Capacitors. February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 1

Transcription

1 Capacitors February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 1

2 Review: Capacitance! The definition of capacitance is Capacitance is a property of the capacitor! The capacitance of a parallel plate capacitor depends on the plate area and the plate separation and is given by A is the area of each plate d is the distance between the plates ε A 0 C q = V C = d Physics for Scientists & Engineers 2 2

3 Circuits! An electric circuit consists of wires that connect circuit elements.! These elements can be capacitors.! Other important elements include resistors, inductors, ammeters, voltmeters, diodes, and transistors.! Circuits usually need a power source.! A battery can provide a fixed potential difference commonly called voltage.! An AC power source provides a time-varying potential difference. February 5, 2014 Chapter 24 3

4 Circuit Symbols! Circuit elements are represented by commonly used symbols. February 5, 2014 Chapter 24 4

5 Capacitors in series and in parallel! The equivalent capacitance for n capacitors in parallel is C eq = n C i i=1 =! The equivalent capacitance for n capacitors in series is 1 C eq = n i=1 1 C i! For two capacitors in series: C eq = C C 1 2 C 1 + C 2 = 2/4/14 Physics for Scientists & Engineers 2, Chapter 23 5

6 System of Capacitors PROBLEM! Consider the circuit shown! What is the combined capacitance of this set of five capacitors?! If each capacitor has a capacitance of 5 nf, what is the equivalent capacitance of the arrangement? SOLUTION! This problem can be solved by sequential steps, using the rules for equivalent capacitances of capacitors February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 8

7 System of Capacitors STEP 1! We can see that C 1 and C 2 are in parallel! and that C 3 is also in parallel with C 1 and C 2! We find C 123 = C 1 + C 2 + C 3! and make a new drawing February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 9

8 System of Capacitors STEP 2! We can see that C 4 and C 123 are in series! We find for the equivalent capacitance: 1 C 1234 = 1 C C 4 C 1234 = C 123 C 4 C C 4! and make a new drawing February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 10

9 System of Capacitors STEP 3! We can see that C 5 and C 1234 are in parallel! We find for the equivalent capacitance C = C C 5 = C 123C 4 ( + C 5 = C + C + C 1 2 3)C 4 + C 5 C C 4 C 1 + C 2 + C 3 + C 4! and make a new drawing February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 11

10 System of Capacitors STEP 4: INSERT THE NUMBERS! So the equivalent capacitance of our system of capacitors ( 5+5+5) nf = 8.75 nf February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 12

11 System of Capacitors Step 5: Calculate the charges Reconstructing the circuit in reverse! C 1234 and C 5 are in parallel so they have the same potential difference across them, 12 V.! The charge on C 5 is then:! C 1234 is composed of C 123 and C 4 in series so C 123 and C 4 must have the same charge q 4 : q 5 = C 5 ΔV = 5 nf ( ) 12 V ( )= 60 nc ΔV = ΔV ΔV 4 = q 4 + q 4 = q 1 C 123 C 4 4 C C 4 February 5, 2014 Chapter 24 13

12 System of Capacitors! The charge on C 4 is then: q 4 = ΔV C C ( = ΔV C + C + C 1 2 3)C 4 = 12 V C C 4 C 1 + C 2 + C 3 + C 4 q 4 = 45 nc ( 15 nf) ( 5 nf) ( ) 20 nf! C 123 is equivalent to three capacitors in parallel and it has the same charge as C 4, 45 nc.! The three capacitors C 1, C 2, and C 3 have the same capacitance and the same potential difference across them and the sum of the charge on these three capacitors is 45 nc.! Therefore the charge on C 1, C 2, and C 3 is: q 1 = q 2 = q 3 = 45 nc 3 =15 nc February 5, 2014 Chapter 24 14

13 Energy Stored in Capacitors! A battery must do work to charge a capacitor! We can think of this work as changing the electric potential energy of the capacitor! The differential work dw done by a battery with a potential difference ΔV to put a differential charge dq on a capacitor with capacitance C is dw = ΔV 'dq' = q' C dq'! The total work required to bring the capacitor to its full charge q is W t = dw = q q' C dq' = q 2 C February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 16

