Download Capacitance and Dielectrics

Capacitance and Dielectrics
AP Physics C
Commercial Capacitor Designs
Section 16.10
Capacitors in Kodak Cameras
Capacitors can be easily purchased at a
local Radio Shack and are commonly
found in disposable Kodak Cameras.
When a voltage is applied to an empty
capacitor, current flows through the
capacitor and each side of the capacitor
becomes charged. The two sides have
equal and opposite charges. When the
capacitor is fully charged, the current
stops flowing. The collected charge is
then ready to be discharged and when
you press the flash it discharges very
quickly released it in the form of light.
Cylindrical Capacitor
Capacitance
In the picture below, the capacitor is symbolized by a set of parallel
lines. Once it's charged, the capacitor has the same voltage as
the battery (1.5 volts on the battery means 1.5 volts on the
capacitor) The difference between a capacitor and a battery is
that a capacitor can dump its entire charge in a tiny fraction of a
second, where a battery would take minutes to completely
discharge itself. That's why the electronic flash on a camera uses
a capacitor -- the battery charges up the flash's capacitor over
several seconds, and then the capacitor dumps the full charge
into the flash tube almost instantly
Measuring Capacitance
Let’s go back to thinking about plates!
V  Ed ,
The unit for capacitance is the FARAD, F.
V E , if d  constant
E Q Therefore
Q V
C  contant of proportion ality
C  Capacitanc e
Q  CV
Q
C
V
Electric Potential for Conducting Sheets
 E  dA 
qenc
Using Gauss’ Law we
V   E dr
derived and equation to
o
b 
define the electric field
Q
V (b)  V (a)   ( )dr
as we move radially
a 
EA 
o
o
away from the charged
a 
sheet or plate. Electric
V (b)  V (a)  ( )dr
Q
A
b 
Potential?
  , EA 
o


A

o

E
o
+
+
E =0
+
+
+
+
+

V (b)  V (a)  (a  b), a  b  d
o

Qd
V  d  Ed 
o
o A
Capacitance

Qd
V  d  Ed 
o
o A
This was derived from integrating the
Gauss’ Law expression for a
conducting plate.
o A
d
V  (
)Q  Q  (
)V
o A
d
Q  CV
C
o A
d
What this is saying is that
YOU CAN change the
capacitance even though it
represents a constant. That
CHANGE, however, can only
happen by physically
changing the GEOMETRY of
the capacitor itself.
These variables represent a
constant of proportionality
between voltage and charge.
That’s what C is !
Capacitor Geometry
The capacitance of a
capacitor depends on
HOW you make it.
1
C A C
d
A  area of plate
d  distance beteween plates
A
C
d
 o  constant of proportion ality
 o  vacuum permittivi ty constant
 o  8.85 x10
C
o A
d
12
C2
Nm 2
Capacitor Problems
What is the AREA of a 1F capacitor that has a plate
separation of 1 mm?
A
C  o
D
1  8.85 x10
A
Is this a practical capacitor to build?
NO! – How can you build this then?
12
A
0.001
1.13x108 m2
Sides 
10629 m
The answer lies in REDUCING the
AREA. But you must have a
CAPACITANCE of 1 F. How can
you keep the capacitance at 1 F
and reduce the Area at the same
time?
Add a DIELECTRIC!!!
Dielectric
Remember, the dielectric is an insulating material placed
between the conductors to help store the charge. In the
previous example we assumed there was NO dielectric and
thus a vacuum between the plates.
A
C  k o
d
k  dielectric constant
for a given material
All insulating materials have a dielectric
constant associated with it. Here now
you can reduce the AREA and use a
LARGE dielectric to establish the
capacitance at 1 F.
Dielectrics


A nonconducting
material is a
dielectric.
When a dielectric is
placed in an electric
field, the molecules of
the dielectric become
polarized, often by the
rotation of dipoles.
This produces a
surface charge on the
dielectric.
©2008 by W.H. Freeman and Company
Dielectrics in Capacitors

When a dielectric is
placed in a capacitor,
it weakens the electric
field.


The electric field
produced at the surface
of the dielectric opposes
the electric field of the
capacitor.
Think of it as some of the
capacitors E-field lines
ending on the surface of
the conductor.
©2008 by W.H. Freeman and Company
Dielectric Constant


The dielectric
constant relates the
strength of the electric
field without a
dielectric to the
strength with a
dielectric.
E0   E
E0  electric field without a dielectric
E  electric field with a dielectric
  dielectric constant
©2008 by W.H. Freeman and Company
Dielectrics in Capacitors



If the electric field is
less, the potential
across the capacitor
is less.
Since C = Q/V , when
V is reduced, the
capacitance
increases.
Dielectrics increase
capacitance.
©2008 by W.H. Freeman and Company
Dielectrics in Parallel Plate Capacitors
C
 0 A
d
©2008 by W.H. Freeman and Company
Capacitance of an Isolated Sphere
kQ
V
R
Q
V
Q R
C    40 R
V k
For a capacitor consisting of 2 concentric spheres go to
http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/capsph.html
Coaxial Capacitors


Consider two long
coaxial conducting
cylinders, with a
charge of +Q on the
inside cylinder and a
charge of –Q on the
outside cylinder.
Find the capacitance
of this capacitor.
©2008 by W.H. Freeman and Company
Coaxial Capacitors



Step 1: Find an
expression for the
electric field in the
region between the
cylinders.
Step 2: Use that
expression to find the
potential between the
cylinders.
Step 3: Use the
expression C=Q/V to
find the capacitance.
©2008 by W.H. Freeman and Company
Coaxial Capacitors

Step 1: Use Gauss’
Law to find the field in
between the
conductors.


