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Announcements
 Exam 1 is Tuesday, February 16, 5:00-6:00 pm.
I need paperwork/notification for any special needs by the
end of lectures today.
 If possible, Test Preparation Homework for Exam 1 will be
available on the Physics 2135 web site (under Handouts) late
Friday (otherwise Sunday). It will be handed out in lecture next
Monday. This is next Tuesday’s homework. Do not ignore it!
 Test 1 Room Assignments (next slide) are available on the
Physics 2135 web site (under Course Information). This
information will be included with Test Preparation Homework 1.
 Physics 2135 Test Rooms, Spring 2016:
Instructor
Dr. Hale
Dr. Kurter
Dr. Madison
Mr. Noble
Dr. Parris
Mr. Upshaw
Sections
G, J
D, F
H, N
B, E
K, M
A, C, L
Special Accommodations
Exam is from
5:00-6:00 pm!
Room
125 BCH
104 Physics
199 Toomey
B-10 Bert.
G-31 EECH
G-3 Schrenk
Testing Center
Know the exam time!
Find your room ahead of time!
If at 5:00 on test day you are lost, go to 104 Physics and check the exam
room schedule, then go to the appropriate room and take the exam there.
More Announcements
 Exam 1 special arrangements:
Test Center students. The Test Center e-mails you confirming your
appointment. The Testing Center will tell you in that email when
your exam starts, and no one more than 15 minutes late will be
admitted, unless you have made other arrangements.
If you have not received a confirming e-mail, you are NOT on the Test
Center list!
Other special cases: you should already have been in correspondence
with me or your recitation instructor.
Anybody else must let me know by the end of the 1:00 lecture today
about special needs for the exam. But there is no longer a 1:00 lecture so let me know by 2:00.
Today’s agenda:
Capacitance.
You must be able to apply the equation C=Q/V.
Capacitors: parallel plate, cylindrical, spherical.
You must be able to calculate the capacitance of capacitors having these geometries, and
you must be able to use the equation C=Q/V to calculate parameters of capacitors.
Circuits containing capacitors in series and parallel.
You must understand the differences between, and be able to calculate the “equivalent
capacitance” of, capacitors connected in series and parallel.
Capacitors and Dielectrics
Capacitance
A capacitor is basically two parallel
conducting plates with air or insulating
material in between.
E
V0
A capacitor doesn’t
have to look like
metal plates.
L
Capacitor for use in
high-performance
audio systems.
V1
The symbol representing a capacitor in an
electric circuit looks like parallel plates.
Here’s the symbol for a battery, or an external
potential.
+-
When a capacitor is connected to an external potential,
charges flow onto the plates and create a potential difference
between the plates.
Capacitor plates
build up charge.
-+ +V
-
The battery in this circuit has some voltage V. We haven’t discussed what
that means yet.
If the external potential is
disconnected, charges remain on the
plates, so capacitors are good for
storing charge (and energy).
+conducting wires
+V
Capacitors are also very good at releasing
their stored charge all at once. The capacitors
in your tube-type TV are so good at storing
energy that touching the two terminals at the
same time can be fatal, even though the TV
may not have been used for months.
High-voltage TV capacitors are supposed to have “bleeder
resistors” that drain the charge away after the circuit is
turned off. I wouldn’t bet my life on it.
On-line “toy” here.
Graphic from http://www.feebleminds-gifs.com/.
assortment of
capacitors
+Q -Q
C
Here’s this V thing again.
It is the potential
difference provided by the
“external potential.” For
example, the voltage of a
battery. V is really a V.
+ V
The magnitude of charge acquired by each plate of a capacitor
is Q=CV where C is the capacitance of the capacitor.
Q
C
V
C is always positive.
V is really
V.
The unit of C is the farad but most capacitors have values
of C ranging from picofarads to microfarads (pF to F).
micro 10-6, nano 10-9, pico 10-12
(Know for exam!)
Today’s agenda:
Capacitance.
You must be able to apply the equation C=Q/V.
Capacitors: parallel plate, cylindrical, spherical.
You must be able to calculate the capacitance of capacitors having these geometries, and
you must be able to use the equation C=Q/V to calculate parameters of capacitors.
Circuits containing capacitors in series and parallel.
You must understand the differences between, and be able to calculate the “equivalent
capacitance” of, capacitors connected in series and parallel.
Parallel Plate Capacitance
-Q
We previously calculated the electric field
between two parallel charged plates:
+Q
E

