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A set of slides from lecture 6 of physics 212, which covers the topic of electric potential. The lecture defines electric potential in terms of a path integral of the electric field and discusses the concept in the context of conducting spheres. The slides include problems and simulations to help illustrate the concepts.
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Physics 212 Lecture 6, Slide Physics 212 Lecture 6, Slide
Defined in terms of Path Integral of Electric Field Defined in terms of Path Integral of Electric Field
Physics 212 Lecture 6, Slide Physics 212 Lecture 6, Slide
2
How confident are you in your understanding of theconcepts presented in the prelecture? A
I am confused by all of it.
B
I understand a little but I am confused by most of it.
C
I understand some parts and I am confused by other parts.
D I understand most of it E
I understand everything
0
60 50 40 30 20 10
Physics 212 Lecture 6, Slide Physics 212 Lecture 6, Slide
4
0
80 60 40 20
100
Its movement is perpendicular to thefield, so no potential change occurs.
dy
E dy
i
EQUIPOTENTIAL SIMULATION EQUIPOTENTIAL SIMULATION
08
Physics 212 Lecture 6, Slide Physics 212 Lecture 6, Slide
5
The electric field does positive work on theThe electric field does positive work on thecharge. Thus, the electric potential energy charge. Thus, the electric potential energymust decrease. must decrease.It is moved along the electric field, which is It is moved along the electric field, which is the direction in which it wants to move.the direction in which it wants to move.Therefore the potential energy decreases. Therefore the potential energy decreases.
0
70 60 50 40 30 20 10
Work(AC) = ELcos
θ
0
L
θ
D D
Work(ADC) = 0 + E
Δ
x > 0
Δ Δ
x x
12
Physics 212 Lecture 6, Slide Physics 212 Lecture 6, Slide
8
Problem Problem
: :
Suppose you have a Suppose you have a
conducting spherical shell conducting spherical shell
of radius of radius
aa
carrying carrying
a net charge a net charge
Q Q
. Find . Find
E E
everywhere. everywhere.
will move will move
if there isif there is
an E field. an E field.
**3) The E field is identical to that of the solid
sphere sphere
everywhere everywhere
. .
Simulation
Physics 212 Lecture 6, Slide Physics 212 Lecture 6, Slide
9
Consider a solid conducting sphere of radius a. Which of the fol Consider a solid conducting sphere of radius a. Which of the fol
lowing graphs best describeslowing graphs best describes
the magnitude of the electric field as a function of distance r the magnitude of the electric field as a function of distance r
from the center of the sphere?from the center of the sphere?
(Notice that they are all identical for r > a) (Notice that they are all identical for r > a)
17
Physics 212 Lecture 6, Slide Physics 212 Lecture 6, Slide
11
a
r
Gauss Law: Gauss Law:
2
0
r
ε
∫
v
0
enclosed
2
0
r
For r > a For r > a
20
Physics 212 Lecture 6, SlidePhysics 212 Lecture 6, Slide
12
Problem Problem
: :
Suppose you have a solid conducting sphere of radius Suppose you have a solid conducting sphere of radius
aa
carrying carrying
a net charge a net charge
Q Q
. Find . Find
V V
everywhere. everywhere.
a
G ∫
G
2 1
2
1
r
r
V r
V r
E dr
We know E everywhere,We know E everywhere,so we can find V everywhere so we can find V everywhere
Physics 212 Lecture 6, SlidePhysics 212 Lecture 6, Slide
14
(Notice that they are all identical for r > a) (Notice that they are all identical for r > a)
Consider a solid conducting sphere of radius Consider a solid conducting sphere of radius
aa
. Which of the following graphs best describes . Which of the following graphs best describes
the electric potential as a function of distance r from the centthe electric potential as a function of distance r from the cent
er of the sphere?er of the sphere?
26
Physics 212 Lecture 6, SlidePhysics 212 Lecture 6, Slide
15
0
50 40 30 20 10
This is the model in the prelecture and the This is the model in the prelecture and the
only one that makes any sense. only one that makes any sense.
Correct Correct
E=0 inside the conductor, so the electric E=0 inside the conductor, so the electric
potential is the same. potential is the same.
Common Misconception:Common Misconception:
Physics 212 Lecture 6, SlidePhysics 212 Lecture 6, Slide
17
Now lets find the change in Now lets find the change in
V V
going fromgoing from
r = ar = a
to to
r > ar > a
: :
∫
r a
V r
V a
E dr
a
r
2
0
E r
r
∫
2
0
0
r a
dr
V r
V a
r
r
a
30
recall for r>a: recall for r>a:
Physics 212 Lecture 6, SlidePhysics 212 Lecture 6, Slide
18
0
V r
V a
a
r
If V(a) = 0 this looks like:
Physics 212 Lecture 6, SlidePhysics 212 Lecture 6, Slide
If V(
) = 0 this looks like:
0
V r
r
Physics 212 Lecture 6, SlidePhysics 212 Lecture 6, Slide
21
When V is chosen to be 0 at r = 0
When V is chosen to be 0 at r =
∞
Same exact curve Same exact curve
- -
just shiftedjust shifted
33