For a long cylinder, the
field must point radially
outward, by symmetry.
The charge in the
Gaussian container is
given by
Q
Qinside  l  l
L
©2008 by W.H. Freeman and Company
Coaxial Capacitors

The field is
Q
E
2L 0 R

The potential is
R1
V    E dR
R2

Note that we integrate
from the negative charge
to the positive charge.
©2008 by W.H. Freeman and Company
Coaxial Capacitor


The capacitance is
C=Q/V
20 L
C
ln( R2 / R1 )
©2008 by W.H. Freeman and Company
Capacitor Practice Problem


A spherical capacitor consists of two
concentric spherical shells of radii R1 and R2,
where R2 > R1.
Show that the capacitance is given by
40 R1 R2
C
( R2  R1 )


Show that for thin capacitors, i.e.
R2 –
R1<<R, the formula is the same as the
parallel plate capacitor with A the area of one
sphere.
The online MIT movie will help you do this
Energy Storage
Potential Energy of a System of
Charges





Consider a system of two equal charges, q1
and q2.
Putting the first charge in place requires no
energy.
Putting the second charge requires q2V,
where V is the potential of the charges.
q2 is ½ the total charge Q, so the energy can
also be written
U  12 QV
Potential Energy of a System of
Charges

The potential energy of a system of charges
qi is given by
U
1
2
qV
i i
i
Energy Stored in a Capacitor

Consider a capacitor
that is charged one
small charge at a
time.


The energy stored in
the capacitor when a
small charge dq is
moved from – to + is V
dq.
V = q/C
©2008 by W.H. Freeman and Company
Energy Stored in a Capacitor

Consider a capacitor
that is charged one
small charge at a
time.


The first small charges
face little potential, so
they store little energy.
The charges moved at
the end face the full
potential of the
charged capacitor.
©2008 by W.H. Freeman and Company
Energy Stored in a Capacitor

dU  Vdq
q
C
V
q
V 
C
dU 
q
dq
C
Q
q
dq
C
0
U 
1 Q2
U
2 C
©2008 by W.H. Freeman and Company
Energy Stored in a Capacitor
1 Q2
U 
2 C
Q
V
C
1Q
1
U
Q  QV
2C
2
Q  CV
1 C 2V 2 1
U
 CV 2
2 C
2
©2008 by W.H. Freeman and Company
Energy Stored in a Capacitor

U  QV  CV
1
2
1
2
2
©2008 by W.H. Freeman and Company
Energy Stored in the Electric Field


Consider a parallel
plate capacitor.
The electric field E is
given by

Q
E

0 0 A

Thus the charge is
Q   0 EA
©2008 by W.H. Freeman and Company
Energy Stored in the Electric Field

Since the electric field
is constant, the
potential is simply
V  Ed

The energy stored in
the capacitor is
U  QV
1
2
This is the energy density
of that space
U  ( 0 EA) Ed   0 E Ad
1
2
1
2
2
©2008 by W.H. Freeman and Company
Ad is the volume of the space
inside the capacitor
Energy Stored in the Electric Field

The energy is stored
in the electric field.
Since the volume that
the electric field
occupies is A d, the
energy density is
given by
U
u  energy density 
volume
2
1
2  0 E Ad
u
 12  0 E 2
Ad
©2008 by W.H. Freeman and Company
Energy Density

The energy density
(energy per unit
volume) in an electric
field is given2 by
u  12  0 E


This result is
applicable to electric
fields in general.
If a dielectric is
2
2
1
1
u


E


E
0
2
2
present,
©2008 by W.H. Freeman and Company
Energy of a Capacitor is affected by adding
or removing a dielectric.





A dielectric is inserted between
the plates of a parallel-plate
capacitor.
The capacitor was connected to a
battery the entire time the
dielectric was inserted.
What happens to the capacitance
of the capacitor, the charge on
the plates, and the energy stored
in the capacitor?
What happens to those factors
when the dielectric is removed
What if a battery wasn’t
connected while the dielectric is
inserted or removed?
©2008 by W.H. Freeman and Company
There are 3 equations for the energy stored in a capacitor involving C, V
and Q. Choose an equation that has the 2 variables you know;
U  QV  CV
1
2
1
2
2
If you are connected to a battery
voltage will be fixed. Changing C by
inserting or removing a dielectric will
change Q, not V.
2
1Q
U
2 C
If you are disconnected from a battery total
charge will be fixed because charge can’t enter
or leave . Changing C by inserting or removing
a dielectric will change V, not q
©2008 by W.H. Freeman and Company
There are 3 equations for the energy stored in a capacitor involving C, V
and Q. Choose an equation that has the 2 variables you know.
If you are connected to a battery voltage will be fixed.
Changing C by inserting or removing a dielectric will change Q,
not V.
U  QV  CV
1
2
1
2
2
Connected to a battery, adding a
dielectric increases C and Q and energy
is gained. Removing it reduces C and Q,
and energy is lost inside the capacitor.
V remains the same
Disconnected from a battery, adding a dielectric increases C and energy is lost.
Removing it reduces C and energy is gained inside the capacitor. This energy
comes from the work you do pulling out the dielectric. Voltage goes up when you
pull it out.
2
1Q
U
2 C
Q remains the same
If you are disconnected from a battery total charge will be fixed
because charge can’t enter or leave . Changing C by inserting or
removing a dielectric will change V, not q
©2008 by W.H. Freeman and Company
Energy of a Capacitor Practice 2