Q
E 
.
0 0 A
This is valid when the separation is small
compared with the plate dimensions.
V0
d
V1
We also showed that E and V are related:
d
d
0
0
V   E  d  E  dx  Ed .
0 A
Q
Q
Q
This lets us calculate C for C 



V Ed  Q 
d
a parallel plate capacitor.

d
 0 A 
A
Reminders:
Q
C
V
Q is the magnitude of the charge on either plate.
V is actually the magnitude of the potential difference
between the plates. V is really |V|. Your book calls it
Vab.
C is always positive.
Parallel plate capacitance depends “only”
on geometry.
-Q
+Q
0 A
C
d
This expression is approximate, and must
be modified if the plates are small, or
separated by a medium other than a
vacuum (lecture 9).
 0 A
C
d
Greek letter Kappa. For
today’s lecture (and for
exam 1), use Kappa=1.
Do not use =9x109!
Because it isn’t!
E
V0
d
V1
A
Coaxial Cylinder Capacitance
We can also calculate the capacitance of a
cylindrical capacitor (made of coaxial
cylinders).
The next slide shows a cross-section view of
the cylinders.

L
b
b
a
a
ΔV = Vb - Va = -  E  d = -  E r dr
Gaussian
surface
b
dr
b
ΔV = - 2k λ  = - 2k λ ln  
r
a
a
b
2kλ
E=
r
r
a
Q
Q
λL
C=
=
=
ΔV
ΔV
λL
b
2k λ ln  
a
2πε 0 L
L
C=
=
b
b
2k ln  
ln  
a
a
E
dl
-Q
This derivation is sometimes needed
for homework problems! (Hint: 24.10, 11, 12.)
Some necessary details are not shown
on this slide! See lectures 4 and 6.
C
Lowercase c is capacitance per unit length: c = =
L
2πε 0
b
ln  
a
This has not been one of those “useless
physics problems that my professor tries to
confuse me with.”
If you are purchasing or specifying coaxial
cables, capacitance per length is a critical part
of the specifications.
example

L
Isolated Sphere Capacitance
An isolated sphere can be thought of as concentric spheres
with the outer sphere at an infinite distance and zero potential.
We already know the potential outside a conducting sphere:
Q
V
.
4 0 r
The potential at the surface of a charged sphere of radius R is
Q
V
4 0 R
so the capacitance at the surface of an isolated sphere is
Q
C
 4 0 R.
V
Capacitance of Concentric Spheres
If you have to calculate the capacitance of a concentric
spherical capacitor of charge Q…
In between the spheres (Gauss’ Law)
Q
E
4 0 r 2
Q
V 
40
b
a

b
a
dr
Q

2
r
40
40
Q
C

V
1 1 
 a  b 
1 1 
 a  b 
+Q
-Q
You need to do this derivation if you have a
problem on spherical capacitors! (not this semester)
If there is spherical capacitor homework, details will be provided in lecture!
Example: calculate the capacitance of a capacitor whose plates
are 20 cm x 3 cm and are separated by a 1.0 mm air gap.
0 A
C
d
C
12
8.85

