Transcript W05D1_Conductors and Insulators_mac_v03_jwb

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W05D1
Conductors and Insulators
Capacitance & Capacitors
Energy Stored in Capacitors
W05D1 Reading Assignment Course
Notes: Sections 3.3, 4.5, 5.1-5.4, 5.6, 5.8
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Announcements
No Math Review this week
PS 4 due W05 Tuesday at 9 pm in boxes outside 32082 or 26-152
W05D2 Reading Assignment
Course Notes: Sections 5.4, 5.6, 5.8-5.9
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Outline
Conductors and Insulators
Conductors as Shields
Capacitance & Capacitors
Energy Stored in Capacitors
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Conductors and Insulators
Conductor: Charges are free to move
Electrons weakly bound to atoms
Example: metals
Insulator: Charges are NOT free to move
Electrons strongly bound to atoms
Examples: plastic, paper, wood
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Charge Distribution and Conductors
The Charged Metal Slab Applet: Non-zero charge
placed in metal slab
http://web.mit.edu/viz/EM/visualizations/electrostatics/CapacitorsAndCondcutors/chargedmetalslab/chargedmetalslab.htm
Charges move to surface
(move as far apart as possible)
Electric field perpendicular
to surface, zero inside slab
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Induced Charge Distribution in
External Electric Field
Charging by Induction, Exterior of a Neutral Metallic
Box
http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/chargebyinductionBox/chargebyinductionBox.htm
Induced charges move
to surface
Electric field perpendicular
to surface, zero inside slab
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Conductors in Equilibrium
Conductor Placed in External Electric Field
1)E = 0 inside
2) E perpendicular to surface
3) Induced surface charge distribution
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Hollow Conductors: Applet
Charge placed OUTSIDE induces charge separation
ON OUTSIDE. Electric field is zero inside.
http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/shielding/shielding.htm
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Electric Field on Surface of Conductor
1) E perpendicular to surface
2) Excess charge on surface
Apply Gauss’s Law
A

Esurface A 
 Esurface 
0
0
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Conductors are Equipotential
Surfaces
1) Conductors are equipotential objects
2) E perpendicular to surface
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Group Problem: Metal Spheres
Connected by a Wire
Two conducting spheres 1 and 2 with radii r1 and
r2 are connected by a thin wire. What is the ratio
of the charges q1/q2 on the surfaces of the
spheres? You may assume that the spheres are
very far apart so that the charge distributions on
the spheres are uniform.
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Concept Question: Point Charge in Conductor
A point charge +q is
placed inside a hollow
cavity of a conductor that
carries a net charge +Q.
What is the total charge on
the outer surface of the
conductor?
1. Q.
2. Q + q.
3. q.
4. Q - q.
5. Zero.
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Concept Q. Ans.: Point Charge in Conductor
Answer 2. Choose
Gaussian surface inside
conductor. Electric field is
zero on Gaussian surface
so flux is zero. Therefore
charged enclosed is zero.
So an induced charge –q
appears on cavity surface.
Hence an additional
charge of +q appears on
outer surface giving a total
charge of Q + q on outer
surface.
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Hollow Conductors: Applet
Charge placed INSIDE induces balancing charge ON
INSIDE. Electric field outside is field of point charge.
http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/shielding/shielding.htm
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Capacitors and Capacitance
Our first of 3 standard electronics devices
(Capacitors, Resistors & Inductors)
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Capacitors: Store Electric Charge
Capacitor: Two isolated conductors
Equal and opposite charges ±Q
Potential difference V between them.
Q
C
V
Units: Coulombs/Volt or
Farads
C is Always Positive
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Parallel Plate Capacitor; Applet
Oppositely charged plates:
Charges move to inner surfaces
Electric field perpendicular to
surface, zero inside plates
http://web.mit.edu/viz/EM/visualizations/electrostatics/CapacitorsAndCondcutors/capacitor/capacitor.htm
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Calculating E (Gauss’s Law)
qin
 E  dA  
S
0
 AGauss
E AGauss 
0

Q
E 
 0 A 0
Note: We only “consider” a single sheet!
Doesn’t the other sheet matter?
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Superposition Principle
Between the plates:
E  E  E  
 ˆ  ˆ

j
j   ˆj
2 0
2 0
0
Above the plates:
E  E  E  
 ˆ  ˆ
j
j0
2 0
2 0
Below plates:
 ˆ  ˆ
E  E  E  
j
j0
2 0
2 0
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Parallel Plate Capacitor
top
Q
V    E  dS  Ed 
d
A 0
bottom
0 A
Q
C

d
V
C depends only on geometric factors A and d
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Group Problem: Spherical Shells
A spherical
conductor of radius
a carries a charge
+Q. A second thin
conducting
spherical shell of
radius b carries a
charge –Q.
Calculate the
capacitance.
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Concept Question: Isolated
Spherical Conductor
What is the capacitance of an isolated spherical
conductor of radius a?
1.Capacitance is not well defined.
2.Capacitance is 4 0 a.
3.Capacitance is infinite.
4.Capacitance is zero.
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Concept Q. Ans. Isolated Spherical
Conductor
Answer 2. Capacitance is 4 0 a
Other equipotential surface
(second conducting surface)
is located at infinity. In
previous calculation set
.
b
4 0
4 0
Q
C