14 Energy Density in Capacitors! This work is stored as electric potential energy U = 1 2! We define the electric energy density, u, as the electric potential energy per unit volume u = q 2 C = 1 2 C ΔV ( )2 = 1 2 q ΔV ( ) U volume! Taking the case of an ideal parallel plate capacitor 1 u = U Ad = 2 C( ΔV )2 Ad = C ΔV ( )2 2Ad February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 17

15 Energy Density in Capacitors! Inserting our formula for the capacitance of a parallel plate capacitor we find u = ε 0 A d ΔV 2Ad! Recognizing that ΔV/d is the magnitude of the electric field, E, we obtain an expression for the electric energy density for parallel plate capacitor u = 1 2 ε 0 E 2 ( ) 2 = 1 2 ε 0 ΔV d 2! This result, which we derived for the parallel plate capacitor, holds for the electric potential energy stored in any electric field per unit volume occupied by that field February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 18

16 Thundercloud! Suppose a thundercloud with horizontal dimensions of 2.0 km by 3.0 km hovers at an altitude of 500 m over a flat area.! The cloud carries a charge of 160 C.! The ground has no charge. PROBLEM 1:! What is the potential difference between the cloud and the ground?! SOLUTION 1:! We approximate the cloud-ground systems as a parallel plate capacitor with capacitance: ( ) 2000 m C = ε A 0 d = F/m 500 m ( )( 3000 m) = 0.11 µf February 5, 2014 Chapter 24 19

17 Thundercloud! A parallel plate capacitor has +q on one plate and q on the other plate.! We take +80 C on the clouds and -80 C on the ground so that q = 80 C. ΔV = q C = PROBLEM 2:! Lightning requires an electric field strength of 2.5 MV/m.! Are the conditions right for lightning? SOLUTION 2:! The electric field is: E = ΔV d 80 C 0.11 µf = V = 730 MV 730 MV = 500 m =1.5 MV/m! So field strength is too low, except for trees, towers, etc. February 5, 2014 Chapter 24 20

18 Thundercloud PROBLEM 3:! What is the total electric potential energy contained in the field between the cloud and ground? SOLUTION 3:! The total electric potential energy is: U = 1 2 q ( ΔV )= 1 ( 2 80 C )( V)= J! This is enough energy to charge 500 Chevy Volt batteries. February 5, 2014 Chapter 24 21

19 Capacitors with Dielectrics! So far, we have discussed capacitors with air or vacuum between the plates! However, most real-life capacitors have an insulating material, called a dielectric, between the two plates! The dielectric serves several purposes: Provides a convenient way to maintain mechanical separation between the plates Provides electrical insulation between the plates Allows the capacitor to hold a higher voltage Increases the capacitance of the capacitor Takes advantage of the molecular structure of the dielectric material! Demo February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 22

20 Capacitors with Dielectrics! Placing a dielectric between the plates of a capacitor increases the capacitance of the capacitor by a numerical factor called the dielectric constant, κ! We can express the capacitance of a capacitor with a dielectric with dielectric constant κ between the plates as C =κc air! Where C air is the capacitance of the capacitor without the dielectric! Placing the dielectric between the plates of the capacitor has the effect of lowering the electric field between the plates and allowing more charge to be stored in the capacitor C = q ΔV February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 23

21 Parallel Plate Capacitor with Dielectric! Placing a dielectric between the plates of a parallel plate capacitor modifies the electric field as E = E air κ =! ε 0 is the electric permittivity of free space! ε is the electric permittivity of the dielectric material ε =κε 0 q κε 0 A = q εa! Note that the replacement of ε 0 by ε is all that is needed to generalize our expressions for the capacitance! The potential difference across a parallel plate capacitor is ΔV = Ed = qd κε 0 A! The capacitance is then C = q ΔV = κε A 0 d February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 24

