#### Transcript Cloudfront.net

```Capacitance
AP Physics B
Capacitors
Consider two separated conductors, like two parallel
plates, with external leads to attach to other circuit
elements. Such a device is called a capacitor.
There is a limit to the amount of charge that a
conductor can hold without leaking to the air.
There is a certain capacity for holding charge.
Capacitance
The capacitance (C) of a conductor is defined
as the ratio of the charge (Q) on the
conductor to the potential (V) produced.
Q
Capacitance: C  ; Units : Coulombs per volt
V
One farad (F) is the capacitance C of a conductor that
holds one coulomb of charge for each volt of potential.
Q
C ;
V
coulomb (C)
volt (V)
Example: When 40 mC of charge are placed on a conductor, the potential is 8 V. What is the capacitance?
Q 40 m C
C 
V
8V
C = 5 mF
Parallel Plate Capacitance
+Q
Area A
-Q
d
For these two
parallel plates:
Q
V
C
and E 
V
d
You will recall from Gauss’ law that E is also:

Q
E

0 0 A
V
Q
E 
d 0 A
Q is charge on either
plate. A is area of plate.
And
Q
A
C   0
V
d
Example 2. The plates of a parallel
plate capacitor have an area of 0.4 m2
and are 3 mm apart in air. What is the
capacitance?
Q
A
C   0
V
d
C
(8.85 x 10
-12
C2
Nm 2
A
0.4 m2
)(0.4 m 2 )
(0.003 m)
C = 1.18 nF
d
3 mm
Applications of Capacitors
A microphone converts sound waves into an
electrical signal (varying voltage) by changing d.
Changing d
Microphone
d
A
C  0
d
Q
V
C
Changing
++
Area
++
-- ++
- + A
---
Variable
Capacitor
The tuner in a radio is a variable capacitor. The changing
area A alters capacitance until desired signal is obtained.
Energy of Charged Capacitor
The potential energy U of a charged
capacitor is equal to the work (qV)
required to charge the capacitor.
If we consider the average potential
difference from 0 to Vf to be V/2:
Work = Q(V/2) = ½QV
U  1 2 QV ;
2
Q
2
1
U  2 CV ; U 
2C
Example 3: In a capacitor, we found its
capacitance to be 11.1 nF, the voltage 200 V,
and the charge 2.22 mC. Find the potential
energy U.
2
1
Capacitor of
U  2 CV
Example 3.
U
1
2
(11.1 nF)(200 V)
2
U = 222 mJ
other formulas for P.E.
U  1 2 QV ;
Q2
U
2C
C = 11.1 nF
200 V
U=?
Q = 2.22 mC
Electrical Circuit Symbols
Electrical circuits often contain two or more
capacitors grouped together and attached
to an energy source, such as a battery.
The following symbols are often used:
Ground
+ - + - + - + -
Battery
+
-
Capacitor
+
+
-
Capacitors in Series
Capacitors or other devices connected
along a single path are said to be
connected in series. See circuit below:
+
+
C1
- +
- +
-+
-+
C2
Battery
-
C3
Series connection
of capacitors.
“+ to – to + …”
Charge inside
dots is induced.
Charge on Capacitors in Series
Since inside charge is only induced, the
charge on each capacitor is the same.
Q1
+
+
C1
Q2
- +
- +
Q3
-+
-+
C2
Battery
-
C3
Charge is same:
series connection
of capacitors.
Q = Q1 = Q2 =Q3
Voltage on Capacitors in Series
Since the potential difference between
points A and B is independent of path, the
battery voltage V must equal the sum of
the voltages across each capacitor.
V1
+
+
C1
•A
V2
- +
- +
V3
-+
-+
C2
Battery
-
C3
•
B
Total voltage V
Series connection
Sum of voltages
V = V1 + V2 + V3
Equivalent Capacitance: Series
V1
+
+
V2
- +
- +
C1
C2
V3
-+
-+
-
Q
Q
C ; V 
V
C
C3
V = V1 + V2 + V3
Q1= Q2 = Q3
Q Q1 Q2 Q3



C C1 C2 C3
1
1
1
1
 

Ce C1 C2 C3
Equivalent Ce
for capacitors
in series:
n
1
1

Ce i 1 Ci
Example 1. Find the equivalent capacitance
of the three capacitors connected in series
with a 24-V battery.
Ce for
series:
n
1
1

Ce i 1 Ci
1
1
1
1



Ce 2 m F 4 m F 6 m F
1
 0.500  0.250  0.167
Ce
1
1
 0.917 or Ce 
Ce
0.917
C1
C2
C3
+ - + -+ + - + -+ 2 mF 4 mF 6 mF
24 V
Ce = 1.09 mF
Example 1 (Cont.): The equivalent circuit can
be shown as follows with single Ce.
C1
C2
C3
+ - + -+ + - + -+ 2 mF 4 mF 6 mF
n
1
1

Ce i 1 Ci
Ce
1.09 mF
24 V
Ce = 1.09 mF
24 V
Note that the equivalent capacitance Ce
for capacitors in series is always less than
the least in the circuit. (1.09 mF < 2 mF)
Example 1 (Cont.): What is the total charge
and the charge on each capacitor?
