Summary (Electric Field and Electric Charge)
Charge: two kinds. Same sign repel, opposite attract
Conductor/insulator? Easiness of electron movement
Atom:proton, neutron, electron; What is an ion?
Coulomb’s Law:
1 q1q2 1
9
2
2
F =
,
=
8.988
×
10
N
•
m
/C
4 πε 0 r 2 4 πε 0
!
1 q
E=
r̂
2
4!" 0 r
Electric field : Force per unit charge. It’s a vector!
€
Total field is the vector sum of the fields of all charges
(superposition)
Electric field line: Direction of the field is the tangent of the line,
the density of the lines is proportional to the magnitude
Calculating electric field of basic charge distributions.
HMWK example
Charge Q is distributed uniformly
along a thin rod (length 2a).
What is the electric field at point A,
at a distance a to the right of the right
end of the rod?
dQ:charges on the segment
From R to R-dR
Q
A
Everything here should
Be defined by the
identified variable
E tot = ∫ dE x =

E tot = E tot xˆ
R ∈[2a + a,a]
2a
a
R-dR
R:distance from dQ to A
The changing parameter(variable)
€
2
dQ
Linear charge
density:
λ=Q/(2a)
Q
(−dR)
2a
a k (−dR)
−kQ a 1
⇒ E tot = ∫ 3a 2
Q=
∫ dR
R 2a
2a 3a R 2
−kQ 1 a
kQ 1 1
E tot =
(− ) 3a =
( − )
2a
R
2a a 3a
kQ
= 2 (3 −1)
6a
kQ
∴ E tot = 2
3a

kQ
E tot = 2 xˆ
3a
dQ = λ(−dR) =
dEx E
xtot
k
∫R
Chapter 22: Gauss’s Law
•  Electric flux: What is it and why is it important
•  Gauss’s law: relating electric flux, and therefore
the electric field, on a closed surface to the total
net charge inside the surface.
•  Using Gauss’s law, the electric field of various
symmetric charge distributions can be calculated
•  Infinite wire, Infinite cylinder, Infinite sheet,
•  Solid insulating sphere (outside, inside)
•  Two oppositely charged conducting plates…
Introduction
•  What is the safest place to be in a lightening storm?
–  You’ll know it after this chapter.
•  Coulumb’s law: Charge Field
•  Gauss’s law: Field Charge
•  Who is Gauss?:
–  Carl Friedrich Gauss (1777-1855), the prince of
mathematics
–  Gauss on Mathematics: Queen of Sciences
–  Child prodigy, Ph.D at 22yo
–  Gaussian distribution and statistics f (x) =
1
2πσ 2
e
−
(x − µ )2
2σ 2
–  Among many, many, other achievements: Gauss’s Law
€
Why would Gauss’s law make your life easy?
If you hated Integrals, you will like Gauss’s law!
Txt book example (Electric field of
Q
1
Ex =
4πε 0 x x 2 + a 2
A thin rod along y-axis, at a location at the x axis)
λ 1
Ex =
When the rod is infinitely long, and λ=Q/(2a)
2πε 0 x
After this chapter, you will learn a much
easier way of calculating electric field for €
this type of charge distributions
Flux as the flow out of an imagined box
•  If we construct a boundary around a charge or
charges, we can think of the flow coming out from
the charge like water through a screen surrounding a
sprinkler, or the light coming out from a bulb.
What happens as I change the conditions?
•  Will the flux change if the charge doubles.
–  Will the luminosity change if two bulbs are enclosed
instead of one
•  Will the flux change if the box dimensions doubles?
–  Imagine: Same bulb, with a bigger box; Does the amount of
light coming out change?
For a fixed charge, the flux can be represented by the number of arrows
going through the surface, figuratively.
A measurement of flux is dependent on the orientation
•  What is the volume flow rate (flux)
through


the rectangle in (a), with v // A, A being the
normal vector of the rectangle
dV
=€vA
dt

A
•  What is the flow rate if the rectangle is tilted
(b)?
€ It’s no longer,
–
–  but
dV
= vA
dt
dV
= vA⊥
dt
€
•  The amount of fluid that flows through the
rectangle (yellow), is the same as that through

€
its projection (purple)
perpendicular to V
•  Therefore: dV = v • A = vAcos φ
dt
φ is the angle between the velocity and
€ the area vector (perpendicular to the surface)
€
Flux in a Uniform Electric Field
•  Just like the fluid flow, the volume flow rate being
dV  
= v • A = vAcos φ
dt
•  The electric flux can be defined similarly, with
dV
⇒ ΦE
dt
 
v⇒E
€Therefore, the electric flux (ΦE) is ΦE = E • A = E⊥A = EAcos φ

A = Anˆ
with the area vector being
Surface is Face-on to E field
Angled
Edge-on
€
φ is the angle between the field and the area vector (perpendicular to the surface)
Do you know any other natural phenomenon that is similar?
Hint: Sun and the titled earth
€
Electric flux through a disk
E=2.0x103 N/C, φ=30o, r=0.10m
What is ΦE through this disk?
 
