PHY 184
Spring 2007
Lecture 11
Title: Capacitors
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Announcements
 Homework Set 3 is due Tuesday, Jan. 30 at 8:00
am.
 Homework Set 4 opened this morning.
 Today we will finish up the electric potential and
start with capacitors.
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Review - Electric Potential, V(x)
► If a charge q moves in an electric field…
U  U f  Ui  We
ΔV  ΔU/q
(potential energy)
(electric potential - definition )
► For reference V=0 at infinity…
V (x)  
 
E  ds

x
► E is the gradient of V…
E
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x
, Ey , Ez 
 V V V 
 
,
,

 x y z 
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Review - Electric Potential (2)
► For a point source Q …
kQ
V (r ) 
r
► For many sources …

V (x) 
n


Vi ( x )
(superposition)
i 1
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Problem-Solving Strategies
 Given a charge distribution,
calculate the electric field and
the electric potential:
(i) Use Gauss Law to derive the
electric field E:
(ii) For the potential (V=0 at
infinity) use
 For an arrangement of charges,
remember the superposition
principle! This holds for the
electric force, the electric field,
the electric potential and the
electric potential energy.
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184 Lecture 11
 
 0  E  dA  qenclosed
 


V ( x )    E  ds
x
Note:
 
a  b  ab cos 
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Capacitors
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Capacitors
 Capacitors are devices that store energy in an
electric field.
 Capacitors are used in many every-day applications
• Heart defibrillators
• Camera flash units
 Capacitors are an essential part of electronics.
• Capacitors can be micro-sized on computer chips
or super-sized for high power circuits such as
FM radio transmitters.
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Capacitance
 Capacitors come in a variety of sizes and
shapes.
 Concept: A capacitor consists of two
separated conductors, usually called
plates, even if these conductors are not
simple planes.
 We will define a simple geometry and
generalize from there.
 We will start with a capacitor
consisting of two parallel conducting
plates, each with area A separated by
a distance d .
 We assume that these plates are in
vacuum (air is very close to a vacuum).
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Parallel Plate Capacitor
q
q
 We charge the capacitor by placing
• a charge +q on the top plate
• a charge -q on the bottom plate
e.g., using a battery
 Because the plates are conductors, the charge will
distribute itself evenly over the surface of the conducting
plates.
 The electric potential, V, is proportional to the amount of
charge on the plates.
More precisely, potential difference V(+)-V(-) = V
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Parallel Plate Capacitor (2)
q
q




The proportionality constant between the charge q and
the electric potential difference V is the capacitance C.
We will call the electric potential difference V the
“potential” or the “voltage” across the plates.
The capacitance of a device depends on the area of the
plates and the distance between the plates, but does not
depend on the voltage across the plates or the charge on
the plates.
The capacitance of a device tells us how much charge is
required to produce a given voltage across the plates.
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q  CV
C q
V
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Clicker Question
q
q
 What is the NET CHARGE on the charged capacitor?
A: +q+(-q)=0
B: |+q|+|-q|=2q
C: q
D: none of the above
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Clicker Question
q
q
What is the NET CHARGE on the charged
capacitor?
A: +q+(-q)=0
Charges are added with their signs.
However, we refer to the charge of a capacitor
as being q (the charge of a capacitor is not the
net charge!)
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Definition of Capacitance
 The definition of capacitance is
q
C
V
 The units of capacitance are coulombs per volt.
 The unit of capacitance has been given the name
farad (abbreviated F) named after British physicist
Michael Faraday (1791 - 1867)
1C
1F 
1V
 A farad is a very large capacitance
• Typically we deal with F (10-6 F), nF (10-9 F),
or pF (10-12 F)
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Charging/Discharging a Capacitor
 We can charge a capacitor by connecting the capacitor to a battery or
to a DC power supply.
 A battery or DC power supply is designed to supply charge at a given
voltage.
 When we connect a capacitor to a battery, charge flows from the
battery until the capacitor is fully charged.
 If we then disconnect the battery or power supply, the capacitor will
retain its charge and voltage.
 A real-life capacitor will leak charge slowly, but here we will assume
ideal capacitors that hold their charge and voltage indefinitely.
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Charging/Discharging a Capacitor (2)
Illustrate the charging processing using a circuit diagram.
Lines represent conductors
The battery or power supply is represented by
The capacitor is represented by the symbol
This circuit has a switch (ab)
(open) When the switch is between positions a and b, the
circuit is open (not connected).
(pos a) When the switch is in position a, the battery is
connected across the capacitor. Fully charged, q = CV.
(pos b) When the switch is in position b, the two plates of the
capacitor are connected. Electrons will move around the
circuit – a current will flow -- and the capacitor will discharge.
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Demo: Big Spark
 Energy stored in this particular capacitor: 90 J
 This is equivalent to the kinetic energy of a
mass of 1 kg moving at a velocity of 13.4 m/s!
E  12 mv 2
2E
2  90 J
v

