Electrostatis

Electrostatics

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Move the balloon towards sweater and then towards wall and observe the phenomenon of electrostatics.

Electrostatics : A study of charges at rest is called as electrostatics.

In our day to day life we come across various object which is made up of atom . Atom consist of positively Charged Protons and Negatively Charged Electrons which revolve around the atom. This charges shows various characteristics when they are at rest as well as when they are in motion. In this chapter we are going to deal with charges at rest.

In earlier class we have seen various method of charging a body such as 

1) Conduction 

2) Induction  

Gauss Law

Gauss Law : The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε0.  . It is a powerful tools which helps us to find total electric flux coming out of the closed surface  .

Mathematically it is given by

                                                        ∅ = q/ε_{0           ---------(1)}

                                                            _OR

                                                                   ∅ = ∫ E .ds ----------(2)

Where  ∅ is the total flux and q is the charge enclosed within that surface .

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)}