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Definitions & Examples

A. + + + +. a . d. - - - - -. b. L. C. 3. C. C. 1. 2. Definitions & Examples. a. º. a. b. C. b. Today…. Calculate E from V Definition of Capacitance Example Calculations- Parallel Plate Capacitor Combinations of Capacitors • Capacitors in Parallel

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Definitions & Examples

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  1. A + + + + a d - - - - - b L C 3 C C 1 2 Definitions & Examples a º a b C b

  2. Today… • Calculate E from V • Definition of Capacitance • Example Calculations-Parallel Plate Capacitor • Combinations of Capacitors • Capacitors in Parallel • Capacitors in Series • Lightning! • Appendices A. Calculate electric field of dipole from potential B. Cylindrical capacitor examples

  3. Last time… • If we know E(x,y,z) we can calculate V(x,y,z): • Potential energy of a test charge in external field: potential x test charge (q) • Conductors in E-fields “become” equipotential surfaces/volumes • E-field always normal to surface of conductor

  4. E from V? V+dV V • Expressed as a vector, E is the negative gradient of V • Spherical coordinates: • Cartesian coordinates: • We can obtain the electric field E from the potential V by inverting our previous relation between E andV: • Note: The units of E[= N/C] can be expressed [V/m].

  5. Preflight 7: This graph shows the electric potential at various points along thex-axis. 2) At which point(s) is the electric field zero? A B C D

  6. E from V: an Example • Consider the following electric potential: • Something for you to try: Can you use the dipole potential to obtain the dipole field? Try it in spherical coordinates ... you should get (see Appendix A): • What electric field does this describe? ... expressing this as a vector:

  7. Lecture 7, ACT 1 1 The electric potential in a region of space is given by The x-component of the electric field Ex at x= 2 is (a)Ex = 0(b)Ex>0(c)Ex< 0

  8. Lecture 7, ACT 1 We know V(x) “everywhere” To obtain Ex “everywhere”, use 1 The electric potential in a region of space is given by The x-component of the electric field Ex at x= 2 is (a)Ex= 0(b)Ex> 0(c)Ex< 0

  9. The Bottom Line If we know the electric field E everywhere, Þ If we know the potential function V everywhere, allows us to calculate the electric field E everywhere • Units for Potential! 1 Joule/Coul = 1 VOLT allows us to calculate the potential function V everywhere (keep in mind, we often define VA = 0 at some convenient place)

  10. Capacitance • A capacitor is a device whose purpose is to store electrical energy which can then be released in a controlled manner during a short period of time. • A capacitor consists of 2 spatially separated conductors which can be charged to +Q and -Q respectively. • The capacitance is defined as the ratio of the charge on one conductor of the capacitor to the potential difference between the conductors. [The unit of capacitance is the Farad: 1 F = 1C/V] • The capacitance belongs only to the capacitor, independent of the charge and voltage.

  11. A + + + + d - - - - - Example: Parallel Plate Capacitor • Calculate the capacitance. We assume+s, -scharge densities on each plate with potential difference V: • Need Q: • Need V: from def’n: • Use Gauss’ Law to find E

  12. - + s s E=0 E=0 - + • Field outside the sheets is zero - + + - • Gaussian surface encloses zero net charge + - A + - + - + - • Field inside sheets is not zero: - + - + • Gaussian surface encloses non-zero net charge - A + - + - E Recall:Two Infinite Sheets(into screen)

  13. A + + + + d - - - - - Þ Example: Parallel Plate Capacitor(see Appendix B for other examples) • Calculate the capacitance: • Assume +Q, -Q on plates with potential difference V. • As hoped for, the capacitance of this capacitor depends only on its geometry (A,d). • Note that C ~ length; this will always be the case!

  14. d Current sensor Battery Moveable plate Fixed plate Practical Application: Microphone (“condenser”) Sound waves incident  pressure oscillations  oscillating plate separation d  oscillating capacitance ( )  oscillating charge on plate  oscillating current in wire ( )  oscillating electrical signal See this in action at http://micro.magnet.fsu.edu/electromag/java/microphone/ !

  15. -Q -Q a2 +Q a1 +Q b b Lecture 7, ACT 2 • In each case below, a charge of +Q is placed on a solid spherical conductor and a charge of -Q is placed on a concentric conducting spherical shell. • LetV1 be the potential difference between the spheres with (a1, b). • Let V2 be the potential difference between the spheres with (a2, b). • What is the relationship between V1 and V2? (Hint – think about parallel plate capacitors.) (a) V1 < V2 (b) V1 = V2 (c) V1 > V2

  16. -Q -Q a2 +Q a1 +Q b b Lecture 7, ACT 2 • In each case below, a charge of +Q is placed on a solid spherical conductor and a charge of -Q is placed on a concentric conducting spherical shell. • LetV1 be the potential difference between the spheres with (a1, b). • Let V2 be the potential difference between the spheres with (a2, b). • What is the relationship between V1 and V2? (Hint – think about parallel plate capacitors.) (a) V1 < V2 (b) V1 = V2 (c) V1 > V2 • What we have here are two spherical capacitors. • Intuition: for parallel plate capacitors: V = (Q/C) = (Qd)/(Ae0). Therefore you might expect that V1 > V2 since(b-a1) > (b-a2). • In fact this is the case as we can show directly from the definition of V!

