          

                                          ∅ = ∫ E .ds cos θ

                                                  ∅ = ∫ E .ds cos0  ( θ=0)

                                                  ∅ = ∫ E .ds  

                                                  ∅ =  E .2πrl -------------(4)

                                                                   

                                                                From (2) and (4)

               q/ε_{0 =} E .2πrl 

                E =  q/ε₀2πrl 

                                                 E =^ll /ε₀^{2πrl    ------------------from(3)}

                                                  E=  l/ε₀2πr  ----------------

   Equation (A)  gives Electric field intensity at point P 

Electric field due to infinite Charged Sheet

Consider a Uniformly charged Infinite sheet having area 'A' charged with a uniform charged density σ in a dielectric medium ε. Consider a point P at a Distance 'r' where electric field intensity is to be found.Consider a gaussian surface in form of a cylinder having cross sectional Area 'A' and length 2r.By symmetry, the End cap is parallel to area vector and curved surface area is perpendicular to Electric Field due to sheet.Thus Electric Field is 0 at curved surface.

The total charge on a sheet is given by 

            q = σA ---------------(1)

By gauss law, net electric flux is given by 

        ∅ = q/ε_{0       -----(2)}

We know that , Electric flux is given by,  

  ∅ = ∫ E .ds cos θ

                            ∅ = ∫ E .ds cos0 + ∫ E .ds cos0  ( θ=0)

           ∅ = ∫ 2E .ds  

                                ∅ =  2EA -------------(3)

 From (2) and (3)

                         q/ε_{0 =} 2EA

                                      E =  q/ε_02A

                                     E =σ A/2Aε₀ 

                     E= σ  /2ε₀ ^{-----------------(A)}

_{│<<< Electrostatics│  │ Concept of potential Energy>>>│} 

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