Transcript Chapter 26
Last time… d d Potential and electric field d V=Vo Capacitors 1 V Q C Friday Honors Lecture: Prof. R. Moss, Physiology Detecting the body’s electrical signals Tues. Oct. 11, 2007 Physics 208 Lecture 12 1 Parallel plate capacitor +Q outer -Q inner Charge Q moved from right conductor to left conductor Each plate has size Length x Width = Area = A Plate surfaces behave as sheets of charge, each producing E-field d Tues. Oct. 11, 2007 Physics 208 Lecture 12 2 Parallel plate capacitor - Charge only on inner surfaces of plates. E-field inside superposition of E-field from each plate. Constant E-field inside capacitor. + d Tues. Oct. 11, 2007 Physics 208 Lecture 12 3 What is the potential difference? Electric field between plates E left E right /2o /2o /o Uniform electric field Potential difference = V+-V= (1/q)x(- work to move + charge from + to minus plate) 1/q qEd d V Ed d /o Q o A Tues. Oct. 11, 2007 Physics 208 Lecture 12 + + + + + + + + + + + + + + + +Q Etotal d - -Q 4 What is the capacitance? -Q d V V V Q o A +Q V Q /C C o A d This is a geometrical factor d Tues. Oct. 11, 2007 Physics 208 Lecture 12 5 -q Quick Quiz pull - + - + - d +q + pull + An isolated parallel plate capacitor with spacing d has charge q. The plates are pulled a small distance further apart. Which of the following describe situation after motion? A) The charge decreases B) The capacitance increases C) The electric field increases D) The voltage between the plates increases C = 0A/d C decreases! E= q/(0A) E constant V= Ed V increases E) None of these Tues. Oct. 11, 2007 Physics 208 Lecture 12 6 Work done to charge a capacitor Requires work to transfer charge dq from one plate: q dW Vdq dq C Total work required = sum of incremental work: Q W 0 Tues. Oct. 11, 2007 Physics 208 Lecture 12 q Q2 dq C 2C 7 Energy stored in a capacitor Energy stored = work done 2 Q 1 1 2 U QV CV 2C 2 2 Example: parallel plate capacitor 1 U Ado E 2 2 1 Energy density U /Ad o E 2 depends only on field strength 2 Tues. Oct. 11, 2007 Physics 208 Lecture 12 8 Quick Quiz An isolated parallel plate capacitor has a charge q. The plates are then pulled further apart. What happens to the energy stored in the capacitor? -q + + - 1) Increases pull - 2) Decreases d + +q pull + 3) Stays the same Tues. Oct. 11, 2007 Physics 208 Lecture 12 9 Human capacitors Cell membrane: 100 µm ‘Empty space’ separating charged fluids (conductors) ~ 7 - 8 nm thick In combination w/fluids, acts as parallel-plate capacitor Extracellular fluid Plasma membrane Cytoplasm Tues. Oct. 11, 2007 Physics 208 Lecture 12 10 Modeling a cell membrane A- K+ Extracellular fluid + + + + + + 7-8 nm V~0.1 V Plasma membrane - - - - - - Cytoplasm Na+ Cl- Ionic charge at surfaces of conducting fluids Capacitance: o A d Tues. Oct. 11, 2007 Charges are +/- ions instead of electrons Charge motion is through cell membrane (ion channels) rather than through wire Otherwise, acts as a capacitor ~0.1 V ‘resting’ potential 100 µm sphere ~ 3x10-4 cm2 surface area 8.8510 12 F /m4 5010 m 6 8 109 m Physics 208 Lecture 12 2 3.5 1011 F 35pF ~0.1µF/cm2 11 Cell membrane depolarization K+ A- Cell membrane can reverse potential by opening ion channels. Potential change ~ 0.12 V Extracellular fluid - + - + - + - + - + + 7-8 nm Ions flow through ion channels V~0.1 V V~-0.02 V Plasma membrane Channel spacing ~ 10x membrane thickness (~ 100 channels / µm2 ) + - + - + - + - + - + - Cytoplasm Na+ Cl- Charge xfer required Q=CV=(35 pF)(0.12V) =(35x10-12 C/V)(0.12V) = 4.2x10-12 Coulombs 1.6x10-19 C/ion -> 2.6x107 ions flow This is an electric current! Tues. Oct. 11, 2007 Physics 208 Lecture 12 12 Charge motion Cell membrane capacitor: ~70 ions flow through each ion channel to depolarize membrane Occurs in ~ 1 ms = 0.001 sec. This is a current, units of Coulombs / sec 1 Coulomb / sec = 1 Amp Tues. Oct. 11, 2007 Physics 208 Lecture 12 13 Electric Current Electric current = I = amount of charge per unit time flowing through a plane perpendicular to charge motion SI unit: ampere 1 A = 1 C / s Depends on sign of charge: + charge particles: current in direction of particle motion is positive - charge particles: current in direction of particle motion is negative Tues. Oct. 11, 2007 Physics 208 Lecture 12 14 Quick Quiz An infinite number of positively charged particles are uniformly distributed throughout an otherwise empty infinite space. A spatially uniform positive electric field is applied. The current due to the charge motion A. increases with time B. decreases with time C. is constant in time D. Depends on field Tues. Oct. 11, 2007 Constant force qE Produces constant accel. qE/m Velocity increases v(t)=qEt/m Charge / time crossing plane increases with time Physics 208 Lecture 12 15 Current in a wire Battery produces E-field in wire Tues. Oct. 11, 2007 Current flows in response to E-field Physics 208 Lecture 12 16 But experiment says… Current constant in time Proportional to voltage 1 I V R Also written J R = resistance (unit Ohm = ) 1 V J = current density = I / (cross-section area) = resistivity = R x (cross-section area) / (length) Resistivity is independent of shape Tues. Oct. 11, 2007 Physics 208 Lecture 12 17 Charge motion with collisions Suppose space not empty, but has various fixed objects. The charged particles would then collide with these objects. Assumption: after collision, charged particle equally likely to move in any direction. x x x x x x x Before collision Tues. Oct. 11, 2007 x After collision Physics 208 Lecture 12 18 Quick Quiz Two cylindrical conductors are made from the same material. They are of equal length but one has twice the diameter of the other. 1 2 A. R1 < R2 B. R1 = R2 C. R1 > R2 Current flow ~ uniform. More cross-sectional area means more current flowing -less resistance. Tues. Oct. 11, 2007 Physics 208 Lecture 12 A R L 19 Resistors Circuits Physical layout Schematic layout Tues. Oct. 11, 2007 Physics 208 Lecture 12 20 Quick Quiz Which bulb is brighter? A. A B. B C. Both the same Current through each must be same Conservation of current (Kirchoff’s current law) Charge that goes in must come out Tues. Oct. 11, 2007 Physics 208 Lecture 12 21 I2 Current conservation Iin I1 I3 I1=I2+I3 I1 I3 Iout Iout = Iin Tues. Oct. 11, 2007 I2 I1+I2=I Physics 208 Lecture 12 3 22 Quick Quiz How does brightness of bulb B compare to that of A? A. B brighter than A B. B dimmer than A C. Both the same Battery maintain constant potential difference Extra bulb makes extra resistance -> less current Tues. Oct. 11, 2007 Physics 208 Lecture 12 23 Resistors in Series I1 = I 2 = I Potentials add V = V1 + V2 = IR1 + IR2 = = I (R1+R2) The equivalent resistance Req = R1+R2 2 resistors in series: RL Like summing lengths R R = Tues. Oct. 11, 2007 2R L R A Physics 208 Lecture 12 24 Quick Quiz What happens to the brightness of the bulb B when the switch is closed? A. Gets dimmer B. Gets brighter C. Stays same D. Something else Battery is constant voltage, not constant current Tues. Oct. 11, 2007 Physics 208 Lecture 12 25 Quick Quiz What happens to the