Your Comments
I really liked how he said "Woah"
I saw someone else's frivolous comment on the screen last lecture. I guess it's true
what they say, the first cut is the deepest. :(
I'm really confused about all of these concepts. I don't understand in words (much
less the formulas) what electric potential and electric potential difference are. I
think it would be a good idea to try to explain what these are first before going into
a lot of examples.
I feel really confused about the difference between electric potential energy and
electric potential, and how these relate to the electric field.
Electric potential energy and electric potential?? what?! For the most part I
understood everything, some topics were a little difficult to grasp at first.
Just unclear on one point, in the example: Charged Spherical Insulator when
integrating the field from infinity to a you integrate with respect to 1/r^2, but
for the field from a to r you integrate with respect to 1/a^3 and I'm just a little
confused about why...
Please go over the change in potential in a uniformly charged insulating sphere slowly
in lecture. Personally, I found the integral for the change for r<a rather confusing.
05
Physics 212 Lecture 6, Slide 1
Charged Spherical Insulator
Electricity & Magnetism Lecture 6, Slide 2
Electricity & Magnetism Lecture 6, Slide 3
Physics 212
Lecture 6
Today’s Concept:
Electric Potential
(Defined in terms of Path Integral of Electric Field)
Electricity & Magnetism Lecture 6, Slide 4
Big Idea
Last time we defined the electric potential energy of charge q in an electric field:
b
 
 
   F  dl    qE  dl
b
U ab
a
a
The only mention of the particle was through its charge q.
We can obtain a new quantity, the electric potential, which is a PROPERTY OF
THE SPACE, as the potential energy per unit charge.
Vab
b 

U ab

   E  dl
q
a
Note the similarity to the definition of another quantity which is also a PROPERTY

OF THE SPACE, the electric field.
 F
E
q
Electricity & Magnetism Lecture 6, Slide 5
Electric Potential is like Height
(E points down hill for positive charge)
The electric potential, however, is difficult to picture because it's
hard to imagine the quantity of energy a particle will have at any
given position.
Physics 212 Lecture 6, Slide 6
Electric Potential from E field
Consider the three points A, B, and C located in a region of constant electric
field as shown.
D
x
What is the sign of VAC  VC  VA ?
A) VAC < 0
B) VAC  0
C) VAC > 0
E points down hill
 
   E  dl
C
Remember the definition: VAC
A
Choose a path (any will do!)
  C 
   E  dl   E  dl
D
VAC
A
D
 
 0   E  dl   Ex < 0
C
VAC
D
Electricity & Magnetism Lecture 6, Slide 7
CheckPoint 2
If the electric field is zero in a region of space, what does that tell you about the electric
potential in that region?
A)
A)
B)
B)
C)
C)
D)
D)
The electric potential is zero everywhere in this region.
The electric potential is zero at at least one point in this region.
The electric potential is constant everywhere in this region.
There is not enough information given to distingush which of the above answers is
correct.
Remember the definition
 
   E  dl
B
VAB
A

E 0
VAB  0
V is constant!
Electricity & Magnetism Lecture 6, Slide 8
E from V
If we can get the potential by integrating the electric field:
 
   E  dl
b
Vab
a
We should be able to get the electric field by differentiating the potential?


E  V
In Cartesian coordinates:
Ex  
V
dx
Ey  
V
dy
V
Ez  
dz
Electricity & Magnetism Lecture 6, Slide 9
CheckPoint 1a
At which point is the magnitude of the electric field greatest?
“A) higher electric potential indicates higher magnitude electric field”
“B) B has the steepest slope so the electrical potential is decreasing the fastest.”
“C) Since voltage is the result of integrating the electric field, E will be the greatest when the
slope has the largest positive slope, which is point c”
How do we get E from V ?


E  V
V
Ex  
dx
Look at slopes!
Electricity & Magnetism Lecture 6, Slide 10
CheckPoint 1b
A
B
C
D
At which point is the electric field pointing in the
negative x direction?
“B) negative slope“
“C) The graph has a positive slope”
How do we get E from V ?


