• 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?

A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

• 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.

A battery establishes a difference of potential that can pump electrons e from a ground (earth) to a conductor
Earth e -
Battery e -
Conductor
--
- - - - -
-
-
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.

The capacitance C of a conductor is defined as the ratio of the charge Q on the conductor to the potential V produced.
Earth e -
Battery e -
Conductor
--
- - - - -
-
-
Capacitance:
C

Q
; :
V

One farad (F) is the capacitance C of a conductor that holds one coulomb of charge for each volt of potential.
C

Q
; farad (F)
V
 coulomb (C) volt (V)
Example:
When 40 m
C of charge are placed on a conductor, the potential is 8 V. What is the capacitance?
C

Q

V 8 V
C = 5 m F

+Q
-Q
Area A d
For these two parallel plates:
C

Q
V
and E

V d
You will recall from Gauss’ law that E is also:
E



0


Q
0
A
Q is charge on either plate. A is area of plate.
E

V d


Q
0
A
And C

Q

V

0
A d

• Capacitance:
1 k

C

Q

V

0
A d
4

0
+Q
• Sometime given as:
C

K

0
A d
-Q
• Where K is the
“dielectric constant”
Area A d


0
• “Epsilon-naught”
• 8.854 x 10 -12 C 2 /N m 2

0

1
4
 k

• 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 m 2 and are 3 mm apart in air. What is the capacitance?
C

Q

V

0
A d
C

(8.85 x 10
-12
C
2
Nm
2
2
)(0.4 m )
(0.003 m)
A
0.4 m 2 d 3 mm
C = 1.18 nF

• List at least 3 electrical quantities
• List at least 3 units
• What is the difference between a variable and a unit/

• 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 m 2 and are 3 mm apart in air. What is the capacitance?
C

Q

V

0
A d
C

(8.85 x 10
-12
C
2
Nm
2
2
)(0.4 m )
(0.003 m)
A
0.4 m 2 d 3 mm
C = 1.18 nF

• What is the formula for kinetic energy?
• How is work related to kinetic energy?

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 V f to be V/2 :
Work = Q(V/2) = ½QV
U

1
2
QV ; U

1
2
CV
2
; U

Q
2 C
2

Example 6: In Ex-4, we found capacitance to be 11.1 nF , the voltage 200 V , and the charge 2.22 m C . Find the potential energy U .
U

1
2
CV
2
U

1
2
(11.1 nF)(200 V)
2
Capacitor of
Example 5 .
C = 11.1 nF
U = 222 m
J
200 V
Verify your answer from the other formulas for P.E.
U = ?
U

1
2
QV ; U

Q
2 C
2
Q = 2.22 m C

• 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:
– jtimson@bpi.edu
• It will save you time in the future if you set up an account in your name

• Pass in your Do Now’s, then pack up and wait for further instructions.

At surface of sphere:
E
 kQ
; V r
2
 kQ r
Recall:
And: k

1
4

0
V
 kQ r

4
Q

0 r
C

Q

V Q
Q
4

0 r
Capacitance, C r
+Q
E and V at surface.
Capacitance:
C

4

0 r
C

Q
V

Example 1: What is the capacitance of a metal sphere of radius 8 cm?
Capacitance, C r
+Q
Capacitance: C = 4
 o r
C
 
-12 C
2
)(0.08 m) r = 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?
C = 8.90 x 10 -12 F Capacitance, C r
+Q r = 0.08 m
C

Q
; 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.

The dielectric strength of a material is that electric intensity E m for which the material becomes a conductor. (Charge leakage.)
E m varies considerably with physical and environmental conditions such as pressure, humidity, and surfaces. r
Q
Dielectric
For air: E m
= 3 x 10 6 N/C for spherical surfaces and as low as 0.8 x 10 6 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
E m
= 3 x 10 6 N/C
E m
Q

 kQ r
2
; Q

E r m k
(3 x 10
6
N
C
)(0.50 m)
2
2
9 x 10
9 Nm 2
C
2
Maximum charge in air:
Q m
= 83.3
m C
This illustrates the large size of the coulomb as a unit of charge in electrostatic applications.

The charge density on a surface is significantly affected by the curvature . The density of charge is greatest where the curvature is greatest.
+
+ + + +
+ + + + +
+ +
+
+ E m
 kQ m r
2
+
+
+
+
+
+
+
+
+
+ ++ ++ +
Leakage (called corona discharge) often occurs at sharp points where curvature r is greatest.

