Given a charge distribution of cylindrical, planar or spherical symmetry Gauss's Law may provide a method to calculate the field.
1) Describe the field lines using the symmetry of the problem
2) Can you draw a Gaussian surface which allows you to easily find the total flux, eg.:
E and normal to A are parallel
EAcos(θ) is constant over some part of the surface and maybe zero over other parts
Usually the flux integral Φ = ? EdA ends up being Φ = E ? dA =EA
3) Calculate the enclosed charge as
Q = ∫ρdV for volume charges if p is constant Q = ρV
Q = ∫σdA for area chargesif a is constant Q = σA
Q = ∫λdL for linear charges if λ is constant Q = λL
4) Plug into ∫EdA = 4ΠkQenclosed and solve for E
Spherical Symmetry - p(r)
What is E for for a boll of charge of radius R ρ(r)= ar2 (a is constont)
1) Describe the field lines using the symmetry of the problem
Field lines ore radial justdepend on r.
2) Calculate the flux Φ = ? EdA
Draw a Gaussion surface as a sphere at radius r where you want to find field. E is constant at this radius so ? EdA = E ? A =
3) Calculate the Qenclosed
r < Qenclosed = ∫ ρ(r)dV =
r > R Qenclosed = ∫ ρ(r)dV =
4) Plug into ? EdA = 4ΠkQenclosed and solve for E
r < R
r > R
Cylindrical Symmetry -ρ(r)
What is E for a Infinitely long cylinder of charge density ρ(r)= arn, radus R?
1) Describe the field lines using the symmetry of the problem
Field lines are radial and just depand on r.
2) Calculate the flux Φ = ? EdA
Draw a Gaussion surface as a cylinder coaxial with charge cylinder at the radius, where you want to find the field. E is constant at this radius so
? EdA = E ? dA = EA = E 2ΠrL
3) Calculate the Qenclosed
r < Qenclosed = ∫ ρdV =
r > R Qenclosed = ∫ ρdV =
4) Plug into ? EdA = 4ΠkQenclosed and solve for E
r < R E(r) =
r > R E(r) =
Charged Wire - λ
What is E for a infineitely long Wire of constant linear charge densily λ?
1) Describe the field lines using the symmetry the problem Field lines are radial and just depand on r.
2) Calculate the flux Φ = ? EdA
Drawn Gaussian surfaceas a cylinder coaxial With the charge cylinder the radius r where you motto find the field.
E is consant at this radius so
? EdA = E ? dA = EA = E 2ΠrL
3) Calculate the Qenclosed
Qenclosed = ∫λdL = λ∫dL =
4) Plug into ∫ EdA = 4ΠkQenclosed and solve for E
E(r) =
Planar Symmetry ρ(x)
What is E for a slab of charg ρ(x)= bx2?
1)Describe the field lines using the symmetry the problem Field lines are radial and just depand on r.
2) Calculate the flux Φ = ? EdA
Draw a Gaussian surface as a cylinder cutting through the slab of length 2x. The flux is the filed going through the ends of
this surface at ± x
? EdA E ? dA= 2A
a) Calculate the Qenclosed
|x| = a/2 Qenclosed = ∫ ρdV = -x∫x bx2Adx = (Nodete dv = Adx)
|x| = a/2 Qenclosed = ∫ ρdV =
4) Plug into ? EdA = 4ΠkQenclosed and solve for E
|x| = a/2 E(x) =
|x| = a/2 E(x) =
E Due to an Infinite sheet of charge by Gauss's Law charge density σ C/m2
What is E for a sheet of charge density σ C/m2
1) Describe the field lines using the symmetry of the problem field lines point pendiculor to the slab of the lenght 2x.
2) Calculate the flux Φ = ? EdA
Draw a Gaussion surface as a cylirlder cutting through the slab of length 2x.
The flux is the filed going through the ends of this surface at ± x
? EdA = E?dA = 2A
3) Calculate the Qenclosed
Qenclosed = ∫σdA = σA
4) Plug into ?EdA = 4ΠkQenclosed and solve for E
E(x) =