1. technology and computing

Do Now (9/23/13):
• What is the voltage of a proton moving at
a constant speed of 3 m/s over 1 s in an
electric field of 300 N/C?
• What does the word “capacity” mean to
you?
Chapter 26A - Capacitance
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
Objectives: After completing this
module, you should be able to:
• Define capacitance in terms of charge and
voltage, and calculate the capacitance for a
parallel plate capacitor given separation and
area of the plates.
• Define dielectric constant and apply to
calculations of voltage, electric field
intensity, and capacitance.
• Find the potential energy stored in capacitors.
Maximum Charge on a Conductor
A battery establishes a difference of potential that can
pump electrons e- from a ground (earth) to a conductor
Battery
Earth
e-
Conductor
- - - - e ----
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.
Battery
Earth
e-
Conductor
- - - - e- - - Q, V - ---
Q
Capacitance: C  ; Units : Coulombs per volt
V
Capacitance in Farads
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)
farad (F) 
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
Parallel Plate Capacitance
• Capacitance:
k
Q
A
C   0
V
d
• Sometime given as:
A
C  K 0
d
• Where K is the
“dielectric constant”
1
4 0
+Q
Area A
-Q
d
Permitivity of free space
• “Epsilon-naught”
• 8.854 x 10-12 C2/N m2
0 
1
4k
0
Practice:
• Work on your homework
• Work on the bonus
• Be ready for an exit question!!
Do Now (9/24/13): 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
Do Now (9/24/13):
• List at least 3 electrical quantities
• List at least 3 units
• What is the difference between a variable
and a unit/
Practice:
• Complete your pre-lab
• Complete your two note sheets! Use your
notes and your peers! You should have
at least six for each sheet!
Example 3. 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
Do Now (9/25/13):
• What is the formula for kinetic energy?
• How is work related to kinetic energy?
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 6: In Ex-4, we found capacitance to
be 11.1 nF, the voltage 200 V, and the
charge 2.22 mC. Find the potential energy U.
U
U
1
2
1
2
CV
2
(11.1 nF)(200 V)
2
U = 222 mJ
Verify your answer from the
other formulas for P.E.
U  1 2 QV ;
Q2
U
2C
Capacitor of
Example 5.
C = 11.1 nF
200 V
U=?
Q = 2.22 mC
Review Challenge
• Go to http://cwx.prenhall.com/giancoli/
• Select Chapter 17, then push Begin
• Select Practice Questions
• Answer the 25 questions and then push Submit
for Grading at that time you can enter your
name and my email address:
– [email protected]
• It will save you time in the future if you set up
an account in your name
Do Now (9/26/13):
• Pass in your Do Now’s, then pack up and
wait for further instructions.
Capacitance of Spherical Conductor
At surface of sphere:
kQ
E 2 ;
r
Recall:
kQ
V
r
k
1
4 0
kQ
Q
And: V 

r
4 0 r
Q
Q
C 
V Q 4 0 r
Capacitance, C
r
+Q
E and V at surface.
Q
Capacitance: C 
V
C  4 0 r
Example 1: What is the capacitance of
a metal sphere of radius 8 cm?
Capacitance, C
r
+Q
r = 0.08 m
Capacitance: C = 4or
C  4 (8.85 x 10-12 C Nm2 )(0.08 m)
C = 8.90 x 10-12 F
Note: The capacitance depends only on physical parameters (the radius r) and is not determined by either
charge or potential. This is true for all capacitors.
Example 1 (Cont.): What charge Q is
needed to give a potential of 400 V?
Capacitance, C
r
+Q
r = 0.08 m
C = 8.90 x 10-12 F
Q
C  ; Q  CV
V
Q  (8.90 pF)(400 V)
Total Charge on Conductor:
Q = 3.56 nC
Note: The farad (F) and the coulomb (C) are
extremely large units for static electricity. The SI
prefixes micro m, nano n, and pico p are often used.
Dielectric Strength
The dielectric strength of a material is that
electric intensity Em for which the material
becomes a conductor. (Charge leakage.)
Em varies considerably with
physical and environmental
conditions such as pressure,
humidity, and surfaces.
r
Q
Dielectric
For air: Em = 3 x 106 N/C for spherical surfaces
and as low as 0.8 x 106 N/C for sharp points.
Example 2: What is the maximum charge
that can be placed on a spherical surface
one meter in diameter? (R = 0.50 m)
Maximum Q
r
Q
Air
Em = 3 x 106 N/C
kQ
Em  2 ;
r
Em r
Q
k
2
(3 x 106 N C)(0.50 m) 2
Q
9 Nm 2
9 x 10
C2
Maximum charge in air: Qm = 83.3 mC
This illustrates the large size of the coulomb as a
unit of charge in electrostatic applications.
Capacitance and Shapes
The charge density on a surface is significantly
affected by the curvature. The density of charge
is greatest where the curvature is greatest.
+ + + ++
+
+
+ + + + + ++
kQm
Em  2
r
+ + + + ++
++
+
+
+
+
+ + +
Leakage (called corona discharge) often occurs
at sharp points where curvature r is greatest.
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 3. 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.
Dielectric Materials
Most capacitors have a dielectric material between
their plates to provide greater dielectric strength and
less probability for electrical discharge.
Eo
+
+
+ Air +
+
+
Co
reduced E
+-+-+ +
+-+-+ +-+-++
+
Dielectric
E < Eo
+
+ +++ -+
++ +
C > Co
The separation of dielectric charge allows more charge
to be placed on the plates—greater capacitance C > Co.
Advantages of Dielectrics
• Smaller plate separation without contact.
• Increases capacitance of a capacitor.
• Higher voltages can be used without
breakdown.
• Often it allows for greater mechanical strength.
Insertion of Dielectric
Air
Dielectric
Field decreases.
E < Eo
+Q
Co Vo Eo o
-Q
+ +
+
+ +
+
Insertion of
a dielectric +Q
C V E 
Same Q
Q = Qo
-Q
Voltage decreases.
V < Vo
Capacitance increases.
C > Co
+ +
Permittivity increases.
 > o
Dielectric Constant, K
The dielectric constant K for a material is the
ratio of the capacitance C with this material as
compared with the capacitance Co in a vacuum.
C
K
C0
Dielectric constant:
K = 1 for Air
K can also be given in terms of voltage V,
electric field intensity E, or permittivity :
V0 E0 
K


