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PHY 184
Spring 2007
Lecture 13
Title: Capacitors
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184 Lecture 13
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Notes
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Homework Set 3 is done!
Homework Set 4 is open and Set 5 opens Thursday morning
Midterm 1 will take place in class Thursday, February 8.
One 8.5 x 11 inch equation sheet (front and back) is allowed.
The exam will cover
Chapters 16 - 19
Homework Sets 1 - 4
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Review
 The capacitance of a spherical capacitor is
r1r2
C  4 0
r2  r1
• r1 is the radius of the inner sphere
• r2 is the radius of the outer sphere
 The capacitance of an isolated spherical conductor is
C  4 0 R
• R is the radius of the sphere
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Review (2)
 The equivalent capacitance for n capacitors in
parallel is
n
Ceq   Ci
i 1
 The equivalent capacitance for n capacitors in
series is
n
1
1

Ceq i 1 Ci
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Example - System of Capacitors
 Let’s analyze a system of five capacitors
 If each capacitor has a capacitance of 5 nF, what is
the capacitance of this system of capacitors?
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System of Capacitors (2)
 We can see that C1 and C2 are in parallel and that C3 is also
in parallel with C1 and C2
 We can define C123 = C1 + C2 + C3 = 15 nF
 … and make a new drawing
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System of Capacitors (3)
 We can see that C4 and C123 are in series
 We can define
1
C1234
1
1
C123C4


 C1234 
C123 C4
C123  C4
= 3.75 nF
 … and make a new drawing
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System of Capacitors (4)
 We can see that C5 and C1234 are in parallel
 We can define
C12345  C1234  C5 
C123C4
(C  C2  C3 )C4
 C5  1
 C5
C123  C4
C1  C2  C3  C4
= 8.75 nF
 And make a new drawing
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System of Capacitors (5)
 So the equivalent capacitance of our system of
capacitors
C12345
 (5  5  5)5


 5  nF  8.75 nF
5555

 More than one half of the total capacitance of this
arrangement is provided by C5 alone.
 This result makes it clear that one has to be
careful how one arranges capacitors in circuits.
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Clicker Question
 Find the equivalent capacitance Ceq
A)
B)
C)
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Clicker Question
 Find the equivalent capacitance Ceq
C)
First Step: C1 and C2 are in series
Second Step: C12 and C3 are in parallel
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A capacitor stores energy.
Field Theory:
The energy belongs to the electric field.
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Energy Stored in Capacitors
 A battery must do work to charge a capacitor.
 We can think of this work as changing the electric potential energy of
the capacitor.
 The differential work dW done by a battery with voltage V to put a
differential charge dq on a capacitor with capacitance C is
q
dW  Vdq  dq
C
 The total work required to bring the capacitor to its full charge q is
q
1 qt2
Wt   dW  
dq 
0 C
2C
 This work is stored as electric potential energy
1 q2 1
1
2
U
 CV  qV
2C 2
2
qt
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Energy Density in Capacitors
 We define the energy density, u, as the electric potential energy per
unit volume
U
u
volume
 Taking the ideal case of a parallel plate capacitor that has no fringe
field, the volume between the plates is the area of each plate times the
distance between the plates, Ad
2
2
1
U
CV
CV
u
 2

Ad
Ad
2Ad
 Inserting our formula for the capacitance of a parallel plate capacitor
we find
 0 A  2
2

