Physics 122B
Electricity and Magnetism
Lecture 12 (Knight: 30.1 to 30.4)
Calculating E from V
April 23, 2007
Martin Savage
Lecture 12 Announcements
 Lecture HW Assignments #4 has been posted on the
Tycho system. Assignment #4 is due at 10 PM, on
Wednesday.
 Requests for regrades of Exam 1 should be written on
a separate sheet (see Syllabus) and taken to Heleb
Gribble in room C136 PAB. They will be accepted until
noon on Wednesday.
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Physics 122B - Lecture 12
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Chapter 29 – Summary (1)
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Chapter 29 – Summary (2)
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The Missing Link
How are E and V connected?
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Finding V from E
V
U q+sources
q
sf
U  W (i  f )    Fs ds; V  U / q; Es  Fs / q
si
sf
V  V ( S f )  V ( Si )    Es ds
si
V   Es s (uniform electric field)
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Potential from the Field
of a Point Charge
V  V ()  V (r )

   Er dr  
r
q

q
1
dr
2

4 0 r r

1
V ( r )  V ( ) 
dr
2

4 0 r r

q  1


4 0  r r
1 q

4 0 r
in
 0
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This is the same result that we obtained
Chapter 29 from energy considerations.
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Example:
The Potential of a Charged Disk
In Chapter 26, we found that the field
on the axis of a charged disk was:


z
Edisk
1  2

2
z R 

Find the corresponding potential Vdisk .
Q

2 R 2 0




Q
z
V ( z )  V ()   Ez ( z )dz  0 
1  2
 dz
2

2
2

R

z R 
0 z 
z


Q 
zdz 
Q 

dz


z  z 2  R2


2
2

2 R  0  z
z 2  R 2  2 R  0 
z
Vdisk 
Q
2 R 2 0
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
z 2  R2  z


z
This is the same expression we previously calculated.
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Finding E from V
V 
U q+sources
q
 Es  
Ex  
W

  Es s
q
V
 dV 
 lim  

s s 0  ds 
dV
dV
dV
; Ey  
; Ez  
dx
dy
dz
 dV ˆ dV ˆ dV ˆ  In other words, the E field components are
E  
i
j
k  determined by how much the potential V
dy
dz  changes in the three coordinate directions.
 dx
dV
d  1 q
1 q
For a point charge: E  Er  
 

dr
dr  4 0 r  4 0 r 2
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Example:
The E Field of a Charged Ring
In Chapter 29, we found that the potential
on the axis of a charged ring was:
1
Q
Vring 
4 0 z 2  R 2
Find the corresponding electric field Ering .
dV
d  1
E  Ez  
 
dz
dz  4 0
Ering 
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1


2
2
z R 
Q
Qz
4 0 z 2  R 2 2


3
This is the same expression
that we previously calculated.
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Example:
Finding E from the Slope of V
An electric potential V in a particular
region of space where E is parallel to the
x axis is shown in the figure to the right.
Draw Ex vs x.
V
Ex  
x
V/x  (20 V)/(0.020 m)  1000 V/m
0<x<2 cm 
 Ex  1000 V/m
V/x  (0 V)/(0.020 m)  0 V/m
2<x<4 cm 
 Ex  0 V/m
V/x  (  20 V)/(0.040 m)  500 V/m
4<x<8 cm 
 Ex  500 V/m
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Question
Which graph of the electric potential V
describes the electric field shown?
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Geometry of Potential and Field
On an equipotential surface, Es  
dV
 0.
ds1
Perpendicular to the surface, E  
V V
dV
V
1

   
ds2
s2
s2
s2
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Example: Finding the E-Field
from Equipotential Surfaces
The figure shows a contour map of a
potential.
Estimate the strength and direction
of the electric field at points 1, 2, and 3.
E1  V1/x  (100 V)/(0.020 m)  5000 V/m
"Downhill" direction  up
E2  V2 /x  (100 V)/(0.040 m)  2500 V/m
"Downhill" direction  left
E3  V3 /x  (100 V)/(0.025 m)  4000 V/m
"Downhill" direction  ~ 30 left of vertical
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Kirchhoff’s Loop Law
Since the electric field is
conservative, any path between
points 1 and 2 finds the same potential
difference. Any path can be approximated
by segments parallel and perpendicular to
equipotential surfaces, and the
perpendicular segments must cross the
same equipotentials.
Since a closed loop starts and ends at
the same point, the potential around the
loop must be zero. This is Kirchhoff’s Loop
Law, which we will use later.
Vloop    V i  0
i
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Question
Which set of equipotential contours
describes the electric field shown?
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A Conductor in
Electrostatic Equilibrium
A conductor is in electrostatic
equilibrium if all charges are at rest and
no currents are flowing. In that case,
Einside=0. Therefore, all of it is at a
single potential: Vinside=constant.
Rules for conductor.
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Sources of Electric Potential
A potential difference can be created
by moving charge from one conductor to
another.
The potential difference on a
capacitor can produce a current
(flow of charge), but this current
cannot be sustained because the
charge separation and potential
difference rapidly disappears.
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The Van de Graaff Generator
A Van de Graaff Generator
“pumps” charge on a moving
belt, creating a large potential
difference.
 Charge is mechanically
transported from the negative
side to the positive side and
sustains a potential difference
between the spherical dome
electrode and its surroundings.
 The electric field of the dome exerts a downward force on the positive
charges moving up the belt, causing the belt motor to do increasing work
as the system charges. The work goes into “lifting” the positive charges
through the rising field.
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Batteries and EMF
A battery is a chemical source of electric
potential difference. Chemical reactions create
potential difference by moving positive ions to
one electrode and negative ions to the other.
The system can be visualized as a chemical
“charge escalator” in which positive charges are
“lifted” through a potential difference.
The potential difference Vbat is determined
by the chemistry of the electrodes (e.g., carbon
and zinc) and remains essentially constant until
the chemicals are exhausted and the battery is
“dead”.
The term EMF (electromotive force),symbol E,
is used to describe the work done per unit charge
by the battery: E = Wchem/q = Vbat. A real
battery has “internal resistance” that increases
as the chemicals are used and limits current flow.
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E Fields and Pointed Objects
On conductors, mobile charge tends to accumulate at locations having
the greatest curvature. This creates very strong electric fields near
the tip of a pointed object. If such an object is negatively charged,
electrons may be “field emitted” from the sharp point.
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Capacitors and Capacitance
Kirschoff: Vbat  Vw1  VC  Vw2  0
Initial: VC  0, Vw1  Vw2  12 Vbat .
Final: Vbat  VC  Ed , Vw1  Vw2  0.
Q
E
0 A
Units: 1 farad  1 F  1 C/V
 0 A 
Q   0 AE  
 VC
 d 
0 A
Q
C

Capacitance
VC
d
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Q  C VC
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Example: Charging a Capacitor
The spacing between the plates of a
1.0 mF capacitor is 1 mm.
(a) What is the surface area A of the
plates?
(b) How much charge is on the plates if
this capacitor is attached to a 1.5 V
battery?
A
dC
0
1.0 mF
1.5 V
.05 mm
 4 (9.0 109 Vm/C)(5.0 105 m)(1.0 10-6 F)  5.65 m 2
Q  C VC  (1.0 106 F)(1.5 V)  1.5 106 C  1.5 m C
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Forming a Capacitor
Any two conductors can form a
capacitor, regardless of their shape.
Q
C
VC
The capacitance depends only on the geometry of the conductors,
not on their present charge or potential difference.
(In fact, one of the conductors can be moved to infinity, so the
capacitance of a single conductor is a meaningful concept.)
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Example: A Spherical Capacitor
A metal sphere of radius R1 is inside and
concentric with a hollow metal sphere
of inner radius R2.
What is capacitance of this spherical
capacitor?
sf
V  V ( s f )  V ( si )    Es ds
si
VC 
Q
Q 1 1 
ds

  
2

4 0 R 1 r
4 0  R1 R2 
1
R2
1
C
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 1
RR
Q
1 
 4 0     4 0 1 2
VC
R2  R1
 R1 R2 
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Combining Capacitors
Parallel: Same V, but different Qs.
Cparallel 
Q  Q2  Q3 
Q
 1
VC
VC
 C1  C2  C3 
Series: Same Q, but different Vs.
Cseries 
Q
Q

VC V1  V2  V3 

1
 V1 / Q    V2 / Q    V3 / Q  

1
1/ C1  1/ C2  1/ C3 
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 C1 || C2 || C3 ||
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Reminder: Combining Resistors
Conducting material that carries current
across its length can form a resistor,
a circuit element characterized by an
electrical resistance R:
R ≡ rL/A
where L is the length of the conductor and A is
its cross sectional area. R has units of ohms.
Multiple resistors may be combined in
series, where resistances add, or in parallel,
where inverse resistances add.
I
Rnet
Rnet
Parallel Connection [(1/A)]:
Series Connection [L]:
Rnet  R1  R2  R3
1
1
1
1
 

Rnet R1 R2 R3
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Example: A Capacitor Circuit
Find the charge and
potential difference
across each capacitor
shown in the figure.
Q
Cparallel  C1  C2
VC
CC
1
Cseries 
 1 2
1/ C1  1/ C2 C1  C2
C
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Energy Stored in a Capacitor
1
U  dqV  qdq
C
Q
2
1
1 Q
U C   qdq  2
C0
C
2
Q
U C  12
 12 C VC 2
C
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End of Lecture 12
 Before the next lecture, read Knight, sections 30.6
and 30.7.
Lecture HW Assignments #4 has been posted on the
Tycho system. Assignment #4 is due at 10 PM, on
Wednesday .
 Requests for regrades of Exam 1 should be written on
a separate sheet (see Syllabus) and taken to Helen
Gribble in room C136 PAB. They will be accepted until
noon on Wednesday.
7/19/2016
Physics 122B - Lecture 12
30