ΦE = E • A = E⊥A = EAcos φ
ΦE = (2.0 × 10 3 )(4 πr 2 )(cos 30 o ) = 54N⋅ m 2 /C
What if φ=0o? What does it mean?
φ=90o? What does it mean?
φ
(This disk is not a closed surface,
The orientation of the disk is arbitrarily specified)
Electric flux through a cube
• (a) The electric field is perpendicular to surface 1 and 2. Flux
through Surface 1 and 2 cancel each other, while it is 0 on all other
surfaces. What is ΦE through the cube?
• (b), what if the cube is rotated?
(a)
(b)
Non-uniform field and Electric flux through the sphere
•  Q=3.0µC
•  Although the direction of E changes, the magnitude is the same
everywhere on the surface
q
4 πε 0 r 2


ΦE = ∫ E • dA =
E=
∫ EdA = E ∫ dA = EA
Why? Why? Why?
E is a constant


E is always in the same direction of dA
ΦE =
€
€
q
q
2
(4
π
r
)
=
4 πε 0 r 2
ε0
€
Gauss’s Law
•  The expression is an alternative to Coulomb’s Law:
•  In fact, Coulomb’s law can be derived from Gauss’s law
•  Gauss’s is a genius mathematician, and a famous physicist as well
•  Gauss’s law: The total electric flux through a closed
surface is equal to the net charge inside
the surface, devided by ε0 (Vacuum permittivity)
•  Can you now derive Coulomb’s law?
Point Charge Inside a Spherical Surface
•  Spheres of different radius enclose the charge. Does the electric
flux dependent on the radius?
•  There are different ways to understand this:
•  Count the field lines
•  Replace the charge with a light bulb
and the electric flux with the light emitted
•  What is the flux anyway through
these spherical surfaces?
ΦE = EA =
1 q
q
2
(4
π
R
)
=
4 πε 0 R 2
ε0
ε 0 = 8.85 × 10C 2 /(N • m 2 )
•  It does not depend on R!
€
•  (E is the same everywhere on the surface)
Point Charge Inside a Nonspherical Surface and the General
Form of Gauss’s law
•  Consider a charge surrounded by a irregular surface
•  Each of the area elements dA projects onto a corresponding
spherical surface element.
Surface integral
•  After integrating, the total electric surface is ΦE =
∫


E • dA
And it it is the same as the total flux through a sphere inside this
surface, which is
ΦE _ Sphere
Therefore
ΦE =
∫
€
q
=
ε0
€

 q
E • dA =
ε0
More than a point charge?
€
ΦE =
∫

 Qencl
E • dA =
ε0
Gauss’s law
Remark: Gaussian surfaces
are imaginary
€
Gauss’s Law
ΦE =
∫

 Qencl
E • dA =
ε 0 The total electric flux through a closed surface
is equal to the total (net) electric charge inside
the surface, divided by ε0
Earlier slide: If you hate integral, you’ll like Gauss’s law….
But what about this “strange” surface integral?
Did we just trade one bad apple with another one that’s even worse?
The answer is obviously NO, we didn’t. More on this on Thursday!
What does Gauss’s law address:
1.  If we measure the electric field on a closed the surface, then
we can determine the charge distribution within that surface!
2. The reverse can be true as well, if symmetry exists
What if the sign of the charge changes?
ΦE =
∫

