Gauss’ Law

Chapter 23: Gauss’s Law
Homework:
• Read Chapter 23
• Questions 2, 5, 10
• Problems 1, 5, 32
Gauss’s Law
• Gauss’s Law is the first of the four
Maxwell Equations which summarize
all of the electromagnetic theory.
• Gauss’s Law gives us an alternative to
Coulomb’s Law for calculating the
electric field due to a given distribution
of charges.
Gauss’s Law: The General Idea
The net number of electric field lines which
leave any volume of space is proportional to
the net electric charge in that volume.
volume V
E lines leaving V
E lines entering V
Flux
∫ Φ = ⋅
S
E dA
r r
The flux Φ of the field E through
the surface S is defined as
The meaning of flux is just the number of
field lines passing through the surface.
Best Statement of Gauss’s Law
The outward flux of the
the electric field through any
closed surface equals the net
enclosed charge divided by ε
0
.
Gauss’s Law: The Equation
∫ ⋅ =
S
A
Qing
E
d
0
/ ε
r
r
• S is any closed surface.
• Qenc is the net charge enclosed within S.
• dA is an element of the area on the surface of S.
• dA is in the direction of the outward normal.
SI units 12
0 .8 85 10− ε = ×
Flux Examples
Assume two charges, +q and –q.
Find fluxes through surfaces.
Remember flux is negative if lines
are entering closed surfaces.
Φ1 = +q/ε
0
Φ2 = –q/ε
0
Φ3 = 0
Φ4 = (q –q )/ε
0 = 0
Another Flux Example
Given uniform field E, find flux through the net.
Must have a closed surface, so let S be (net+circle).
Then Qenc = 0 so Gauss’s law says ΦS = 0.
But circle an E2 Φ = π
net a E2 ∴ Φ = −π
Gauss ⇒ Coulomb
Given a point charge,
draw a concentric sphere
and apply Gauss’s Law:
∫ Φ = ⋅ = 2 E dA E(r) 4π r r r
E = E(r)rˆ
r
Gauss ⇒
2
0
2 ∴ E(r) = q /(4π r ε ) = kq /r !
0 Φ = q / ε
Q.23-1
If a vector field v passes through an area A at an
angle θ, what is the name for the product
v ⋅ A = vAcosθ ?
r
r
(1) curl (2) energy (3) flux (4) gradient
Q.23-1 What is the name for the product
v ⋅ A = vAcosθ ?
r
r
1. Curl
2. Energy
3. Flux
4. Gradient
Q.23-1 Φ = v ⋅ A = vAcosθ
r
r
(1) curl (2) energy (3) flux (4) gradient
= flux from the Latin “to flow”.
Q.23-2
( )
( )
2
0
1 2
2
0
1 2
2
1 2
1 2 0
4 )5(
4 )4(
)3( /
)2( /
)1( 0
a
Q Q
a
Q Q
k Q Q a
Q Q
πε
πε
ε − Φ = + Φ = Φ = + Φ = +
Φ =
Two charges Q1 and Q2 are inside
a closed cubical box of side a.
What is the net outward flux
through the box?
Q.23-2
( )
( )
2
0
1 2
2
0
1 2
2
1 2
1 2 0
4
4
/
/
0
a
Q Q
a
Q Q
k Q Q a
Q Q
πε
πε
ε − Φ = + Φ = Φ = + Φ = +
1. Φ =
2.
3.
4.
5.
Two charges Q1 and Q2 are inside a closed
cubical box of side a. What is the net
outward flux through the box?
Q.32-2
The outward flux of the
electric field through any
closed surface equals the net
enclosed charge divided by
ε
0
.
Gauss:
Two charges
Q
1 and
Q
2 are inside
a closed cubical box of side
a.
What is the net outward flux
through the box?
( )
( )
2
0
1
2
2
0
1
2
2
1
2
1
2
0
4
4
/
/
0
a
Q
Q
a
Q
Q
k
Q
Q
a
Q
Q
πε
πε
ε − Φ = + Φ = Φ = + Φ = +
Φ =
Application of Gauss’s Law
• We want to compute the electric field at
the surface of a charged metal object.
• This gives a good example of the
application of Gauss’s Law.
• First we establish some facts about good
conductors.
• Then we can get a neat useful result:
0
E = σ
/ ε
Fields in Good Conductors
Fact: In a steady state the electric field inside a
good conductor must be zero.