 4 0 a
 1 1  1 1 
V
 a  b   a   
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Capacitance of Earth
For an isolated spherical conductor of radius a:
C  40 a
 0  8.85 10
12
Fm
a  6.4 10 m
6
4
C  7 10 F  0.7mF
A Farad is REALLY BIG! We usually use pF (10-12) or nF (10-9)
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Energy To Charge Capacitor
+q
-q
1. Capacitor starts uncharged.
2. Carry +dq from bottom to top.
Now top has charge q = +dq, bottom -dq
3. Repeat
4. Finish when top has charge q = +Q, bottom -Q
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Stored Energy in Charging Capacitor
At some point top plate has +q, bottom has –q
Potential difference is V = q / C
Change in stored energy done lifting another dq
is dU = dq V
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Stored Energy in Charging Capacitor
So change in stored energy to move dq is:
q 1
dU  dqV  dq  q dq
C C
Total energy to charge to Q
Q
2
1
1Q
U   dU   q dq 
C0
C 2
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Energy Stored in Capacitor
Q
Since C 
V
2
Q
1
1
U
 Q V  C V
2C 2
2
2
Where is the energy stored???
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Energy Stored in Capacitor
Energy stored in the E field!
C
Parallel-plate capacitor:
o A
d
and V  Ed
2
2

A

E
1
1 o
2
U  CV 
Ed   o  ( Ad )  u E  (volume)

2
2 d
2
Energy density [J/m3]
uE 
o E
2
2
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Demonstration:
Changing Distance Between
Circular Capacitor Plates E4
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=E%204&show=0
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Concept Question: Changing Dimensions
A parallel-plate capacitor is charged until the plates have equal
and opposite charges ±Q, separated by a distance d, and then
disconnected from the charging source (battery). The plates
are pulled apart to a distance D > d. What happens to the
magnitude of the potential difference V and charge Q?
1.
2.
3.
4.
5.
6.
7.
8.
9.
V, Q increases.
V increases, Q is the same.
V increases, Q decreases.
V is the same, Q increases.
V is the same, Q is the same.
V is the same, Q decreases.
V decreases, Q increases.
V decreases, Q is the same.
V decreases, Q decreases.
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Concept Q. Answer: Changing Dimensions
Answer: 2. V increases, Q is the same
With no battery connected to the plates the charge on
them has no possibility of changing.
In this situation, the electric field doesn’t change when
you change the distance between the plates, so:
V=Ed
As d increases, V increases.
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Concept Question: Changing Dimensions
A parallel-plate capacitor is charged until the plates have equal
and opposite charges ±Q, separated by a distance d. While
still connected to the charging source, the plates are pulled
apart to a distance D > d. What happens to the magnitude of
the potential difference V and charge Q?
1. V, Q increases.
2. V increases, Q is the same.
3. V increases, Q decreases.
4. V is the same, Q increases.
5. V is the same, Q is the same.
6. V is the same, Q decreases.
7. V decreases, Q increases.
8. V decreases, Q is the same.
9. V decreases, Q decreases.
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Concept Q. Answer: Changing Dimensions
Answer: 6. V is the same, Q decreases
With a charging source (battery) connected to the plates
the potential V between them is held constant
In this situation, since
V=Ed
As d increases, E must decrease.
Since the electric field is proportional to the charge on the
plates, Q must decrease as well.
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Concept Question: Changing Dimensions
A parallel-plate capacitor, disconnected from a battery,
has plates with equal and opposite charges, separated
by a distance d.
Suppose the plates are pulled apart until separated by
a distance D > d.
How does the final electrostatic energy stored in the
capacitor compare to the initial energy?
1. The final stored energy is smaller
2. The final stored energy is larger
3. Stored energy does not change.
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Concept Q. Answer: Changing Dimensions
Answer: 2. The stored energy increases
As you pull apart the capacitor plates you increase
the amount of space in which the E field is non-zero
and hence increase the stored energy. Where
does the extra energy come from? From the work
you do pulling the plates apart.
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Demonstration: Charging Up a Capacitor
A 100 microfarad
oil-filled capacitor is
charged to 4 KV
and discharged
through a wire
Stored Energy:
1
1
2
4
3
2
U  CV  (1  10 F)(4  10 V)  800 J
2
2
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=E%206&show=0
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