+Q
-Q
Area A d
For these two parallel plates:
C

Q
V
and E

V d
You will recall from Gauss’ law that E is also:
E



0


Q
0
A
Q is charge on either plate. A is area of plate.
E

V d


Q
0
A
And C

Q

V

0
A d

Example 3.
The plates of a parallel plate capacitor have an area of 0.4 m 2 and are 3 mm apart in air. What is the capacitance?
C

Q

V

0
A d
C

(8.85 x 10
-12
C
2
Nm
2
2
)(0.4 m )
(0.003 m)
A
0.4 m 2 d 3 mm
C = 1.18 nF

A microphone converts sound waves into an electrical signal (varying voltage) by changing d .
Changing d
Microphone d
C
 
0
A d
V

Q
C
+
+
-
-
Changing
+
+
+
+
+
Area
--
-
-
-
Variable
A
Capacitor
The tuner in a radio is a variable capacitor . The changing area A alters capacitance until desired signal is obtained.

Most capacitors have a dielectric material between their plates to provide greater dielectric strength and less probability for electrical discharge.
E
+
+
+
+
+
+
Air
C o o
-
-
-
-
-
reduced E
+
+
+
+
+
+
- + - +
- + - +
- + - +
-
-
-
-
-
-
Dielectric
+
+
+
+
+
+
E < E
-
+
+
+
C > C o o
-
-
-
-
-
-
The separation of dielectric charge allows more charge to be placed on the plates— greater capacitance C > C o
.

• Smaller plate separation without contact.
• Increases capacitance of a capacitor.
• Higher voltages can be used without breakdown.
• Often it allows for greater mechanical strength.

Air
Dielectric
+Q
C o
V o
E o
 o
-Q
Insertion of a dielectric
+Q
C V E

Same Q
Q = Q o
-Q
+
+ +
+
+
+
+
Field decreases.
E < E o
Voltage decreases.
V < V o
Capacitance increases.
C > C o
Permittivity increases.

>
 o

The dielectric constant K for a material is the ratio of the capacitance C with this material as compared with the capacitance C o in a vacuum.
K

C
C
0
Dielectric constant:
K = 1 for Air
K can also be given in terms of voltage electric field intensity
V ,
E , or permittivity  :
K

V
0

E
V E
0



0

The capacitance of a parallel plate capacitor with a dielectric can be found from:
C

KC
0
or C

K

0
A
or C d
 
A 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 C
2
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 -12 C/Nm 2 )
 o

44.25 x 10 -12 C/Nm 2
C
 
A d

(44.25 x 10
-12 C
2
Nm
2
2
)(0.5 m )
0.002 m
C = 11.1 nF
Q if connected to V = 200 V?
Q = CV = (11.1 nF)(200 V) d
Q
  K
0
A
0.5 m 2
2 mm
= 2.22 m C

Example 4 (Cont.): Find the field the plates. Recall Q = 2.22 m C;
E between
V = 200 V .
' : E





Q
A
 
44.25 x 10 -12 C/Nm 2
  K
0
A
0.5 m 2
E

2.22 x 10 C
(44.25 x 10
-12 C
2
Nm
2
-6
2
)(0.5 m ) 200 V
E = 100 N/C 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 6 m F 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 ?
Air dielectric
K

C

V
C V
0
0 ; V

V
0
K
V

400 V
;
5
V = 80.0 V
C = Kc o
= 5(6 m F)
V o
= 400 V
Mica dielectric
C = 30 m F
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 ?
Q
0
= C
Q
0
0
V
0
= (6 m F)(400 V)
= 2400 m C
Air C
V o o
= 6
= 400 V m F
Q = CV = (30 m F)(400 V)
Mica C = 30 m F
Q = 12,000 m C
Mica, K = 5
D Q = 12,000 m C – 2400 m C
D Q = 9600 m C D Q = 9.60 mC

Energy density u is the energy per unit volume
( J/m 3 ). For a capacitor of area A and separation d , the energy density u is found as follows:
Energy Density u for an E-field: A d u

U

U
Vol .
Ad
Recall C
U

1
2
CV
2


0
A
and V

Ed d
 
0
A

 d 
( Ed )
2
: u

U
 u

1
2

2
0

0
E
AdE
2

C

Q
; farad (F)
V
 coulomb (C) volt (V)
K
C

Q

V
K

0
A d

C

V
0

E
0

C V E
0


0
C

4

0 r u

1
2

0
E
2
U

1
2
QV ; U

1
2
CV
2
; U

Q
2 C
2