10/25/2005 Section 5_2 Conductors empty.doc 1/ Conductors. We have been studying the electrostatics of freespace (i.e., a vacuum).

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1 10/25/2005 Section 5_2 Conductors empty.doc 1/3 5-2 Conductors Reding Assignment: pp We hve been studying the electrosttics of freespce (i.e., vcuum). But, the universe is full of stuff! Q: Does stuff (mteril) ffect our electrosttics knowledge? A: HO: Dielectrics nd Conductors A. Ohm s Lw So, in conductors, pplying n electric field E will cuse current J to form. Q: A: HO: Ohm s Lw

2 10/25/2005 Section 5_2 Conductors empty.doc 2/3 B. Resistnce Q: I thought Ohm s Lw ws R=V/I? A: HO: Resistors C. Perfect Conductors Consider now perfect conductor (i.e., σ = ). Q: Does this men tht current density J is likewise infinite? A: HO: Perfect Conductors D. Kirchoff s Voltge Lw Recll since sttic field E is conservtive: E C d = 0

3 10/25/2005 Section 5_2 Conductors empty.doc 3/3 v V = n 0 HO: Kirchoff s Voltge Lw E. Joule s Lw Conducting mteril will bsorb energy it will het up! HO: Joule s Lw

4 10/25/2005 Dielectrics nd Conductors.doc 1/3 Dielectrics nd Conductors Consider very simple model of n tom: = electron (negtive chrge) = nucleus (positive chrge) Sy n electric field is pplied to this tom. Note the field will pply force on both the positively chrged nucleus nd the negtively chrged electron. However, these forces will move these prticles in opposite directions! Two things my occur. In the first cse, the tom my stretch, but the electron will remin bound to the tom:

5 10/25/2005 Dielectrics nd Conductors.doc 2/3 p - + E Note, n electric dipole hs been creted! For the second cse, the electron my be brek free from the tom, creting positive ion nd free electron. We cll these free chrges, nd the electric field will cuse them to move in opposite directions. u - E + u + Moving chrge! We know wht moving chrge is. Moving chrge J r. is electric current ( ) These two exmples provide simple demonstrtion of wht occurs when n electric field is pplied to some mteril (e.g., plstic, copper, wter, oxygen).

6 10/25/2005 Dielectrics nd Conductors.doc 3/3 1) Mterils where the chrges remin bound (nd thus dipoles re creted) re clled insultor (or dielectric) mterils. 2) Mterils where the electrons re free to move re clled conductors. Of course, mterils consists of molecules with mny electrons, nd in generl some electrons re bound nd some re free. As result, there re no perfect conductors or perfect insultors, lthough some mterils re very close! Additionlly, some mterils re lie between being good conductor or good insultor. We cll these mterils semiconductors (e.g., Silicon).

7 10/25/2005 Ohms Lw.doc 1/3 Ohm s Lw Recll tht positively chrge prticle will move in the direction of n electric field, wheres negtive chre will move in the opposite direction. Both types of chrge, however, result in current moving in the sme direction s the electric field: E u - + u+ J Q: So, the direction of current density J nd electric field E re the sme. The question then is, how re their mgnitudes relted? A: They re relted by Ohm s Lw: ( r) = σ ( r) ( r) J E The sclr vlue σ is clled the mteril s conductivity. Note the units of conductivity re:

8 10/25/2005 Ohms Lw.doc 2/3 σ ( ) ( ) ( ) ( ) ( r) ( ) ( r) J r Amps Volts = 2 E r m m J r Amps m = 2 E r m Volts J Amps = E Volts m J r 1 = E Ohm m 1 In other words, the unit of conductivity is conductnce/unit length. We emphsize tht conductivity σ is mteril prmeter. For exmple, the conductivity of copper is: σ copper 7 1 = 58x10. Ω m nd the conductivity of polyethylene ( plstic) is: σ polyethylene = 15x10. Ω m Note the vst difference in conductivity between these two mterils. Copper is conductor nd polyethylene is n insultor.

