Capacitors

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Exam One Regrade Policy
Exam One will be handed back at the end of class
on W05D1 Mon and Tuesday. Students can take the
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W05D1Conductors and InsulatorsCapacitance
CapacitorsEnergy 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 32-082 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/electrost
  atics/CapacitorsAndCondcutors/chargedmetalslab/cha
  rgedmetalslab.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/electrost
  atics/ChargingByInduction/chargebyinductionBox/cha
  rgebyinductionBox.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
• 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/electrost
atics/ChargingByInduction/shielding/shielding.htm
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Electric Field on Surface of Conductor

• E perpendicular to surface
• 2) Excess charge on surface
• Apply Gausss Law

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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/electrost
atics/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
between them.
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/electrost
atics/CapacitorsAndCondcutors/capacitor/capacitor.
htm
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Calculating E (Gausss Law)
Note We only consider a single sheet!
Doesnt the other sheet matter?
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Superposition Principle
Between the plates Above the plates Below
plates
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Parallel Plate Capacitor
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?
• Capacitance is not well defined.
• Capacitance is .
• Capacitance is infinite.
• Capacitance is zero.

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Concept Q. Ans. Isolated Spherical Conductor

• Answer 2. Capacitance is Other
  equipotential surface (second conducting surface)
  is located at infinity. In previous calculation
  set .

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Capacitance of Earth
For an isolated spherical conductor of radius a
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
Total energy to charge to Q
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Energy Stored in Capacitor
Since
Where is the energy stored???
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Energy Stored in Capacitor
Energy stored in the E field!
Parallel-plate capacitor
Energy density J/m3
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DemonstrationChanging Distance Between
Circular Capacitor Plates E4
http//tsgphysics.mit.edu/front/?pagedemo.phplet
numE204show0
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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 gt 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 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 doesnt
  change when you change the distance between the
  plates, so
• V E d
• 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 gt 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 E d
• 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 gt 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 wireStored
Energy
http//tsgphysics.mit.edu/front/?pagedemo.phplet
numE206show0