  17. a Q1 Q2 V b Þ Parallel Combination: Equivalent Capacitor: Þ Capacitors in Parallel a Q º V C C C 1 2 -Q1 -Q2 -Q b • Find “equivalent” capacitance C in the sense that no measurement at a,b could distinguish the above two situations. • Aha! The voltage across the two is the same….

  18. Preflight 7: Two identical parallel plate capacitors are shown in an end-view in A) of the figure. Each has a capacitance of C. 4) If the two are joined together as shown in B), forming a single capacitor, what will be the final capacitance? a) C/2 b) C c) 2C

  19. +Q -Q +Q -Q º a b a b +Q -Q RHS: Þ LHS: Capacitors in Series C 2 C C 1 • Find “equivalent” capacitance C in the sense that no measurement at a,b could distinguish the above two situations. • The charge on C1must be the same as the charge on C2 since applying a potential difference across ab cannot produce a net charge on the inner plates of C1 and C2 • assume there is no net charge on node betweenC1andC2

  20. a º a b b Þ Examples:Combinations of Capacitors C 3 C C C 1 2 • How do we start?? • Recognize C3 is in series with the parallel combination on C1 and C2. i.e.,

  21. Preflight 7: C C C C C Configuration A Configuration B Configuration C Three configurations are constructed using identical capacitors 6) Which of these configurations has the lowest overall capacitance? a) Configuration A b) Configuration B c) Configuration C

  22. Preflight 7: A circuit consists of three unequal capacitorsC1,C2, andC3which areconnected to a battery of emfE. The capacitors obtain chargesQ1Q2, Q3,and have voltages across their platesV1, V2, andV3. Ceqis the equivalent capacitance of the circuit. 8) Check all of the following that apply: a) Q1= Q2 b) Q2= Q3 c) V2= V3 d) E = V1 e) V1 < V2 f) Ceq > C1

  23. o C Ceq C C o (b) Ceq= (2/3)C (c) Ceq= 3C (a) Ceq= (3/2)C Lecture 7, ACT 3 • What is the equivalent capacitance, Ceq, of the combination shown?

  24. o C Ceq C C o (b) Ceq= (2/3)C (c) Ceq= 3C (a) Ceq= (3/2)C C C1 C C C Lecture 7, ACT 3 • What is the equivalent capacitance, Ceq, of the combination shown?

  25. _ _ _ + + + Lightning (a.k.a. the atmosphere is a BIG capacitor!!) Collisions produce charged particles. The heavier particles (-) sit near the bottom of the cloud; the lighter particles (+) near the top. Stepped Leader Negatively charged electrons begin zigzagging downward. Attraction As the stepped leader nears the ground, it draws a streamer of positive charge upward. Flowing Charge As the leader and the streamer come together, powerful electric current begins flowing Contact! Intense wave of positive charge, a “return stroke,” travels upward at 108 m/s Factoids:

  26. -Q A r a + + + + +Q d b - - - - - L -Q a +Q b Summary • A Capacitor is an object with two spatially separated conducting surfaces. • The definition of the capacitance of such an object is: • The capacitance depends on the geometry : Parallel Plates Spherical Cylindrical

  27. z r 1 +q r r 2 a q a -q the dipole moment Þ Appendix A: Electric Dipole • In Lecture 5 (appendix), we found the potential of an electric dipole to be: • Now use (in spherical coordinates) to find the electric field everywhere.

  28. Appendix A: Dipole Field z +q r q a a -q

  29. (b) qmax= 45° (c) qmax= 90° (a) qmax= 0 • The expression for the electric field of a dipole (r >> a) is: • The polar component of E is maximum when sinq is maximum. • Therefore, Eq has its maximum value when q= 90°. z Appendix A: Sample Problem r • Consider the dipole shown at the right. • Fixr = r0>> a • Define qmax such that the polar component of the electric field has its maximum value (for r = r0). 1 +q r r 2 a q a -q What is qmax?

  30. r a b L Appendix B: Cylindrical Capacitor Examples • Calculate the capacitance: • Assume +Q, -Q on surface of cylinders with potential difference V. • Gaussian surface is cylinder of radius r (a < r < b) and length L Þ • Apply Gauss' Law: If we assume that inner cylinder has +Q, then the potentialV is positive if we take the zero of potential to be defined at r = b: Þ

  31. -Q +Q -Q +Q b a c d Þ Note:This is just the result for 2 cylindrical capacitors in series! Appendix B: Another example • Suppose we have 4 concentric cylinders of radii a,b,c,d and charges +Q, -Q, +Q, -Q • Question:What is the capacitance between a and d? • Note: E-field betweenb and c is zero! WHY?? • A cylinder of radius r1: b < r1< c encloses zero charge!

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