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

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1 Chapter 6A - Capacitance A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 007

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 e - Battery e - Conductor There is is a limit to to the amount of of charge that a conductor can hold without leaking to to the air. There is is a certain capacity for holding charge.

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

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. C ; farad (F) V coulomb (C) volt (V) Example: When 40 C of charge are placed on a conductor,, the potential is 8 V. What is the capacitance? 40 C C C = 5 F V 8 V

6 Capacitance of Spherical Conductor At surface of sphere: E k ; r Recall: And: V V k r 1 k 4 0 k r 4 r C V 4 r 0 0 Capacitance, C r + E and V at surface. Capacitance: C 4 r 0 C V

7 Example 1: What is the capacitance of a metal sphere of radius 8 cm? Capacitance, C r + r = 0.08 m Capacitance: C = 44 r C 4 (8.85 x 10 )(0.08 m) -1 C Nm C = 8.90 x F Note: The capacitance depends only on physical para- meters (the radius r) r) and is is not determined by either charge or or potential. This is is true for all capacitors.

8 Example 1 (Cont.): What charge is needed to give a potential of 400 V? Capacitance, C r + r = 0.08 m C = 8.90 x F C ; CV V (8.90 pf)(400 V) Total Charge on Conductor: = 3.56 nc Note: The farad (F) and the coulomb (C) are extremely large units for static electricity. The SI SI prefixes micro,, nano n, n, and pico p are often used.

9 Dielectric Strength 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 Dielectric For air: E m = 3 x 10 6 N/C for spherical surfaces and as as low as as 0.8 x 10 6 N/C for sharp points.

10 Example : What is the maximum charge that can be placed on a spherical surface one meter in diameter? (R = 0.50 m) Maximum r Air E m = 3 x 10 6 N/C E m k r Er k ; m (3 x 10 )(0.50 m) 9 x 10 6 N C 9 Nm C Maximum charge in air: m = 83.3 C This illustrates the large size of of the coulomb as as a unit of of charge in 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 E m k r m Leakage (called corona discharge) often occurs at at sharp points where curvature r is is greatest.

12 Parallel Plate Capacitance + - Area A d C For these two parallel plates: V and E V d You will recall from Gauss law that E is also: E A 0 0 V E And d A 0 is charge on either plate. A is area of plate. A C V d And 0

13 C Example 3. The plates of a parallel plate capacitor have an area of 0.4 m and are 3 mm apart in air. What is the capacitance? A C 0 V d (8.85 x 10 )(0.4 m ) -1 C Nm (0.003 m) C = 1.18 nf d A 0.4 m 3 mm

14 Applications of Capacitors A microphone converts sound waves into an electrical signal (varying voltage) by changing d. Microphone Changing d d A C 0 d 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.

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

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

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

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 C o in a vacuum. K C C 0 Dielectric constant: K = 1 for Air K can also be given in terms of voltage V, electric field intensity E,, or permittivity : V E 0 0 K V E 0

19 The Permittivity of a Medium The capacitance of a parallel plate capacitor with a dielectric can be found from: A A C KC0 or C K0 or C d d The constant is the permittivity of the medium which relates to the density of field lines. K -1 ; 8.85 x 10 C 0 0 Nm

20 Example 4: Find the capacitance C and the charge if connected to 00-V battery. Assume the dielectric constant is K = (8.85 x 10-1 C/Nm ) 44.5 x 10-1 C/Nm A C A d (44.5 x 10 )(0.5 m ) -1 C Nm 0.00 m 0.5 m C = 11.1 nf if connected to V = 00 V? d mm = CV = (11.1 nf)(00 V) =. C

21 Example 4 (Cont.): Find the field E between the plates. Recall =. C; V = 00 V. V Gauss ' law : E A 44.5 x 10-1 C/Nm E -6. x 10 C (44.5 x 10 )(0.5 m ) -1 C Nm A 0.5 m 00 V E = 100 N/C d mm Since V = 00 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 6F 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( K = 5) is inserted? What is new capacitance C? K V C V0 V0 ; V C V K V ; 5 V = 80.0 V C = Kc o = 5(6 F) C = 30 F Air dielectric V o = 400 V Mica dielectric Mica, K = 5

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? Air C o = 6 F 0 = C 0 V 0 = (6 F)(400 V) 0 = 400 C V o = 400 V = CV = (30 F)(400 V) = 1,000 C = 1,000 C 400 C = 9600 C Mica C = 30 F Mica, K = 5 = 9.60 mc

24 Energy of Charged Capacitor The potential energy U of a charged capacitor is equal to the work (qv( qv) required to charge the capacitor. If we consider the average potential difference from 0 to V f to be V/: Work = (V/) = ½V U V U CV U 1 1 ; ; C

25 Example 6: In Ex-4, we found capacitance to be 11.1 nf,, the voltage 00 V, and the charge. C.. Find the potential energy U. U 1 CV U 1 (11.1 nf)(00 V) Capacitor of Example 5. 5 C = 11.1 nf U = J Verify your answer from the other formulas for P.E. 00 V U =? U V U 1 ; C =. C

26 Energy Density for Capacitor 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: E 0 Recall C and V Ed : A U CV ( Ed) d u U A d Vol. U Ad A 1 d Energy U Density u: u u Ad 1 0 AdE 0 Ad E

27 Summary of Formulas C V ; farad (F) coulomb (C) volt (V) C V K 0 A d C 4 0 r V E C 0 0 K C V E 0 0 u 1 0 E U V U CV U 1 1 ; ; C

28 CONCLUSION: Chapter 5 Capacitance

Episode 126: Capacitance and the equation C =Q/V

Episode 126: Capacitance and the equation C =Q/V Having established that there is charge on each capacitor plate, the next stage is to establish the relationship between charge and potential difference