A parallel plate capacitor
of area A and separation
d is charged to a potential
difference V and then
disconnected from the
voltage source. The
plates are plates are then
pulled apart to a distance
3d. How much work was
require to pull them
apart?
©2008 by W.H. Freeman and Company
Using MORE than 1 capacitor
Let’s say you decide that 1
capacitor will not be
enough to build what
you need to build. You
may need to use more
than 1. There are 2
basic ways to assemble
them together
 Series – One after
another
 Parallel – between a set
of junctions and parallel
to each other.
Capacitors in Series
Capacitors in series each charge each other by INDUCTION. So
they each have the SAME charge. The electric potential on the
other hand is divided up amongst them. In other words, the sum
of the individual voltages will equal the total voltage of the battery
or power source.
Capacitors in Parallel
In a parallel configuration, the voltage is the same
because ALL THREE capacitors touch BOTH ends
of the battery. As a result, they split up the charge
amongst them.
Circuit Reduction Example
Section 16.8
Finding V and Q

First find Ceq.

Then work
backwards,
remembering that


In parallel, voltages are
constant; charges add
up.
In series, voltages add
up, charges are
constant.
Remember;
Par-V
Seri-q
Capacitor Circuit Example 1
Remember what’s the
same for caps;
Find the total capacitance, then total charge (which is the same
on all caps in series)
1
1
1/C = 6 + 12
3
= 12
Seri-q
Par-V
𝑠𝑜 𝐶 = 4𝑢𝐹
𝑞 = 𝐶𝑉 = 4𝑢𝐹 𝑥 12𝑉 = 48 𝑚𝑖𝑐𝑟𝑜𝐶𝑜𝑢𝑙𝑜𝑚𝑏𝑠.
𝑇ℎ𝑒𝑦 𝑔𝑒𝑡 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 48 𝑢𝐶 but NOT the same voltage.
+48 𝑢𝐶 𝑑𝑖𝑟𝑒𝑐𝑡𝑙𝑦 𝑓𝑟𝑜𝑚 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 𝑟𝑒𝑎𝑐ℎ
𝑡ℎ𝑒 𝑡𝑜𝑝 𝑝𝑙𝑎𝑡𝑒

The voltage across
each capacitor is
then found by using
V =q/C.
𝒒
𝑪
=
𝟒𝟖𝒖𝑪
𝟔𝒖𝑭

𝑽𝒕𝒐𝒑 =

𝑽𝒃𝒐𝒕 = 𝑪 = 𝟏𝟐𝒖𝑭 = 𝟒V
𝒒
Induced opposite
charges appear on
the isolated middle
plates
= 𝟖𝑽
𝟒𝟖𝒖𝑪
−48 𝑢𝐶 𝑟𝑒𝑎𝑐ℎ 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 𝑝𝑙𝑎𝑡𝑒 𝑓𝑟𝑜𝑚 𝑏𝑎𝑡𝑡𝑒𝑟𝑦
An youtube example clip with 3 capacitors in series is at
https://www.youtube.com/watch?v=xMbXlhhrXf8
Capacitor Circuit Example 2
V=12μC/6μF=2V

This circuit is
connected to a 6 V
battery. Find the
charge on and
potential across each
capacitor.
Remember;
Seri-q
Para-V
6μF
1/6μF +1/3μF
=3/6 μF
Flip 2
μF
Qtot=Ctot Vtot= 2μF(6V)= 12μC
Both series elements in blue circle get 12 μC.
Now use V=C/q at each cap.
Finally in the red parallel branch, the 12μC is divided
unevenly by capacitance into 8 and 4 μC .
Qbottom =CV= 4 μF (2V) = 8 μC
Qtop =CV= 2 μF (2V) = 4 μC
V=12μC/3μF=4V
Example, Capacitors in Parallel and in Series:
Example, Capacitors in Parallel and in Series:
Example, Capacitors in Parallel and in Series:
Example, Capacitors in Parallel and in Series:
Example, Capacitors in Parallel and in Series:
Example, Capacitors in Parallel and in Series:
Example, Capacitors in Parallel and in Series:
Example, Capacitors in Parallel and in Series:
Example, Capacitors in Parallel and in Series:
Example, One Capacitor Charging up Another Capacitor: