Phys102
General Physics II
Gauss’ Law
Chapter 24: Gauss’s Law
•Flux
•Electric Flux
•Gauss’ Law
•Coulombs Law from Gauss’ Law
•Isolated conductor and Electric field outside conductor
•Application of Gauss’ Law
•Charged wire or rod
•Plane of charge
•Conducting Plates
•Spherical shell of charge
1
Flux
Φ = (normal component) x Area
θ =0
Φ = vA
Φ = v cosθ A
Φ = v. A
A = Anˆ
θ = 45
Φ = 0.707vA
Gauss’s Law
ƒ Gauss’s law makes it possible to find
the electric field easily in highly
symmetric situations.
ƒ Drawing electric field lines around
charges leads us to Gauss’ Law
ƒ The idea is to draw a closed surface like a balloon
around any charge distribution, then some field line will
exit through the surface and some will enter or renter. If
we count those that leave as positive and those that
enter as negative, then the net number leaving will give
a measure of the net positive charge inside.
2
Flux of Electric Field lines
ƒ
The flux φ through a plane surface of area A
due to a uniform field E is a simple product:
n̂
Α
φ = E A where E is normal to the area A .
• φ = En A = 0 x A = 0 because the
normal component of E is 0
E
E
n̂
Α
E
n̂
• φ = En A =E cos Q A
θ
Α
Approximate Flux
Φ = ∑ E • ∆A
Exact Flux
Φ=
∫ E • dA
dA = nˆdA
Circle means you integrate
over a closed surface.
3
Find the electric flux through a cylindrical
surface in a uniform electric field E
∫ E • dA = ∫ E cosθdA
Φ = ∫ E cos180dA = − ∫ EdA = −EπR
Φ = ∫ E cos90dA = 0
Φ = ∫ E cos180dA = ∫ EdA = EπR
dA = nˆ dA
Φ=
a.
b.
c.
2
2
Flux from
a. + b. + c. = 0
What is the flux if the
cylinder were vertical ?
Flux of Electric Field lines and
Derivation of Gauss’ Law using Coulombs law
ƒ Consider a sphere drawn around a positive point charge.
Evaluate the net flux through the closed surface.
Net Flux =
Φ=
∫ E • dA
=
∫ E cosθdA = ∫ EdA
E II n
Cos 0 = 1
For a Point charge E=kq/r2
Φ=
∫ EdA = ∫ kq /r dA
2
nˆ
Φ = kq /r 2 ∫ dA = kq /r 2 (4 πr 2 )
dA
Φ = 4 πkq
4 πk = 1/ε0 where ε0 = 8.85x10−12
Φ net =
qenc
ε0
C2
Nm 2
dA = nˆdA
Gauss’ Law
4
Gauss’ Law
Φ net =
qenc
ε0
ε 0 ∫ E.dA = qenc
This result can be extended to any shape surface
with any number of point charges inside and
outside the surface as long as we evaluate the
net flux through it.
Gauss’ Law
Φ net =
qenc
ε0
ε 0 ∫ E.dA = qenc
This result can be extended to any shape surface with any
number of point charges inside and outside the surface as
long as we evaluate the net flux through it.
5
Gauss’ Law
Φ net =
qenc
ε0
ε 0 ∫ E.dA = qenc
This result can be extended to any shape surface with any
number of point charges inside and outside the surface as
long as we evaluate the net flux through it.
Applications of Gauss’s Law
ƒ Find electric filed of an infinite long uniformly charged wire of
of
negligible radius.
ƒ Find electric field of a large thin flat plane or sheet of charge
charge
ƒ Find electric field around two parallel flat planes
ƒ Find E inside and outside of a long solid cylinder of charge
density ρ and radius r.
ƒ Find E for a thin cylindrical shell of surface charge density σ
ƒ Find E inside and outside a solid charged sphere of charge
density ρ
6
Wire: Cylindrical symmetry
Spherical symmetry
7
Electric field in and around
conductors
ƒ Inside a conductor in electrostatic
equilibrium the electric field is zero
( averaged over many atomic volumes).
The electrons in a conductor move
around so that they cancel out any
electric field inside the conductor
resulting from free charges
anywhere including outside the
conductor. This results in a net force of
F = eE = 0 inside the conductor.
Electric field in and around
conductors
ƒ Any net electric charge resides
on the surface of the conductor
within a few angstroms (10-10 m).
Draw a gaussian surface just
inside the conductor. We know
E=0
everywhere on this surface.
Hence , the net flux is zero.
Hence, the net charge inside is
zero.
8
Electric field in and around
conductors
ƒ The electric field just outside a conductor has
magnitude σ /ε0 and is directed
perpendicular to the surface.
– Draw a small pill box that extends
into the conductor. Since there is
no field inside, all the flux comes
out through the top.
– EA=q/ε
EA=q/ε0= σA/ ε0,
– so E= σ / ε0
9
Nonconducting sheet
Two conducting
plates
10
Two Nonconducting sheets
11