 Qencl
q
E • dA =
=+
ε0
ε0


E is in the opposite direction of dA
€
€
€
ΦE =
∫

 Qencl
q
E • dA =
=−
ε0
ε0


E is in the opposite direction of dA
Q22.1
A spherical Gaussian surface (#1)
encloses and is centered on a point
charge +q. A second spherical
Gaussian surface (#2) of the same
size also encloses the charge but is
not centered on it.
Compared to the electric flux
through surface #1, the flux
through surface #2 is
+q
Gaussian
surface #1
A. greater.
B. the same.
C. less, but not zero.
D. zero.
Gaussian
surface #2
E. not enough information given to decide
Q22.2
Two point charges,
+q (in red) and –q
(in blue), are
arranged as shown.
Through which
closed surface(s) is
the net electric flux
equal to zero?
A. surface A
B. surface B
C. surface C
D. surface D
E. both surface C and surface D
Electric Charge and Field of a solid conductor: What s special?
•  In electrostatics, the E field is always zero in a conductor.
–  Otherwise, the charges will still be moving.
•  All excess charges placed on the conductor are on the surface.
Electric Field of a conducting sphere
Qencl
"0
!E #0
$Qencl # 0
EA =
This Gaussian surface can be
!
infinitely small
any anywhere
inside conductor:
The field of an infinite line or plane of charge
The direction of the field can be determined using symmetry
Electric field of an infinite
line of charge
Electric field of an plane
line of charge
σ: 2d charge density
λ: Linear charge density
Dependent on r!
Q
E"A = EA = encl
#0
! Qencl = $l
A = (2%r)l
$l
& E (2%r)l =
'
#0
$
E =
2%r# 0
E"Atot = E (2 A) =
Qencl
#0
! Qencl = $A
% E (2 A) =
E =
$
2# 0
$A
&
#0
Dependent on r?
No
Electric field between parallel plates of opposite charge:
Capacitor
Real capacitor
Idealized model
E=
"
2# 0
!
• Away from the two plates,
the fields cancel each other
• Between the plates, the fields
from both sides are equal
"
E1 = E 2 =
2# 0
"
"
E = 2( ) =
2# 0
#0
Previous slide
Electric field of a uniformly charged (insulating) sphere
•  This is very similar to the gravitation of a planet (inside & outside)
Inside the sphere, at the distance of r from center:
Qencl
#0
QV
= $V =
4
%R 3
3
E"A = EA =
! Qencl
4 3
%r , A = 4 %r 2
3
4
Q %r 3
3
& E (4 %r 2 ) =
'
4
3
%R # 0
3
Qr
E =
4 %# 0 R 3
V =
Outside the sphere, at the distance of r from center:
!
!
E=
Q
4 "# 0 r 2
Are these two the same at r=R?
Charges on conductors (Solid or with cavity)
Key word: electrostatic situation
E=0 everywhere within conductor!
Charges reside only on
the surface!
No net motion of charge
Example: Using Gauss s law to find charge distribution
A solid conductor with a cavity carries a total charge of +7nC.
A point charge of -5nC resides inside the cavity.
How much charge is on the inner/outer surface of the conductor?
!
E "0#
!
!
Qencl
% E • dA = 0 = $
0
# Qencl = (&5nC) + Qinner = 0
'Qinner = +5nC
Qtot = +7nC = Qinner + Qouter
# Qouter = (+7nC) & (+5nC) = +2nC
!
How do we know Gauss s law is correct?
Faraday s icepail experiment
Why do we care?
1 q1q2
F=
Gauss s law can derive Coulomb s law:
4 "# 0 r 2
We now know
it doesn t
differ from 2
more than 10-16
!
Charge won t go anywhere
If Gauss s law is correct,
the ball must lose all charges!
The Van de Graaff electrostatic generator
•  Can increase the charge on the shell and the electric field around it
rapidly
•  Used for accelerator of charged particles
•  And of course, crazy hairdos!
A Faraday cage and electric shielding
The field modifies the charge distribution on the conducting box!
The induced charge produces a fields that cancels the original one!
Again, where is the safest place to be in a lightning storm?
(if you don t have a faraday cage)
Q22.3
A conducting spherical shell with inner
radius a and outer radius b has a
positive point charge Q located at its
center. The total charge on the shell is –
3Q, and it is insulated from its
surroundings. In the region a < r < b,
A. the electric field points radially outward.
B. the electric field points radially inward.
C. is zero.
D. not enough information given to decide
Q22.5
There is a negative surface charge density in a certain region on
the surface of a solid conductor.
Just beneath the surface of this region, the electric field
A. points outward, toward the surface of the conductor.
B. points inward, away from the surface of the conductor.
C. points parallel to the surface.
D. is zero.
E. not enough information given to decide
Q22.6
For which of the following charge distributions would
Gauss s law not be useful for calculating the electric
field?
A. a uniformly charged sphere of radius R
B. a spherical shell of radius R with charge uniformly
distributed over its surface
C. a right circular cylinder of radius R and height h with
charge uniformly distributed over its surface
D. an infinitely long circular cylinder of radius R with
charge uniformly distributed over its surface
E. Gauss s law would be useful for finding the electric
field in all of these cases.
Summary: Gauss’s Law
• Electric Flux: flow of electric field through a surface
•
(Just like the light flow from your flashlight)
• Gauss’s Law: total electric flux through a closed surface (Gaussian
Surface) is proportional to the total charge enclosed.
"E =
#
!
! q
E • dA =
$0
• Calculating Electric Field using Gauss’s law:
•
!
useful when charge distribution is highly symmetric
• Special properties of a conductor:
•
Field within an electrostatic conductor is always 0
•
The charges are distributed on the surface (not inside)
•  What’s real: Electric Field Line/Gaussian Surface/Electric Flux/
Electric Field