1. Chapter 26A - Capacitance
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

2. 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.

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

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

5. Capacitance in Farads
One farad (F) is the capacitance C of a conductor that holds one coulomb of charge for each volt of potential.
Example: When 40 mC of charge are placed on a con- ductor, the potential is 8 V. What is the capacitance?
C = 5 mF

6. Capacitance of Spherical Conductor
At surface of sphere: +Q r
E and V at surface.
Capacitance, C
Recall:
And:
Capacitance:

7. Example 1: What is the capacitance of a metal sphere of radius 8 cm?
r = 0.08 m
Capacitance, C +Q r
Capacitance: C = 4peor
C = 8.90 x F
Note: The capacitance depends only on physical para- meters (the radius r) and is not determined by either charge or potential. This is true for all capacitors.

8. Example 1 (Cont.): What charge Q is needed to give a potential of 400 V?
r = 0.08 m
Capacitance, C +Q r
C = 8.90 x F
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.

9. Dielectric Strength
The dielectric strength of a material is that electric intensity Em for which the material becomes a conductor. (Charge leakage.) r Q
Dielectric
Em varies considerably with physical and environmental conditions such as pressure, humidity, and surfaces.
For air: Em = 3 x 106 N/C for spherical surfaces and as low as 0.8 x 106 N/C for sharp points.

10. Example 2: What is the maximum charge that can be placed on a spherical surface one meter in diameter? (R = 0.50 m) r Q
Em = 3 x 106 N/C
Maximum Q
Air
Maximum charge in air:
Qm = 83.3 mC
This illustrates the large size of the coulomb as a unit of charge in electrostatic applications.

11. 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. + +
Leakage (called corona discharge) often occurs at sharp points where curvature r is greatest.

12. Parallel Plate Capacitanced
Area A +Q -Q
For these two parallel plates:
You will recall from Gauss’ law that E is also:
Q is charge on either plate. A is area of plate.
And

13. Example 3. The plates of a parallel plate capacitor have an area of 0
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?
3 mm d A
0.4 m2
C = 1.18 nF

14. Applications of Capacitors
A microphone converts sound waves into an electrical signal (varying voltage) by changing d. d
Changing d
Microphone + - A
Variable Capacitor
Changing Area
The tuner in a radio is a variable capacitor. The changing area A alters capacitance until desired signal is obtained.

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

16. 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.

17. Insertion of Dielectric
Field decreases.
E < Eo
Air +
Co Vo Eo eo +Q -Q
Dielectric
Voltage decreases.
V < Vo
Insertion of a dielectric
Same Q
Q = Qo
C V E e
Capacitance increases.
C > Co + +Q -Q
Permittivity increases.
e > eo

18. 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.
Dielectric constant: K = 1 for Air
K can also be given in terms of voltage V, electric field intensity E, or permittivity e:

19. The Permittivity of a Medium
The capacitance of a parallel plate capacitor with a dielectric can be found from:
The constant e is the permittivity of the medium which relates to the density of field lines.

20. Example 4: Find the capacitance C and the charge Q if connected to 200-V battery. Assume the dielectric constant is K = 5.0.
e = Ke0= 5(8.85 x 10-12C/Nm2)
2 mm d A
0.5 m2
e = Ke0
eo = x C/Nm2
C = 11.1 nF
Q if connected to V = 200 V?
Q = CV = (11.1 nF)(200 V)
Q = 2.22 mC

21. Example 4 (Cont. ): Find the field E between the plates. Recall Q = 2
Example 4 (Cont.): Find the field E between the plates. Recall Q = 2.22 mC; V = 200 V.
2 mm d A
0.5 m2
e = Ke0
200 V
e = x C/Nm2
E = 100 N/C
Since V = 200 V, the same result is found if E = V/d is used to find the field.

22. 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 ?
Vo = 400 V
Mica, K = 5
Air dielectric
Mica dielectric
V = 80.0 V
C = Kco = 5(6 mF)
C = 30 mF

23. 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?
Vo = 400 V
Mica, K = 5
Air Co = 6 mF
Mica C = 30 mF
Q0 = C0V0 = (6 mF)(400 V)
Q0 = 2400 mC
Q = CV = (30 mF)(400 V)
Q = 12,000 mC
DQ = 12,000 mC – 2400 mC
DQ = 9600 mC
DQ = 9.60 mC

24. 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

25. Verify your answer from the other formulas for P.E.
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.
C = 11.1 nF
200 V
Q = 2.22 mC
U = ?
Capacitor of Example 5.
U = 222 mJ
Verify your answer from the other formulas for P.E.

26. 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: A d
Energy Density u:

27. Summary of Formulas

28. CONCLUSION: Chapter 25 Capacitance