C1
C2
C3
+ - + -+ + - + -+ 2 mF 4 mF 6 mF
Ce
1.09 mF
24 V
24 V
QT = CeV = (1.09 mF)(24 V);
For series circuits:
QT = Q1 = Q2 = Q3
Ce = 1.09 mF
Q
C
V
Q  CV
QT = 26.2 mC
Q1 = Q2 = Q3 = 26.2 mC
Example 1 (Cont.): What is the voltage across
each capacitor?
Q
Q
C ; V 
C1 C2 C3
V
C
+ - + -+ + - + -+ Q1 26.2 m C
2 mF 4 mF 6 mF
V1 

 13.1 V
C1
2 mF
24 V
Q2 26.2 m C
V2 

 6.55 V
C2
4 mF
Q3 26.2 m C
V3 

 4.37 V
VT = 24 V
C3
6 mF
Note: VT = 13.1 V + 6.55 V + 4.37 V = 24.0 V
Short Cut: Two Series Capacitors
The equivalent capacitance Ce for two series
capacitors is the product divided by the sum.
1
1
1
  ;
Ce C1 C2
C1C2
Ce 
C1  C2
Example:
(3 m F)(6 m F)
Ce 
3 m F  6m F
C1
C2
+ - + + - + 3 mF 6 mF
Ce = 2 mF
Parallel Circuits
Capacitors which are all connected to the
same source of potential are said to be
connected in parallel. See below:
Parallel capacitors:
“+ to +; - to -”
- -
C3
+
+
- -
C2
+
+
+
+
C1
- -
Voltages:
VT = V1 = V2 = V3
Charges:
QT = Q1 + Q2 + Q3
Equivalent Capacitance: Parallel
Q
C  ; Q  CV
V
Parallel capacitors
in Parallel:
- -
Ce = C1 + C2 + C3
C3
+
+
- -
C2
+
+
+
+
C1
- -
Q = Q1 + Q2 + Q3
Equal Voltages:
CV = C1V1 + C2V2 + C3V3
Equivalent Ce
for capacitors
in parallel:
n
Ce   Ci
i 1
Example 2. Find the equivalent capacitance
of the three capacitors connected in parallel
with a 24-V battery.
Ce for
parallel:
n
Ce   Ci
VT = V1 = V2 = V3
Q = Q1 + Q2 + Q3
i 1
24 V
Ce = (2 + 4 + 6) mF
2 mF
C1 C2
C3
4 mF
6 mF
Ce = 12 mF
Note that the equivalent capacitance Ce for
capacitors in parallel is always greater than
the largest in the circuit. (12 mF > 6 mF)
Example 2 (Cont.) Find the total charge QT
and charge across each capacitor.
Q = Q1 + Q2 + Q3
24 V
2 mF
C1 C2
C3
4 mF
6 mF
Ce = 12 mF
V1 = V2 = V3 = 24 V
Q
C  ; Q  CV
V
QT = CeV
Q1 = (2 mF)(24 V) = 48 mC
QT = (12 mF)(24 V)
Q1 = (4 mF)(24 V) = 96 mC
QT = 288 mC
Q1 = (6 mF)(24 V) = 144 mC
Example 3. Find the equivalent capacitance
of the circuit drawn below.
24 V
C1
4 mF
24 V
4 mF
C1
C3,6
C2
3 mF
C3
6 mF
C3,6
(3m F)(6m F)

 2m F
3m F  6m F
Ce = 4 mF + 2 mF
Ce = 6 mF
24 V
2 mF
Ce
6 mF
Example 3 (Cont.) Find the total charge QT.
Ce = 6 mF
24 V
C1
4 mF
24 V
4 mF
C1
C2
3 mF
C3
6 mF
C3,6
Q = CV = (6 mF)(24 V)
QT = 144 mC
24 V
2 mF
Ce
6 mF
Example 3 (Cont.) Find the charge Q4 and
voltage V4 across the the 4-mF capacitor.
V4 = VT = 24 V
24 V
4 mF
C1
C2
3 mF
C3
6 mF
Q4 = (4 mF)(24 V)
Q4 = 96 mC
The remainder of the charge: (144 mC – 96 mC)
is on EACH of the other capacitors. (Series)
Q3 = Q6 = 48 mC
This can also be found from
Q = C3,6V3,6 = (2 mF)(24 V)
Example 3 (Cont.) Find the voltages across
the 3 and 6-mF capacitors.
Q3 = Q6 = 48 mC
24 V
4 mF
C1
C2
3 mF
C3
6 mF
48m C
V3 
 16.0V
3m F
48m C
V6 
 8.00V
6m F
Note: V3 + V6 = 16.0 V + 8.00 V = 24 V
Use these techniques to find voltage and
capacitance across each capacitor in a circuit.
Summary: Series Circuits
n
1
1

Ce i 1 Ci
Q = Q1 = Q2 = Q3
V = V1 + V2 + V3
For two capacitors at a time:
C1C2
Ce 
C1  C2
Summary: Parallel Circuits
n
Ce   Ci
i 1
Q = Q1 + Q2 + Q3
V = V1 = V2 =V3
For complex circuits, reduce the circuit in steps
using the rules for both series and parallel
connections until you are able to solve problem.
AP Physics HW 3/26
• Read Chapter 16.3 and 16.5 (that’s it!)
• Do Problems…#63, 65, 67, 69, 71, 93, 95
• Finish ALL other HW problems sets
• Yosemite \$ DUE.
• T-Shirt Design? ...25 points up for grabs!
```