Why? If there were a field, charges would move.
Charges will move around until they find the
arrangement that makes the electric field zero in
the interior.
Charges in Good Conductors
Fact: In a steady state, any net charge on a good
conductor must be entirely on the surface.
Why? If there were a charge in the interior, then by
Gauss’s Law there would be a field in the interior,
which we know cannot be true.
Field at the Surface of a Conductor
• Flux through left face is zero
because E=0.
• Construct closed gaussian
surface, sides perpendicular to
metal surface, face area = A.
• Field is perpendicular to
surface or charges would
move, therefore flux through
sides is 0.
• So net outward flux is EA.
Field at the Surface of a Conductor
• Then
Qenc = σ A.
• Let σ stand for the surface
charge density (C/m
2
)
• Now Gauss’s Law gives us
0
0
0
/
/
/
σ ε
σ ε
ε
=
=
Φ =
E
EA
A
Qenc
Large Sheet of Charge
Q
A
Q
=
σA
σ = charge /area = surface charge density
0
ε
Q
Φ
=
Φ
= ⋅ d
=
2EA
∫
E
A
0
2
ε
σA EA
=
2
0
ε
σ
E
=
Gauss’ law
Long Line of Charge
Φ
= ⋅ d
= EA
∫
E
A
0
ε
Q
Φ
= Gauss’ law
Q
=
l
λ
A
=
2
πrl
0
2
π
ε
λ
r
E
=
λ = charge/length = linear charge density
l
r
λ
E
Top view
/(4 )
4 /
2
0
0
2
E Q r
r E Q
πε
π ε
=
Φ = =
/(2 )
2 /
0
0
E r
rlE l
λ πε
π λ ε
=
Φ = =
/(2 )
2 /
0
0
σ ε
σ ε
=
Φ = =
E
AE A
Point charge, d=0
Line charge, d=1
Surface charge, d=2
Summary for different dimensions
d r E ∝ 2−1
Potential
Outline for today
• Potential as energy per unit charge.
• Third form of Coulomb’s Law.
• Relations between field and potential.
Potential Energy per Unit Charge
The SI unit for potential is the volt
. (1V=1J/C)
• Potential is often casually called “voltage”.
• As with potential energy, it is really the
potential difference which is important.
F =
qE and
U = qV
r
r
Just as the field is defined as force per unit
charge, the potential is defined as potential
energy per unit charge:
Potential Energy Difference
If a charge
q is originally at point A, and we
then move it to point B, the potential
energy will increase by the amount of work
we have done in carrying the charge from A
to B.
∫ − = = ⋅
B
A
B
A AB ext
U
U
W
F sd
r
r
A sd
B
r
q
dW
F sd ext
r
r
= ⋅
Potential Difference
The potential difference between point A and
point B is this work per unit charge.
∫ ∫ = = ⋅ = − ⋅
B
A
B
A
AB AB ext
W qV
F sd
qE sd
r
r
r
r
∫ ∴ = − = − ⋅
B
A
AB
B
A
V
V
V
E sd
r
r
Relation between E and V
x
E
r
etc
.
dx
dV And
E
x
= −
V
E dx
∆
=
−
∫
x
Potential Relative to Infinity
• We have defined potential difference ∆V.
But to have a value for V itself we need to
decide on a zero point.
• In circuits, we define the earth to have
V=0, and often ground the circuit.
• In electrostatics we normally define V=0
far away from all charges. Then at point P
we write V(P) to mean V(P) – V(∞).
Potential at a point
The potential at point P is the work
required to bring a one-coulomb test
charge from far away to the point P.
∫
∫
∞
∞
=
⋅
⋅
=
P
P
E sd
q
F sd
V
P
r
r
r
r
(
)
P sd
r
q
Example 1: Uniform Field
x
E
r
d
V
V
d Ed
V
Edx Ed
−
=
∆
=
−
=
−
∫
)0(
(
)
Work required to push a
1-coulomb test charge
against the field from
x=d to x=0.
“Work = force X distance”
Q.24-1 Suppose, using an xyz coordinate system, in some region of space,
we find the electric potential is
2 V(x) = Ax where A is a constant.
What is the x-component of the
electric field in this region?
E Ax x = −2
2 E Ax x =
E Ax A. x =
B.
C.
