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Advanced Physics

Advanced Physics. Capacitance . Chapter 25 – Problems 1, 3, 8, (17), 19, (33), 39, 40 & 49. The definition of capacitance. Capacitor: two conductors (separated by an insulator) usually oppositely charged

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Advanced Physics

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  1. Advanced Physics Capacitance • Chapter 25 – Problems 1, 3, 8, (17), 19, (33), 39, 40 & 49.

  2. The definition of capacitance • Capacitor: two conductors (separated by an insulator) • usually oppositely charged • The capacitance, C, of a capacitor is defined as a ratio of the magnitude of a charge on either conductor to the magnitude of the potential difference between the conductors a +Q b -Q

  3. A capacitor is basically two parallel conducting plates with insulating material in between. The capacitor doesn’t have to look like metal plates. • When a capacitor is connected to an external potential, charges flow onto the plates and create a potential difference between the plates. Capacitor for use in high-performance audio systems. • Capacitors in circuits • symbols • analysis follow from conservation of energy • conservation of charge - - - +

  4. Units of capacitance • The unit of C is the farad (F), but most capacitors have values of C ranging from picofarads to microfarads (pF to F). • Recall, micro 10-6, nano 10-9, pico 10-12 • If the external potential is disconnected, charges remain on the plates, so capacitors are good for storing charge (and energy).

  5. 16.7 The parallel-plate capacitor • The capacitance of a device depends on the geometric arrangement of the conductors • where A is the area of one of the plates, d is the separation, e0 is a constant called the permittivity of free space, e0= 8.85´10-12 C2/N·m2 A +Q d A -Q

  6. Problem: parallel-plate capacitor • A parallel plate capacitor has plates 2.00 m2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine • the capacitance • the charge on each plate

  7. A parallel plate capacitor has plates 2.00 m2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine • the capacitance • the charge on each plate Solution: Given: DV=10,000 V A = 2.00 m2 d = 5.00 mm Find: C=? Q=? Since we are dealing with the parallel-plate capacitor, the capacitance can be found as Once the capacitance is known, the charge can be found from the definition of a capacitance via charge and potential difference:

  8. C1 C5 C3 C2 C4 16.8 Combinations of capacitors • It is very often that more than one capacitor is used in an electric circuit • We would have to learn how to compute the equivalent capacitance of certain combinations of capacitors C2 C1 C3

  9. a C1 +Q1 C2 +Q2 V=Vab -Q1 -Q2 b a. Parallel combination Connecting a battery to the parallel combination of capacitors is equivalent to introducing the same potential difference for both capacitors, A total charge transferred to the system from the battery is the sum of charges of the two capacitors, By definition, Thus, Ceq would be

  10. Parallel combination: notes • Analogous formula is true for any number of capacitors, • It follows that the equivalent capacitance of a parallel combination of capacitors is greater than any of the individual capacitors (parallel combination)

  11. Problem: parallel combination of capacitors A 3 mF capacitor and a 6 mF capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited.

  12. a C1 +Q1 C2 +Q2 V=Vab -Q1 -Q2 b A 3 mF capacitor and a 6 mF capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited. Given: V = 18 V C1= 3 mF C2= 6 mF Find: Ceq=? Q=? First determine equivalent capacitance of C1 and C2: Next, determine the charge

  13. a +Q1 C1 -Q1 V=Vab c +Q2 C2 -Q2 b b. Series combination Connecting a battery to the serial combination of capacitors is equivalent to introducing the same charge for both capacitors, A voltage induced in the system from the battery is the sum of potential differences across the individual capacitors, By definition, Thus, Ceq would be

  14. Series combination: notes • Analogous formula is true for any number of capacitors, • It follows that the equivalent capacitance of a series combination of capacitors is always less than any of the individual capacitance in the combination (series combination)

  15. Problem: series combination of capacitors A 3 mF capacitor and a 6 mF capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance.

  16. a +Q1 C1 -Q1 V=Vab c +Q2 C2 -Q2 b A 3 mF capacitor and a 6 mF capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance and total charge deposited. Given: V = 18 V C1= 3 mF C2= 6 mF Find: Ceq=? Q=? First determine equivalent capacitance of C1 and C2: Next, determine the charge

  17. Series / Parallel Combination Example

  18. Energy stored in a capacitor • C = Q / V Q = V * C • V = U / Q U = V * Q • U = V * V * C = CV2 This is not quite right do to averaging, really • U = ½ CV2 = ½ QV

  19. Dielectrics A non-conducting material used in capacitors to increase capacitance. Dielectric constant K = C / C0 Some are polarized

  20. Dielectrics C = K0A / x

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