9 10/25/2005 Ohms Lw.doc 3/3 Georg Simon Ohm ( ) ws the Germn physicist who in 1827 discovered the lw tht the current flow through conductor is proportionl to the voltge nd inversely proportionl to the resistnce. Ohm ws then professor of mthemtics in Cologne. His work ws coldly received! The Prussin minister of eduction nnounced tht " professor who preched such heresies ws unworthy to tech science." Ohm resigned his post, went into cdemic exile for severl yers, nd then left Prussi nd becme professor in Bvri. From:

10 10/25/2005 Resistors.doc 1/7 Resistors Consider uniform cylinder of mteril with mediocre to poor to σ r = σ. pthetic conductivity ( ) S I J = J ˆ z b z This cylinder is centered on the z-xis, nd hs length. The surfce re of the ends of the cylinder is S. Sy the cylinder hs current I flowing into it (nd thus out of J r. it), producing current density ( ) By the wy, this cylinder is commonly referred to s resistor! Q: Wht is its resistnce R of this resistor, given length, cross-section re S, nd conductivity σ? A: Let s first begin with the circuit form of Ohm s Lw: V R = I

11 10/25/2005 Resistors.doc 2/7 where V is the potentil difference between the two ends of the resistor (i.e., the voltge cross the resistor), nd I is the current through the resistor. From electromgnetics, we know tht the potentil difference V is: nd the current I is: I V = V = ( r b E ) d S b J = ds Thus, we cn combine these expressions nd find resistnce R, E r within the resistor, expressed in terms of electric field ( ) nd the current density J within the resistor: V R = = I b S E J d ds Lets evlute ech integrl in this expression to determine the resistnce R of the device described erlier!

12 10/25/2005 Resistors.doc 3/7 1) The voltge V is the potentil difference V b between point nd point b : V = V = ( r b E ) d Q: But, wht is the electric field E? b A: The electric field within the resistor cn be determined from Ohm s Lw: E = J σ We cn ssume tht the current density is pproximtely constnt cross the cross section of the cylinder: J = J ˆ z Likewise, we know tht the conductivity of the resistor mteril is constnt: = σ σ As result, the electric field within the resistor is: E J = = σ σ J z ˆ

13 10/25/2005 Resistors.doc 4/7 Therefore, integrting in stright line long the z-xis from point to point b, we find the potentil difference V to be: b 1 b V = E d z = Jˆ ˆ z dz z σ J = σ z z z J = σ b dz 2) We likewise know tht the current I through the resistor is found by evluting the surfce integrl: S I = J ds = Jˆ ˆ ds S = J S = JS z z z ds z Therefore, the resistnce R of this prticulr resistor is:

14 10/25/2005 Resistors.doc 5/7 V R = I J 1 = σ JS = σs An interesting result! Consider resistor s sort of clogged pipe. Incresing the cross-sectionl re S mkes the pipe bigger, llowing for more current flow. In other words, the resistnce of the pipe decreses, s predicted by the bove eqution. Likewise, incresing the length simply increses the length of the clog. The current encounters resistnce for longer distnce, thus the vlue of R increses with incresing length. Agin, this behvior is predicted by the eqution shown bove. For exmple, consider the cse where we dd two resistors together: R 1 1 = σs R 2 2 = σs 1 2

15 10/25/2005 Resistors.doc 6/7 We cn view this cse s single resistor with length 1 + 2, resulting in totl resistnce of: R totl = σs 1 2 = + σs σs = R + R 1 2 But, this result is not the lest bit surprising, s the two resistors re connected in series! Now let s consider the cse where two resistors re connected in different mnner: S 1 R1 = σs 1 S 2 R2 = σs 2

16 10/25/2005 Resistors.doc 7/7 We cn view this s single resistor with totl cross sectionl re of S 1 +S 2. Thus, its totl resistnce is: R totl = σ ( S1 + S2) ( S + S ) 1 σ 1 2 = σs1 σs2 = = + R1 R2 Agin, this should be no surprise, s these two resistors re connected in prllel. IMPORTANT NOTE: The result R = σs is vlid only for the resistor described in this hndout. Most importntly, it is vlid σ r = σ ). only for resistor whose conductivity is constnt ( ( ) 1 1 If the conductivity is not constnt, then we must evlute the potentil difference cross the resistor with the more generl expression: V b b = E d b J = d σ