 13.4 m/s
m
1 kg
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Parallel plate capacitor –
-- a simple ideal model
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Parallel Plate Capacitor
 Consider two parallel conducting plates separated by a distance d
 This arrangement is called a parallel plate capacitor.
 The upper plate has +q and the lower plate has –q.
 The electric field between the plates points from the positively charged
plate to the negatively charged plate.
 We will assume ideal parallel plate capacitors in which the electric field
is constant between the plates and zero elsewhere.
 Real-life capacitors have fringe field near the edges.
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Parallel Plate Capacitor (2)
We can calculate the electric field
between the plates using Gauss’ Law
 
 0  E  dA  qenclosed
We take a Gaussian surface shown
by the red dashed line
Flux (through bottom surface of the top plate only!) = EA
Enclosed charge = q
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q
E
0A
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Parallel Plate Capacitor (3)
 Now we calculate the electric
potential across the plates of the
capacitor in terms of the electric
field.
 We define the electric potential
across the capacitor to be V and
we carry out the integral in the
direction of the blue arrow.
V  Vf  Vi   
f
i
 
E  ds
 E  ds cos(180)  E d
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Note: V = V(+)  V()
V = Ed
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Parallel Plate Capacitor (4)
q
C
V
Remember the definition of capacitance…
… so the capacitance of a parallel plate capacitor is
Variables:
A is the area of each plate
d is the distance between the plates
q
q
0A
C


Ed  q 
d

d
 0A 
Note that the capacitance depends only on the geometrical factors
and not on the amount of charge or the voltage across the
capacitor.
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C
0A
d
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Demo: Parallel Plate Capacitor
 The voltage of a charged capacitor as function of
the distance between the plates
q qd
V 
C 0A
 For constant charge, the voltage is proportional to
the distance between the plates
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Example - Capacitance of a Parallel Plate Capacitor


We have a parallel plate capacitor
constructed of two parallel plates,
each with area 625 cm2 separated by
a distance of 1.00 mm.
What is the capacitance of this
parallel plate capacitor?
0A
8.85  10-12 F/m  0.0625 m 2
C

d
0.001 m
 5.53  10 10 F
C = 0.553 nF
Result: A parallel plate capacitor constructed out of square
conducting plates 25 cm x 25 cm separated by 1 mm produces a
capacitor with a capacitance of about 0.5 nF.
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Example 2 - Capacitance of a Parallel Plate Capacitor


We have a parallel plate capacitor
constructed of two parallel plates
separated by a distance of 1.00 mm.
What area is required to produce a
capacitance of 1.00 F?
1 F  0.001 m
A

 0 8.85  10-12 F/m
Cd
 1.13  108 m 2
Result: A parallel plate capacitor constructed out of square
conducting plates 10.6 km x 10.6 km (6 miles x 6 miles) separated
by 1 mm produces a capacitor with a capacitance of 1 F.
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Example - Capacitance, Charge and Electrons …
 A storage capacitor on a random access memory
(RAM) chip has a capacitance of 55 nF. If the
capacitor is charged to 5.3 V, how many excess
electrons are on the negative plate?
Idea: We can find the number of excess electrons
on the negative plate if we know the total charge q on
the plate. Then, the number of electrons n=q/e,
where e is the electron charge in coulomb.
Second idea: The charge q of the plate is related to
the voltage V to which the capacitor is charged:
q=CV.
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