22 Dielectric Strength! The dielectric strength of a material measures the ability of that material to withstand potential difference! If the electric field strength in the dielectric exceeds the dielectric strength, the dielectric will break down and begin to conduct charge between the plates via a spark, which usually destroys the capacitor! A useful capacitor must contain a dielectric that not only provides a given capacitance but also enables the device to hold the required potential difference without breaking down! Capacitors are usually specified in terms of their capacitance and by the maximum potential difference that they are designed to handle February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 25

23 Dielectric Constant! The dielectric constant of vacuum is defined to be 1! The dielectric constant of air is close to 1 and we will use the dielectric constant of air as 1 in our problems! The dielectric constants of common materials are listed below (more are listed in the book in Table 24.1) February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 26

24 Microscopic Perspective on Dielectrics! Let s consider what happens at the atomic and molecular level when a dielectric is placed in an electric field! A polar dielectric material is composed of molecules that have a permanent electric dipole moment! Normally the directions of the electric dipoles are random! When an electric field is applied to these polar molecules, they tend to align with the electric field February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 27

25 Microscopic Perspective on Dielectrics! A nonpolar dielectric material is composed of atoms or molecules that have no inherent electric dipole moment! These atoms are molecules can be induced to have a dipole moment by an external electric field! The electric field acts on the positive and negative charges in the atom or molecule and produce and induced dipole moment February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 28

26 Microscopic Perspective on Dielectrics! In both polar and non-polar dielectrics, the fields resulting from aligned electric dipole moments tend to partially cancel the original electric field + -! The resulting electric field inside the capacitor then is the original field minus the induced field E r = E E d February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 29

27 Example I! A parallel plate capacitor whose capacitance C is 13.5pF is charged by a battery to a potential difference of V =12.5V between its plates. The battery is now disconnected and material with κ=6.5 is slipped between the plates. (a) What is the potential energy before the material is inserted? (b) What is U after the material has been inserted? (a) (b) Key Idea: Because the battery is disconnected, the charge on the capacitor cannot change! February 4, 2014 Physics for Scientists&Engineers 2 30

28 Example II! A parallel plate capacitor whose capacitance C is 13.5pF is charged by a battery to a potential difference of V =12.5V between its plates. The battery is now disconnected and material with κ=6.5 is slipped between the plates. (a) What is the potential energy before the material is inserted? (b) What is U after the material has been inserted? (b) Key Idea: Because the battery is disconnected, the charge on the capacitor cannot change, but the capacitance does change (C--> κc)! February 4, 2014 Physics for Scientists&Engineers 2 31

29 Example III! A parallel plate capacitor whose capacitance C is 13.5pF is charged by a battery to a potential difference of V =12.5V between its plates. The battery is now disconnected and material with κ=6.5 is slipped between the plates. (b) What is U after the material has been inserted? The potential energy decreased by a factor κ. The missing energy, in principle, would be apparent to the person inserting the material. The capacitor would exert a tiny tug on the material and would do work on it, in amount February 4, 2014 Physics for Scientists&Engineers 2 32

30 Capacitance of a Coaxial Cable! Coaxial cables are used to transport signals between devices with minimum interference! A 20.0 m long coaxial cable is composed of a conductor and a coaxial conducting shield around the conductor! The space between the conductor and the shield is filled with polystyrene! The radius of the conductor is mm and the radius of the shield is 2.00 mm! PROBLEM! What is the capacitance of the coaxial cable? February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 35

31 Capacitance of a Coaxial Cable SOLUTION! We can think of the coaxial cable as a cylinder! The dielectric constant of polystyrene is 2.6! We can treat the coaxial cable as a cylindrical capacitor with r 1 = mm and r 2 = 2.00 mm, filled with a dielectric with κ = 2.6! The capacitance of the coaxial cable is C =κ 2πε L 0 ln r 2 r 1 ( ) 2π F/m = 2.6 C = F=1.39 nf ( ) ( ) 20.0 m ln m m February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 36

32 Supercapacitor / Ultracapacitor! Supercapacitors (ultracapacitors) are made using a material with a very large surface area between the capacitor plates! Two layers of activated charcoal are given opposite charge and are separated by an insulating material! This produces a capacitor with capacitance millions of times larger than ordinary capacitors! However, the potential difference can only be 2 to 3 V February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 37