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
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/ε0 ---------(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/ε0 ------------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/ε0 = E .4πr2
E = q/ε04πr2
E =σ 4πR2 /ε04πr2 ------------------from(3)
E= σ R2 /ε0r2 -----------------(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)