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12/4/2016 Advanced Physics Capacitance  Chapter 25 – Problems 1, 3, 8, (17), 19, (33), 39, 40 & 49.

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Presentation on theme: "12/4/2016 Advanced Physics Capacitance  Chapter 25 – Problems 1, 3, 8, (17), 19, (33), 39, 40 & 49."— Presentation transcript:

1 12/4/2016 Advanced Physics Capacitance  Chapter 25 – Problems 1, 3, 8, (17), 19, (33), 39, 40 & 49.

2 22/4/2016 a +Q b -Q 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

3 32/4/2016 1.A capacitor is basically two parallel conducting plates with insulating material in between. The capacitor doesn’t have to look like metal plates. Capacitor for use in high-performance audio systems. 2.When a capacitor is connected to an external potential, charges flow onto the plates and create a potential difference between the plates. + - - - 3.Capacitors in circuits symbols analysis follow from conservation of energy conservation of charge

4 42/4/2016 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 52/4/2016 A A +Q -Q d 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,  0 is a constant called the permittivity of free space,  0 = 8.85  10 -12 C 2 /N·m 2  0 = 8.85  10 -12 C 2 /N·m 2

6 62/4/2016 A parallel plate capacitor has plates 2.00 m 2 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 Problem: parallel-plate capacitor

7 72/4/2016 A parallel plate capacitor has plates 2.00 m 2 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 Given:  V=10,000 V A = 2.00 m 2 d = 5.00 mm Find: C=? Q=? Solution: 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 82/4/2016 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 C1C1 C2C2 C3C3 C5C5 C1C1 C2C2 C3C3 C4C4

9 92/4/2016 +Q 1 Q1Q1 C1C1 V=V ab a b +Q 2 Q2Q2 C2C2 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, C eq would be

10 102/4/2016 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)

11 112/4/2016 A 3  F capacitor and a 6  F capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited. Problem: parallel combination of capacitors

12 122/4/2016 A 3  F capacitor and a 6  F capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited. +Q 1 Q1Q1 C1C1 V=V ab a b +Q 2 Q2Q2 C2C2 Given: V = 18 V C 1 = 3  F C 2 = 6  F Find: C eq =? Q=? First determine equivalent capacitance of C 1 and C 2 : Next, determine the charge

13 132/4/2016 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, C eq would be +Q 1 Q1Q1 C1C1 +Q 2 Q2Q2 C2C2 V=V ab a c b

14 142/4/2016 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)

15 152/4/2016 A 3  F capacitor and a 6  F capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance. Problem: series combination of capacitors

16 162/4/2016 A 3  F capacitor and a 6  F capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance and total charge deposited. +Q 1 Q1Q1 C1C1 +Q 2 Q2Q2 C2C2 V=V ab a c b Given: V = 18 V C 1 = 3  F C 2 = 6  F Find: C eq =? Q=? First determine equivalent capacitance of C 1 and C 2 : Next, determine the charge

17 Series / Parallel Combination Example 172/4/2016

18 Energy stored in a capacitor C = Q / V Q = V * C V = U / Q U = V * Q U = V * V * C = CV 2 This is not quite right do to averaging, really U = ½ CV 2 = ½ QV 182/4/2016

19 Dielectrics A non-conducting material used in capacitors to increase capacitance. Dielectric constant K = C / C 0 Some are polarized 192/4/2016

20 Dielectrics C = K  0 A / x 202/4/2016


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