Your Comments
"Wow." <- Best part of the prelecture
Could you explain what exactly an equipotential is? And also how the del
operator illustrates the realationship between the electric potential and the
electric field?
Do we need to learn the spherical and cylindrical representations of the potential
gradient that were given in the prelecture? Because those formulas seem like they
could get very tricky.
i don't understand how you get the direction of the electric field from the graph of
the electric potential. the magnitude makes sense, that's just the slope, but
where does the direction come from...?
I just really want to get on the screen at the beginning of lecture...it's my goal in life
So what's the difference between electric potential energy and electric potential?
More examples and hopefully less homework this week!
05
But overall it seems easier than Gauss's law.
Physics 212 Lecture 6, Slide 1
Physics 212
Lecture 6
Today’s Concept:
Electric Potential
(Defined in terms of Path Integral of Electric Field)
Electricity & Magnetism Lecture 6, Slide 2
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 3
Electric Potential is like Height
(E points down hill for positive charge)
While I understand the logic of the math here, I'm having some trouble understanding
electric potential intuitively. Could you help explain what exactly the value shows us?
Physics 212 Lecture 6, Slide 4
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 5
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 6
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 7
CheckPoint 1a
At which point is the magnitude of the electric field
greatest?
“A) The The higher the potential the higher the electric field at that point “
“B) The slope of the line is the steepest at point B.”
“C) C has the greatest slope “
How do we get E from V?


E  V
V
Ex  
dx
Look at slopes!
Electricity & Magnetism Lecture 6, Slide 8
CheckPoint 1b
A
B
C
D
At which point is the electric field pointing in the
negative x direction?
“B) Negative slope of the line “
“C) There is a positive slope. “
How do we get E from V?


E  V
V
Ex  
dx
Look at slopes!
Electricity & Magnetism Lecture 6, Slide 9
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 10
CheckPoint 3a
A
B
C
D
At which point is the magnitude of the electric field
the smallest?
“The electric field lines are the least dense at D “
Electricity & Magnetism Lecture 6, Slide 11
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 12
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 13
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 14
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 15
Calculation for Potential
cross-section
a4
a3
+Q
a2
a1
+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 16
Calculation: Quantitative Analysis
cross-section
a4
a3
r > a4 : What is E(r)?
+Q
A) 0
a2
a1
+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 17
Calculation: Quantitative Analysis
cross-section
a4
a3
a3 < r < a4 : What is E(r)?
+Q
1
a2
a1
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 18
Calculation: Quantitative Analysis
cross-section
a4
a3
+Q
Continue on in…
a2 < r < a3 :
a2
a1
a1 < r < a2 :
+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 19
Calculation: Quantitative Analysis
cross-section
a4
a3
r > a4 :
+Q
V
Q+q
4  0 r
1
a3 < r < a4 : V  1 Q + q
a2
a1
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 20