 V
1 V
d
u
 0  
2Ad
2  d
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Energy Density in Capacitors (2)
 Recognizing that V/d is the magnitude of the electric field,
E, we obtain an expression for the electric potential energy
density for parallel plate capacitor
1
u  0E 2
2
 This result, which we derived for the parallel plate
capacitor, is in fact completely general.
 This equation holds for all electric fields produced in any
way
• The formula gives the quantity of electric field energy per unit
volume.
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Example - isolated conducting sphere
 An isolated conducting sphere whose radius R is 6.85 cm has a charge of
q=1.25 nC.
a) How much potential energy is stored in the electric field of the
charged conductor?
Key Idea: An isolated sphere has a capacitance of C=40R (see previous
lecture). The energy U stored in a capacitor depends on the charge and
the capacitance according to
… and substituting C=40R gives
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Example - isolated conducting sphere
 An isolated conducting sphere whose radius R is 6.85 cm has a
charge of q=1.25 nC.
b) What is the field energy density at the surface of the sphere?
Key Idea: The energy density u depends on the magnitude of the
electric field E according to
1
u  0E 2
2
so we must first find the E field at the surface of the sphere.
Recall:
(Why?)
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Example: Thundercloud
 Suppose a thundercloud with horizontal dimensions of 2.0 km by 3.0 km
hovers over a flat area, at an altitude of 500 m and carries a charge of
160 C.
 Question 1:
• What is the potential difference between
the cloud and the ground?
 Question 2:
• Knowing that lightning strikes require
electric field strengths of approximately
2.5 MV/m, are these conditions sufficient
for a lightning strike?
 Question 3:
• What is the total electrical energy contained in this cloud?
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Example: Thundercloud (2)
 Question 1
 We can approximate the cloud-ground system as a parallel plate
capacitor whose capacitance is
0 A
(8.85·10-12 F/m)(2000 m)(3000 m)
C

 0.11 F
d
500 m
 The charge carried by the cloud is 160 C, which means that the “plate
surface” facing the earth has a charge of 80 C
V
q
80 C

 7.2 10 8 V
C 0.11 F
…++++++++++++ …
++++++++++++
 720 million volts
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Example: Thundercloud (3)
 Question 2
 We know the potential difference between the cloud and ground so we
can calculate the electric field
V 7.2 10 8 V
E 
 1.5 MV/m
d
500 m
 E is lower than 2.5 MV/m, so no lightning cloud to ground
• May have lightning to radio tower or tree….
 Question 3
 The total energy stored in a parallel place capacitor is
1
U  qV  0.5(80 C)(7.2 10 8 V)  2.9 1010 J
2
• Enough energy to run a 1500 W hair dryer for more than 5000 hours
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Clicker Question
 A 1.0 F capacitor and a 3.0 F capacitor are connected in
parallel across a 500 V potential difference V. What is the
total energy stored in the capacitors?
A) U=0.5 J
B) U=0.27 J
C) U=1.5 J
D) U=0.02 J
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Hint: Use
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Clicker Question
 A 1.0 F capacitor and a 3.0 F capacitor are connected in
parallel across a 500 V potential difference V. What is the
total energy stored in the capacitors?
A) U=0.5 J
Ceq  4.0  F
q2 1 2
U
 2 CV  0.5  4  F  500 V 2
2C
U = 0.5 J
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Capacitors with Dielectrics
 We have only discussed capacitors with air or vacuum
between the plates.
 However, most real-life capacitors have an insulating
material, called a dielectric, between the two plates.
 The dielectric serves several purposes:
• Provides a convenient way to maintain mechanical
separation between the plates.
• Provides electrical insulation between the plates.
• Allows the capacitor to hold a higher voltage
• Increases the capacitance of the capacitor
• Takes advantage of the molecular structure of the dielectric
material
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Capacitors with Dielectrics (2)
 Placing a dielectric between the plates of a capacitor
increases the capacitance of the capacitor by a numerical
factor called the dielectric constant, 
 We can express the capacitance of a capacitor with a
dielectric with dielectric constant  between the plates as
C   Cair
 … where Cair is the capacitance of the capacitor without the
dielectric
 Placing the dielectric between the plates of the capacitor
has the effect of lowering the electric field between the
plates and allowing more charge to be stored in the
capacitor.
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Parallel Plate Capacitor with Dielectric
 Placing a dielectric between the
plates of a parallel plate capacitor
modifies the electric field as
Eair
q
q
E



 0 A  A
 The constant 0 is the electric
permittivity of free space
 The constant  is the electric
permittivity of the dielectric
material
   0
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