E  V
V
Ex  
dx
Look at slopes!
Electricity & Magnetism Lecture 6, Slide 11
Equipotentials
Equipotentials are the locus of points having the same potential.
Equipotentials produced
by a point charge
Equipotentials are
ALWAYS
perpendicular to the electric field lines.
The SPACING of the equipotentials indicates
The STRENGTH of the electric field.
Electricity & Magnetism Lecture 6, Slide 12
CheckPoint 3a
A
B
C
D
At which point is the magnitude of the electric field
the smallest?
“This is where the density of the lines is the smallest.“
Electricity & Magnetism Lecture 6, Slide 13
CheckPoint 3b
A
B
C
D
Compare the work needed to move a NEGATIVE charge from A to B, with that required
to move it from C to D
A) More work from A to B
A)
B)
B) More work from C to D
C)
D)
C) Same
D) Can not determine w/o performing calculation
Less than ½ got this correct!
Electricity & Magnetism Lecture 6, Slide 14
Hint
What are these?
A
B
C
D
ELECTRIC FIELD LINES!
What are these?
EQUIPOTENTIALS!
What is the sign of WAC  work done by E field to move negative charge from A to C ?
A) WAC < 0
B) WAC  0
A and C are on the same equipotential
C) WAC > 0
WAC  0
Equipotentials are perpendicular to the E field: No work is done along an equipotential
Electricity & Magnetism Lecture 6, Slide 15
CheckPoint 3b Again?
A
B
C
D
Compare the work needed to move a NEGATIVE charge from A to B, with that required
to move it from C to D
A)
B)
A) More work from A to B
C)
B) More work from C to D
D)
C) Same
D) Can not determine w/o performing calculation
 A and C are on the same equipotential
 B and D are on the same equipotential
 Therefore the potential difference between A and B is the SAME as the potential between C
Electricity & Magnetism Lecture 6, Slide 16
and D
CheckPoint 3c
A
B
C
D
Compare the work needed to move a NEGATIVE charge from A to B, with that required
to move it from A to D
A)
B)
A) More work from A to B
C)
B) More work from D to D
D)
C) Same
D) Can not determine w/o performing calculation
Electricity & Magnetism Lecture 6, Slide 17
Calculation for Potential
cross-section
a4
a3
a2
a1
+Q
+q
Point charge q at center of concentric
conducting spherical shells of radii a1, a2, a3, and
a4. The inner shell is uncharged, but the outer
shell carries charge Q.
What is V as a function of r?
metal
metal
Conceptual Analysis:
 Charges q and Q will create an E field throughout space
 
 V (r )   E  d 

r
r0
Strategic Analysis:
 Spherical symmetry: Use Gauss’ Law to calculate E everywhere
 Integrate E to get V
Electricity & Magnetism Lecture 6, Slide 18
Calculation: Quantitative Analysis
cross-section
r > a4 : What is E(r) outside spheres?
+Q
a4
a3
a2
a1
A) 0
+q
metal
r
B)
Q
4 0 r 2
1 Q+q
D)
4 0 r 2
1
C)
1 Q+q
2 0 r
1 Qq
E) 4 r 2
0
metal
Why?
Gauss’ law:
  Qenclosed
 E  dA 
0
E 4 r 
2
E
Q+q
0
1 Q+q
4 0 r 2
Electricity & Magnetism Lecture 6, Slide 19
Calculation: Quantitative Analysis
cross-section
a3 < r < a4 : What is E(r) Inside outer metal sphere?
+Q
a4
a3
a2
a1
1
A) 0
+q
metal
B)
q
4 0 r 2
1 q
D) 4 0 r 2
r
1 q
C) 2 0 r
1 Qq
E) 4 0 r 2
metal
Applying Gauss’ law, what is Qenclosed for red sphere shown?
A) q
B) q
C) 0
How is this possible?
q must be induced at r  a3 surface
3 
q
4 a32
charge at r  a4 surface  Q + q
4 
Q+q
4 a42
Electricity & Magnetism Lecture 6, Slide 20
Calculation: Quantitative Analysis
cross-section
a4
a3
+Q
Continue on in…
a2 < r < a3 :
a2
a1
a1 < r < a 2 :
+q
metal
E
r
r < a1 :
1
q
4 0 r 2
E 0
E
1
q
4 0 r 2
metal
To find V:
1) Choose r0 such that V(r0)  0 (usual: r0  infinity)
2) Integrate!
1 Q+q
V

r > a4 :
4  0 r
a3 < r < a4 : A)
V 0
B)
V
C)
V
Q+q
4  0 a4
1
Q+q
4  0 a3
1
Electricity & Magnetism Lecture 6, Slide 21
Calculation: Quantitative Analysis
cross-section
+Q
a4
a3
a2
a1
r > a4 :
V
Q+q
4  0 r
1
a3 < r < a4 : V  1 Q + q
4  0 a4
+q
metal
metal
a2 < r < a3 :
a1 < r < a2 :
V (r )  V (  a4 ) + 0 + V (a3  r )
V (r ) 
Q+q
q 1 1 
  
+0+
4 0 a4
4 0  r a3 
V (r ) 
1 Q+q q q 

+  
4 0  a4
a2 a3 
V (r ) 
1 Q+q q q 

+  
4 0  a4
r a3 
1 Q+q q q q q 

V
(
r
)

+  +  
0 < r < a1 :
4 0  a4
a2 a3 r a1 
Electricity & Magnetism Lecture 6, Slide 22