17 10/25/2005 Perfect Conductors.doc 1/2 Perfect Conductors Consider now some current with density J, flowing within some mteril with perfect conductivity (i.e., σ = )! σ = J E =?? Q: Wht is the electric field E within this perfectly conducting mteril? A: Well, we know from Ohm s Lw tht the electric field is to the mteril conductivity nd current density s: E = J σ Thus, s the mteril conductivity pproches infinity, we find: J lim E = = 0 σ σ

18 10/25/2005 Perfect Conductors.doc 2/2 The electric field in within perfectly conducting mteril is lwys equl to zero! This mkes since when you think bout it! Since the mteril offers no resistnce, we cn move chrges through it without hving to pply ny force (i.e., nd electric field). This is just like skter moving cross frictionless ice! I cn continue to move with gret velocity, even though no force is being pplied! Consider wht this mens with regrds to wire mde of perfectly conducting mteril (n often pplied ssumption). The electric potentil difference between either end of perfectly conducting wire is zero! + V = ( r) d = 0 - E C Since the electric field within perfect wire is zero, the voltge cross ny perfect wire is lso zero, regrdless of the current flowing through it.

19 10/25/2005 Kirchoffs Voltge Lw.doc 1/7 Kirchoff s Voltge Lw Consider simple electricl circuit: + v 1 + V i + _ f b R 1 c R 2 d e + v 2 We find tht if the voltge source is on (i.e., V 0), then there will be electric potentil differences (i.e., voltge) between different points of the circuit. This cn only be true if electric fields re present! The electric field in this circuit will look something like this: + _

20 10/25/2005 Kirchoffs Voltge Lw.doc 2/7 So, insted of using circuit theory, let s use our new electromgnetic knowledge to nlyze this circuit. First, consider contour C 1 tht follows the circuit pth. + v 1 + V i + _ f b R 1 C 1 c R 2 d e + v 2 Using this pth, let s evlute the contour integrl: E C 1 d This is most esily done by breking the contour C 1 into six sections: section 1 extends from point to point b, section 2 extends from point b to point c, etc. Thus, the integrl becomes: 1 b c d ( ) ( ) ( ) ( ) E r d = E r d+ E r d+ E r d+ C b c e f ( ) ( ) ( ) E r d+ E r d+ E r d d e f

21 10/25/2005 Kirchoffs Voltge Lw.doc 3/7 Let s evlute ech term individully: Section 1 ( to b) In this section, the contour follows the wire from the voltge source to the first resistor. We know tht the electric field in perfect conductor is zero, nd likewise in good conductor it is very smll. Assuming the wire is in fct mde of good conductor (e.g. copper), we cn pproximte the electric field within the wire (nd thus t every point long section 1) s zero (i.e., Ε = 0). Therefore, this first integrl equls zero! b Ε d = 0 This of course mkes sense! We know tht the electric potentil difference cross wire is zero volts. Section 2 (b to c) In this section, the contour moves through the first resistor. The contour integrl long this section therefore llows us to determine the electric potentil difference cross this resistor. Let s denote this potentil difference s v 1 : c Ε d = V ( Pb ) V ( Pc ) = v1 b

22 10/25/2005 Kirchoffs Voltge Lw.doc 4/7 Section 3 (c to d) Just like section 1, the contour follows wire, nd thus the electric field long this section of the contour is zero, s is the potentil difference between point c nd point d. d Ε c d = 0 Section 4 (d to e) Just like section 2, the contour moves through resistor. The contour integrl for this section is thus equl to the potentil difference cross this second resistor, which we denote s v 2 : e Ε d d = V ( P ) V ( Pe ) = v2 d Section 5 (e to f) Agin, the contour follows conducting wire nd gin, the electric field long the contour nd the potentil difference cross it re both zero: f Ε e d = 0