10

  0.2  0.03
0.001
C  53 1012 F
d = 0.001
area = 0.2 x 0.03
C  53 pF
If you keep everything in SI (mks) units, the result is “automatically” in SI units.
Example: what is the charge on each plate if the capacitor is
connected to a 12 volt* battery?
Q  CV
0V
Q   53 1012  12 
V= 12V
Q  6.4 1010 C
+12 V
*Remember, it’s the potential difference that matters.
If you keep everything in SI (mks) units, the result is “automatically” in SI units.
Example: what is the electric field between the plates?
V
E
d
0V
12V
E
0.001 m
E
V= 12V
d = 0.001
V
E  12000
,"up."
m
+12 V
If you keep everything in SI (mks) units, the result is “automatically” in SI units.
Demo: Professor Tries to Avoid
Spot-Welding His Fingers
to the Terminals of a Capacitor
While Demonstrating Energy Storage
Asynchronous lecture students: we’ll try to make a video of this.
Today’s agenda:
Capacitance.
You must be able to apply the equation C=Q/V.
Capacitors: parallel plate, cylindrical, spherical.
You must be able to calculate the capacitance of capacitors having these geometries, and
you must be able to use the equation C=Q/V to calculate parameters of capacitors.
Circuits containing capacitors in series and parallel.
You must understand the differences between, and be able to calculate the “equivalent
capacitance” of, capacitors connected in series and parallel.
Capacitors in Circuits
Recall: this is the symbol representing a
capacitor in an electric circuit.
And this is the symbol for a battery…
…or this…
…or this.
+-
Circuits Containing Capacitors in Parallel
Vab
Capacitors connected in parallel:
C1
a
C2
b
C3
+ V
The potential difference (voltage drop) from a to b must equal V.
Vab = V = voltage drop across each individual capacitor.
Note how I have introduced the idea that when circuit components are connected in parallel, then the voltage
drops across the components are all the same. You may use this fact in homework solutions.
C1
Q=CV
 Q1 = C1 V
& Q2 = C2 V
& Q3 = C3 V
Q1
a
C2 -
+
Q2
C3
Q3
+ V
Now imagine replacing the parallel
combination of capacitors by a single
equivalent capacitor.
a
By “equivalent,” we mean “stores the same
total charge if the voltage is the same.”
Qtotal = Ceq V = Q1 + Q2 + Q3
Important!
Ceq
Q
+ V
Summarizing the equations on the last slide:
Q1 = C1 V
Q2 = C2 V
Q3 = C3 V
C1
C2
a
Q1 + Q2 + Q3 = Ceq V
C3
+ -
Using Q1 = C1V, etc., gives
V
C1V + C2V + C3V = Ceq V
C1 + C2 + C3 = Ceq
Generalizing:
(after dividing both sides by V)
Ceq = Ci
(capacitors in parallel)
b
Circuits Containing Capacitors in Series
Capacitors connected in series:
C1
C2
C3
+ -
+Q V -Q
An amount of charge +Q flows from the battery to the left plate
of C1. (Of course, the charge doesn’t all flow at once).
An amount of charge -Q flows from the battery to the right
plate of C3. Note that +Q and –Q must be the same in
magnitude but of opposite sign.
The charges +Q and –Q attract equal and opposite charges to
the other plates of their respective capacitors:
C1
+Q -Q
A
C2
+Q -Q
B
C3
+Q -Q
+ V
These equal and opposite charges came from the originally
neutral circuit regions A and B.
Because region A must be neutral, there must be a charge +Q
on the left plate of C2.
Because region B must be neutral, there must be a charge -Q
on the right plate of C2.
Vab
a
C1
+Q -Q
V1
C2
A
+Q -Q
V2
C3
B
b
+Q -Q
V3
+ V
The charges on C1, C2, and C3 are the same, and are
Q = C1 V1
Q = C2 V2
Q = C3 V3
But we don’t know V1, V2, and V3 yet.
We do know that Vab = V and also Vab = V1 + V2 + V3.
Note how I have introduced the idea that when circuit components are connected in series, then the voltage
drop across all the components is the sum of the voltage drops across the individual components. This is
actually a consequence of the conservation of energy. You may use this fact in homework solutions.
Let’s replace the three capacitors by a single equivalent
capacitor.
Ceq
+Q -Q
V
+ V
By “equivalent” we mean V is the same as the total voltage
drop across the three capacitors, and the amount of charge Q
that flowed out of the battery is the same as when there were
three capacitors.
Q = Ceq V
Collecting equations:
Q = C1 V1
Q = C2 V2
Q = C3 V3
Important!
Vab = V = V1 + V2 + V3.
Q = Ceq V
Substituting for V1, V2, and V3:
Q
Q
Q
V=
+
+
C1 C 2 C 3
Substituting for V:
Q
Q
Q
Q
=
+
+
Ceq C1 C2 C3
Dividing both sides by Q:
1
1
1
1
=
+
+
Ceq C1 C2 C3
Generalizing:
OSE:
1
1
=
Ceq
Ci
i
(capacitors in series)
Summary (know for exam!):
Series
C1
C2
same Q, V’s add
C3
1
1