Q. 24-1 Given 2 V(x) = Ax
where A is has the constant value 25 / . 2 A = V m
What is the x-component of the electric field?
( )1 Ex Ax )2( Ex Ax )3( Ex 2Ax 2 = = = − Solution: Ax dx dx V A dxd Ex 2 2 = − = − = −
Furthermore
What are the y and z components of the
electric field in the previous question?
0
2
= − = − =
dy
dx V A
dy
d Ey
= − V = 0
dz
d Ez and also
Case of a Single Point Charge
P sd
r
q
Q
R
r
∫ ∫
∞ ∞
= ⋅ =
P
R
V
(
P
)
E sd
E
(
r
)dr
r
r
R
kQ
R kQ
r
kQ
r
dr
V
R
E
r dr kQ
R
R
R
=  − ∞ = −   = −= = ∞
∞ ∞
∫ ∫
1
1
(
)
(
)
2
Coulomb’s Law for V
So we now have a third form of Coulomb’s Law:
V kQ
r
E kQ
r
F kQq
r
.3
/
.2
/
.1
/
2
2
=
=
=
Potential is not a Vector
Adding forces and fields means adding vectors:
finding the resultant vector.
Adding potentials means adding numbers, and
taking account of their signs. But it is much
simpler than adding vectors.
Thus the third form of Coulomb’s Law is the simplest!
Example 1: Adding Potentials
(
)
1
2  =
0   = + = + −
a
kQ
a
kQ
V
P
V
V
Note that the field at point P is not zero!
P
Q
–
Q
a a
a
Q. 24-2 Uniformly charged rod with
charge of –Q bent into arc of 120°
with radius R.
What is V(P), the electric
potential at the center?
R
kQ
R
kQ
R
kQ
R
kQ
2 )4( 2 )1( + )2( − )3( −
Q.24-2
A. +kQ/R
B. -kQ/R
C. +kQ/2R
D. -kQ/2R
Uniformly charged rod with charge
of –Q bent into arc of 120° with
radius R.
What is V(P), the electric
potential at the center?
Q. 24-2 What is V(P), the electric
potential at the center?
Solution: All bits of charge are
at the same distance from P.
Thus
R
kQ
R
kQ
R
kQ
R
kQ
2 )4( 2 )1( + )2( − )3( − ( ) Q Rk dq Rk r kdq V = = = − ∫ ∫
Potential Energy of Some Charges
The potential energy U of a group
of charges is the work W required
to assemble the group, bringing
each charge in from infinity.
We can show that the result is
12
1
2 12
U = kq
q
/
r
U =
U12 +
U13 +
U23 +
L
Where the potential energy of each pair is
of the form
Binding Energy
If the total potential energy U of a group of
charges is negative that means we have to do
work to pull them apart. The magnitude of
this negative potential energy is called the
binding energy.
Examples:
• Removing an electron from an atom to form a positive ion.
• Removing a space probe from earth’s gravitational field.
Example 3: Charged Ring
• In Ch. 22 the
E field of a
charged ring is calculated.
• Here we compute
V first and
then use it to get
E.
• Get
V at
P on axis of circular
ring, a distance
z from center:
∫
=
r
kdQ
V
(
P
)
Example 3 Continued
Key point: All bits of charge are
the same distance from point P!
∫ ∫ + = = =
2
2
(
)
R
z
kQ dQ
r
k
r
kdQ
V
P
Note we have no struggling with angles or
adding many little vectors as we do if we
compute E.
Using V to get E
Now that we have V on the axis we can get
Ez
on the axis by differentiation:
dz
dV
E
z
= − This is slightly messy, but we just need to
remember that
dz
df nf dz df
n
n −
1
=
where here
2
2 2/1
(
)−
f =
R +
z
n
2 2 2/3 2 2 ( ) ( ) R z kQz R z kQ dzd E z + = + = − which is exactly the result the textbook gets in
Ch. 22 by direct integration of the field.
So finally we get
And also:
0
0
2
= − =
= − = − =
V
dz
d E
dy
dx V A
dy
d E
z
y
Furthermore
What are the y and z components of the
electric field in the previous question?
0
2
= − = − =
dy
dx V A
dy
d Ey
= − V = 0
dz
d Ez and also
Summary of Basics
• U = qV
• V = kQ/r
•
WAB = q ( V
B – VA
)
V
E dx ∆ = −∫
x
etc
.
dx
dV
E
x
= −