Physics 212 Lecture 6, Slide 1

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Physics 212 Lecture 6
Today's Concept Electric Potential Defined in
terms of Path Integral of Electric Field
2
How confident are you in your understanding of
the concepts 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 itE  I
understand everything
3
Things you wanted to hear more about
How electric potential is determined when there
is an insulator vs a conductor?
The idea behind setting the zero of potential at
infinity.
How a hollow sphere would affect the potential
energy field.
More problems solved fully like in the previous
lecture . That was great.
More problems concerning a conducting sphere,
solid and hollow, and the electric potential.
It would be good to go over most of these
concepts in lecture.
The integral stuff concerning electric potential
fields. Gausses Law.
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4
Preflight 8
EQUIPOTENTIAL SIMULATION
08
5
Preflight 10
The electric field does positive work on the
charge. Thus, the electric potential energy must
decrease. It is moved along the electric field,
which is the direction in which it wants to move.
Therefore the potential energy decreases.
12
6
Potential due to point charges
Single Charge Q
Many charges Qi use superposition
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Problem Suppose you have a solid conducting
sphere of radius a carrying a net charge Q. Find
E everywhere.
Important Facts
Q
1) Charges in a conductor will move if there is
an E field.
a
2) The E field in a conductor is always ZERO.
3) Gauss law says that if E 0 everywhere inside
the sphere then all points in the sphere are
neutral.
4) All of the extra charge must be uniformly
spread on the outside surface of the sphere.
Simulation
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8
Problem Suppose you have a conducting spherical
shell of radius a carrying a net charge Q. Find
E everywhere.
Important Facts
Q
1) Charges in a conductor will move if there is
an E field.
2) Just like in the case of the the solid sphere,
all of the extra charge must be uniformly spread
on the outside surface of the sphere.
3) The E field is identical to that of the solid
sphere everywhere.
Simulation
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Preflight 2
Consider a solid conducting sphere of radius a.
Which of the following graphs best describes the
magnitude of the electric field as a function of
distance r from the center of the sphere?
A
C
B
(Notice that they are all identical for r gt a)
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10
Preflight 2
Consider a solid conducting sphere of radius a.
Which of the following graphs best describes the
magnitude of the electric field as a function of
distance r from the center of the sphere?
Correct
the electric field inside of a conductor is
always zero. Therefore, there is not going to be
a field until the radius is greater than a
Common Misconception
The field increases as you get closer to the
outside surface where charge is stored and then
drops off as you get further away.
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Gauss Law
For r gt a
Q
a
r
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12
Problem Suppose you have a solid conducting
sphere of radius a carrying a net charge Q. Find
V everywhere.
Important Facts
Q
1) V can be determined once we know E
a
We know E everywhere, so we can find V everywhere
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Q
Lets find the change in V going from r 0 to r
a
a
0
E 0
DV 0 between any two points in a conductor
V does not change inside a conductor
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Preflight 4
Consider a solid conducting sphere of radius a.
Which of the following graphs best describes the
electric potential as a function of distance r
from the center of the sphere?
A
C
B
(Notice that they are all identical for r gt a)
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15
Preflight 4
Correct
E0 inside the conductor, so the electric
potential is the same.
Common Misconception
This is the model in the prelecture and the only
one that makes any sense.
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Preflight Solid Conductor Charge all on surface
Prelecture Solid Insulator Charge spread
throughout solid
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17
Now lets find the change in V going from r a to
r gt a
recall for rgta
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18
If V(a) 0 this looks like
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Now lets find the change in V going from r 8 to
r gt a
0
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20
If V(8) 0 this looks like
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When V is chosen to be 0 at r 8
When V is chosen to be 0 at r 0
Same exact curve just shifted
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22
Preflight 6
conducting shell
solid conductor
Correct
Many people picked 1 3
In both the cases, the charge resides only on the
surface. therefore, it does not matter if it is a
shell or a sphere.
Common Misconception
the solid sphere has the charge evenly
distributed inside it,
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23
Recall a few slides ago
Q
The E field is identical to that of the solid
sphere everywhere. Therefore V is also identical
Simulation
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Homework Problem
cross-section
a4
a3
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?

a2
a1
q
metal
metal

• Conceptual Analysis
• Charges q and Q will create an E field throughout
  space

• Strategic Analysis
• Spherical symmetry Use Gauss Law to calculate E
  everywhere
• Integrate E to get V

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Homework Problem Quantitative Analysis
cross-section
a4
a3
Q

• r gt a4 What is E(r)?
• (A) 0 (B) (C)

a2
a1
q
(D) (E)
metal
metal
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Homework Problem Quantitative Analysis
cross-section
a4
a3
Q

• a3 lt r lt a4 What is E(r)?
• (A) 0 (B) (C)

a2
a1
q
(D) (E)
metal
metal
How is this possible??? -q must be
induced at ra3 surface
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Homework Problem Quantitative Analysis
cross-section
a4
a3
Q

• Continue on in.
• a2 lt r lt a3

a2
a1
q
metal
metal

• To find V
• Choose r0 such that V(r0) 0 (usual r0
  infinity)
• Integrate !!

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Homework Problem Quantitative Analysis
cross-section
a4
a3
r gt a4
Q
a2
a1
a3 lt r lt a4
q
metal
metal