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 4k 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 = 4or C 4 (8.85 x 10-12 C Nm2 )(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. K0 5(8.85 x 10-12C/Nm2) K0 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 K0 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 K0 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

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