V
E 0
The Permittivity of a Medium
The capacitance of a parallel plate capacitor
with a dielectric can be found from:
C  KC0
A
A
or C  K  0
or C  
d
d
The constant  is the permittivity of the medium
which relates to the density of field lines.
  K 0 ;
 0  8.85 x 10
-12 C2
Nm 2
Example 4: Find the capacitance C and the
charge Q if connected to 200-V battery.
Assume the dielectric constant is K = 5.0.
  K0 5(8.85 x 10-12C/Nm2)
  K0
o  44.25 x 10-12 C/Nm2
-12 C2
Nm2
A
0.5 m2
2
)(0.5 m )
A (44.25 x 10
C  
d
0.002 m
C = 11.1 nF
Q if connected to V = 200 V?
Q = CV = (11.1 nF)(200 V)
d
2 mm
Q = 2.22 mC
Example 4 (Cont.): Find the field E between
the plates. Recall Q = 2.22 mC; V = 200 V.
 Q
Gauss ' law : E  
 A
  K0
  44.25 x 10-12 C/Nm2
A
0.5 m2
-6
2.22 x 10 C
E
-12 C 2
(44.25 x 10 Nm2 )(0.5 m2 )
E = 100 N/C
200 V
d
2 mm
Since V = 200 V, the same result is found
if E = V/d is used to find the field.
Example 5: A capacitor has a capacitance of 6mF
with air as the dielectric. A battery charges the
capacitor to 400 V and is then disconnected. What
is the new voltage if a sheet of mica (K = 5) is
inserted? What is new capacitance C ?
V0
C V0
K
 ; V
C0 V
K
400 V
V = 80.0 V
V
;
5
Air dielectric
Vo = 400 V
C = Kco = 5(6 mF)
Mica dielectric
C = 30 mF
Mica, K = 5
Example 5 (Cont.): If the 400-V battery is
reconnected after insertion of the mica, what
additional charge will be added to the plates
due to the increased C?
Air Co = 6 mF
Q0 = C0V0 = (6 mF)(400 V)
Vo = 400 V
Q = 2400 mC
0
Q = CV = (30 mF)(400 V)
Mica C = 30 mF
Q = 12,000 mC
Mica, K = 5
DQ = 12,000 mC – 2400 mC
DQ = 9600 mC
DQ = 9.60 mC
Energy Density for Capacitor
Energy density u is the energy per unit volume
(J/m3). For a capacitor of area A and separation
d, the energy density u is found as follows:
Energy Density
u for an E-field:
Recall C 
0 A
d
A d
and V  Ed :
2
2
1
1  0 A 
U  2 CV  2 
(
Ed
)

 d 
U
U
u

Vol. Ad
U
Energy
u
uAd
 AdE
u:
1
Density
2 0

1
2
0
2
Ad
E
2
Summary of Formulas
Q
C ;
V
coulomb (C)
farad (F) 
volt (V)
Q
A
C   K0
V
d
C V0 E0 
K
 

C0 V
E 0
U  1 2 QV ;
C  4 0 r
u  0E
1
2
2
Q
U  1 2 CV 2 ; U 
2C
2
CONCLUSION: Chapter 25
Capacitance