Ceq
i Ci
C1
Parallel
C2
C3
same V, Q’s add
Ceq   Ci
i
Example: determine the
capacitance of a single capacitor
that will have the same effect as
the combination shown. Use
C1 = C2 = C3 = C.
C2
C1
C3
I don’t see a series combination of capacitors, but I do see a
parallel combination.
C23 = C2 + C3 = C + C = 2C
Now I see a series combination.
1
1
1
=
+
Ceq C1 C23
1
1
1
2
1
3
= +
=
+
=
Ceq C 2C 2C 2C 2C
C eq
2
= C
3
C23 = 2C
C 1= C
Example: for the capacitor circuit shown, C1 = 3F, C2 = 6F, C3
= 2F, and C4 =4F. (a) Find the equivalent capacitance. (b) if
V=12 V, find the potential difference across C4.
C1
I’ll work this at the blackboard.
C2
C4
C3
V
Homework Hint: each capacitor has associated
with it a Q, C, and V. If you don’t know what to do
next, near each capacitor, write down Q= , C= ,
and V= . Next to the = sign record the known
value or a “?” if you don’t know the value. As soon
as you know any two of Q, C, and V, you can
determine the third. This technique often provides
visual clues about what to do next.
(a) Find Ceq. (b) if V=12 V, find V4.
C1=3F
C2=6F
C4=4F
C3=2F
C1 and C3 are not in parallel. Make
sure you understand why!
C2 and C4 are not in series. Make
sure you understand why!
V=12 V
C1 and C2 are in series. Make sure you use the correct equation!
1
1
1
1 1
2 1
3
1
=
+
=
+
=
+
=
=
C12
C1 C 2
3 6
6 6
6
2
Don’t forget to invert: C12 = 2 F.
(a) Find Ceq. (b) if V=12 V, find V4.
C12=2F
C4=4F
C12 and C4 are not in series. Make
sure you understand why!
C3=2F
V=12 V
C12 and C3 are in parallel. Make sure you use the correct
equation!
C123 = C12 + C3 = 2 + 2 = 4μF
(a) Find Ceq. (b) if V=12 V, find V4.
C123=4F
C4=4F
C123 and C4 are in series. Make
sure you understand why!
Combined, they make give Ceq.
V=12 V
Make sure you use the correct equation!
1
1
1
1 1
2
1
=
+
=
+
=
=
Ceq
C123 C24
4 4
4
2
Don’t forget to invert: Ceq = 2 F.
(a) Find Ceq. (b) if V=12 V, find V4.
Ceq=2F
Ceq = 2 F.
V=12 V
If you see a capacitor circuit on the test, read the problem first.
Don’t go rushing off to calculate Ceq. Sometimes you are asked
to do other things.
Truth in advertising: there’s a high probability you will need to calculate Ceq at some point in the problem.
(a) Find Ceq. (b) if V=12 V, find V4.
Q1=?
C1=3F
V1=?
Q2=?
C2=6F Q =?
4
V2=? C =4F
4
V4=?
Q3=?
C3=2F
V3=?
Homework Hint: each capacitor has associated
with it a Q, C, and V. If you don’t know what to do
next, near each capacitor, write down Q= , C= ,
and V= . Next to the = sign record the known
value or a “?” if you don’t know the value. As soon
as you know any two of Q, C, and V, you can
determine the third. This technique often provides
visual clues about what to do next.
V=12 V
We know C4 and want to find V4. If we know Q4 we can
calculate V4. Maybe that is a good way to proceed.
(a) Find Ceq. (b) if V=12 V, find V4.
Q123=?
C123=4F
V123=?
Q4=?
C4=4F
V4=?
C4 is in series with C123 and
together they form Ceq.
Therefore Q4 = Q123 = Qeq.
V=12 V
Qeq = Ceq V =
C=
2 12 =
Q
Q
 V=
V
C
24μC = Q4
 V4 =
Q4
24
=
= 6V
C4
4
You really need to know this:
Capacitors in series…
all have the same charge
add the voltages to get the total voltage
Capacitors in parallel…
all have the same voltage
add the charges to get the total charge
(and it would be nice if you could explain why)
Homework Hint!
C1
What does our text mean by Vab?
C2
C4
b
a
Our text’s convention is Vab = Va – Vb.
This is explained on page 759. This is
in contrast to Physics 1135 notation,
where Vab = Vb – Va.
C3
V
In the figure on this slide, if Vab = 100 V then point a is at a
potential 100 volts higher than point b, and Vab = -100 V;
there is a 100 volt drop on going from a to b.
A “toy” to play with…
http://phet.colorado.edu/en/simulation/capacitor-lab
(You might even learn something.)
For now, select
“multiple
capacitors.”
Pick a circuit.