23 10/25/2005 Kirchoffs Voltge Lw.doc 5/7 V Section 6 (f to ) This finl section of contour C 1 extends through the voltge source, thus the contour integrl of this section provides the electric potentil difference between the two V P V P ). By terminls of the this voltge source (i.e., ( ) ( ) definition, the potentil difference between points nd f is vlue of V volts (i.e., V ( P ) V ( P ) = V ). Therefore, we find tht the contour integrl of section 6 is : Ε f d = V ( P ) V ( P ) = V f f ( V ( P ) V ( Pf )) = Whew! Now let s combine these results to determine the contour integrl for the entire contour C 1. f 1 b c d ( ) ( ) ( ) ( ) E r d = E r d+ E r d+ E r d+ C b c e f ( ) ( ) ( ) E r d+ E r d+ E r d d e f = 0+ v + 0+ v + 0 V = v + v V

24 10/25/2005 Kirchoffs Voltge Lw.doc 6/7 Q: Wit; I ve forgotten, Why re we evluting these contour integrls? A: Remember, since the electric field is sttic, we lso know tht integrl round ny closed contour is zero. Thus, we cn conclude tht: 0 = E d = v1 + v2 V C 1 In other words, we find by performing n electromgnetic nlysis of the circuit, the voltges cross ech circuit element re relted s: v1 + v2 V = 0 Q: You hve wsted my time! Using only Kirchoff s Voltge Lw (KVL), I rrived t precisely the sme result ( v 1 + v 2 V = 0 ). I think the bove eqution is true becuse of KVL, not becuse of your fncy electromgnetic theory! A: It is true tht the result we obtined by integrting the electric field round the circuit contour is likewise pprent from KVL. However, this result is still ttributble to electrosttic physics, becuse KVL is direct result of electrosttics!

25 10/25/2005 Kirchoffs Voltge Lw.doc 7/7 The electrosttic eqution : Ε C d = 0 when pplied to the closed contour of ny circuit, results in Kirchoff s Voltge Lw, i.e.: n v = n 0 where v n re the electric potentil differences cross ech element of circuit loop (i.e., closed contour). Gustv Robert Kirchhoff ( ), Germn physicist, nnounced the lws tht llow clcultion of the currents, voltges, nd resistnces of electricl networks in 1845, when he ws only twenty-one! His other work estblished the technique of spectrum nlysis tht he pplied to determine the composition of the Sun. From

26 10/25/2005 Joules Lw.doc 1/3 Joule s Lw Recll tht the work done on chrge Q by n electric field in moving the chrge long some contour C is: W = Q E d C Q: Sy insted of one chrge Q, we hve stedy strem of chrges (i.e., electric current) flowing long contour C? A: We would need to determine the rte of work per unit time, i.e., the power pplied by the field to the current. Recll lso tht the time derivtive of work is power! dw d P = = Q d dt dt E C Since the electric field is sttic, we cn write: dq P = d dt E C C = I E d

27 10/25/2005 Joules Lw.doc 2/3 But look! The contour integrl we know is equl to the potentil difference V between either end of the contour. Therefore: Look fmilir!? C P = I E d = IV The power delivered to chrges by the field is equl to the current I flowing long the contour, times the potentil difference (i.e., voltge V ) cross the contour. Consider now the power delivered in some volume V, sy the volume of resistor. Recll the electric field hs units of volts/m, nd the current density hs units of mps/m 2. We find therefore tht the dot product of the electric field nd the current density is sclr vlue with units of Wtts/m 3. We cll this sclr vlue the power density: V A W power density = E J = 2 3 m m m Integrting power density over some volume V gives the totl power delivered by the field within tht volume:

28 10/25/2005 Joules Lw.doc 3/3 P = E = = V V V σ ( r) J( r) 2 E dv [ W] 1 σ J dv 2 dv Jmes Prescott Joule ( ), born into well-to-do fmily prominent in the brewery industry, studied t Mnchester under Dlton. At ge twenty-one he published the "I-squred-R" lw which bers his nme. Two yers lter, he published the first determintion of the mechnicl equivlent of het. He becme collbortor with Thomson nd they discovered tht the temperture of n expnding gs flls. The "Joule-Thomson effect" ws the bsis for the lrge refrigertion plnts constructed in the 19th century (but not used by the British brewery industry). Joule ws ptient, methodicl nd devoted scientist; it becme known tht he hd tken thermometer with him on his honeymoon nd spent time ttempting to mesure wter temperture differences t the tops nd bottoms of wterflls. From