brightness of the bulb A when the switch is closed? A. Gets dimmer B. Gets brighter C. Stays same D. Something else Tues. Oct. 11, 2007 Physics 208 Lecture 12 26 Resistors in Parallel V = V1 = V2 I = I 1 + I 2 (lower resistance path has higher current) R R Equivalent Resistance R/2 Add areas L R A Tues. Oct. 11, 2007 Physics 208 Lecture 12 27 Quick Quiz Calculate the voltage across each resistor if the battery has potential V0= 22 V. R1=1 V0 R2=10 R1 and R2 are in series •R12 = R1 + R2 •V12 = V1 + V2 •I12 = I1 = I2 •V1 = I1R1 •V2 = I2R2 = 11 = V0 = 22 Volts = V12/R12 = 2 Amps = 2 x 1 = 2 Volts = 2 x 10 = 20 Volts V0 R12 Check: V1 + V2 = V12 Tues. Oct. 11, 2007 Physics 208 Lecture 12 28 Quick Quiz What happens to the current through resistor 2 when the switch is closed? Switch open: 1. Increases Ibattery = /R2 = I2 2. Remains Same 3. Decreases Follow Up:What happens to the current through the battery? 1. Increases Switch closed: Ibattery = /Req = (1/R2+1/R3) 2. Remains Same 3. Decreases I =I +I battery Tues. Oct. 11, 2007 Physics 208 Lecture 12 2 3 29 How did the charge get transferred? Battery has fixed electric potential difference across its terminals Conducting plates connected to battery terminals by conducting wires. Vplates = Vbattery across plates Electrons move V from negative battery terminal to -Q plate from +Q plate to positive battery terminal This charge motion requires work The battery supplies the work 1 V Q C Tues. Oct. 11, 2007 Physics 208 Lecture 12 30 Capacitors in Parallel 0A C d Both ends connected together by wire • Same voltage: V1 = V2 = Veq Add Areas: Ceq = C1+C2 Share Charge: Qeq = Q1+Q2 15 V 15 V C1 C2 10 V 10 V Tues. Oct. 11, 2007 Physics 208 Lecture 12 15 V Ceq 10 V 31 Capacitors in Series Same Charge: Q1 = Q2 = Qeq Share Voltage:V1+V2=Veq Add d: +Q + + + ++ -Q +Q -Q --++ + + ++ - + -+ + + + Tues. Oct. 11, 2007 0A C d 1 1 1 Ceq C1 C2 C1 +Q + Ceq C2 -Q Physics 208 Lecture 12 - 32 Energy density The energy stored per unit volume is U/(Ad) = 1/2 oAdE2 This is a fundamental relationship for the energy stored in an electric field valid for any geometry and not restricted to capacitors Tues. Oct. 11, 2007 Physics 208 Lecture 12 33 Current Density Alternative expression of Ohm’s law 2 qE nq I nqvd A nq AE A m m I nq 2 E J E A m J = current density, = conductivity Independent of sample geometry Local relation between J and E Tues. Oct. 11, 2007 Physics 208 Lecture 12 34 What is the drift velocity? Copper wire: A = 3.31 x 10-6 m2 and = 8.95 g/cm3 Copper molar mass m = 63.5 g/mol The volume occupied by a mol V=m/ = 7.09 cm3 1 mol contains NA = 6.02 x 1023 atoms Density of electrons n = NA/V = 8.49 x 1028 m-3 I Hence for I = 10 A vd 2.22 104 m / s nqA Even if the drift velocity is so low the effect of flipping a switch is instantaneous since electrons do not have to travel from the light switch to the light bulb in order for the light to operate but they are already in the light filament Tues. Oct. 11, 2007 Physics 208 Lecture 12 35 Electric Current Electric current rate of flow of charge through some region of space Charge moves in response to a force Usually electric field, corresponding potential difference SI unit: ampere 1 A = 1 C / s Average current: Q charges moving perpendicular to a surface of area A cross it in time t Instantaneous value Convention: the current has the direction of the positive charge carriers Tues. Oct. 11, 2007 Physics 208 Lecture 12 36 Charge Carrier Motion in a Conductor Electric field should produce acceleration. Experiment: I = V/R (Ohm’s law) Despite the collisions with the atoms of the conductor the electrons drift in the opposite direction of the field with a small net velocity Tues. Oct. 11, 2007 Physics 208 Lecture 12 37 Current Density Ohm’s Law: for many materials, the current density is proportional to the electric field producing the current J=E = conductivity of the conductor Materials that obey Ohm’s law are said to be ohmic When an electric field is applied free electrons experience an acceleration (opposite to E): a = F/m = eE/m Calling t the mean time between 2 collisions electron-atoms electrons achieve a drift velocity: Hence we obtain for the conductivity: Tues. Oct. 11, 2007 eE eV vd at m mL Q nALe nAe2 V I E t L / vd m L Physics 208 Lecture 12 38 Resistance The potential difference maintained across the wire of length L creates an electric field V = EL Hence J = E = V/L = I/A And L V I RI A The constant of proportionality is called the resistance of the conductor and SI units of resistance are ohms () 1=1V/A Resistance in a circuit arises due to collisions between the electrons carrying the current with fixed atoms Tues. Oct. 11, 2007 Physics 208 Lecture 12 39 Quick Quiz 1 Two cylindrical resistors are made from the same material. They are of equal length but one has twice the1 diameter 2 of the other. 1. R1 > R2 2. R1 = R2 3. R1 < R2 Smaller diameter not as easy for carriers to flow through Tues. Oct. 11, 2007 Physics 208 Lecture 12 40 Electrical Power A battery is used to establish an electric current in a conductor As a charge moves from a to b, the electric potential energy of the system increases by U=QV The chemical energy of the battery decreases by the same amount As the charge moves through the resistor (c to d), the system loses the same amount of energy due collisions with atoms of the resistor This energy goes into vibrational motion of the atoms in the resistor and energy irradiated and heat transfer to air Tues. Oct. 11, 2007 Physics 208 Lecture 12 41 Cell membrane as dielectric K+ A- Extracellular fluid + + + + + + 7-8 nm Plasma membrane Membrane is not really empty It has molecules inside that respond to electric field. The molecules in the membrane can be polarized - - - - - Cytoplasm Na+ Cl- Dielectric: insulating materials can respond to an electric field by generating an opposing field. Tues. Oct. 11, 2007 Physics 208 Lecture 12 42 Effect of E-field on insulators If the molecules of the dielectric are non-polar molecules, the electric field produces some charge separation This produces an induced dipole moment + E=0 Tues. Oct. 11, 2007 + E Physics 208 Lecture 12 43 Dielectrics in a capacitor An external field can polarize the dielectric The induced electric field is opposite to the original field The total field and the potential are lower than w/o dielectric E = E0/ kand V = V0/ k The capacitance increases C = k C0 Tues. Oct. 11, 2007 Physics 208 Lecture 12 Eind E0 44 Cell membrane as dielectric K+ A- Extracellular fluid + + + + + + 7-8 nm Plasma membrane - - - - - - Without dielectric, we found 7 ions/channel were needed to depolarize the membrane. Suppose lipid bilayer has dielectric constant of 10. How may ions / channel needed? Cytoplasm Na+ A. 70 Cl- B. 7 C. 0.7 C increases by factor of 10 10 times as much charged needed to reach potential Tues. Oct. 11, 2007 Physics 208 Lecture 12 45 Charge distributions -Q arranged on inner/outer surfaces of outer sphere. + Charge enclosed by Guassian + surface = -Q+Q=0 + + Flux through Gaussian surface=0, -> E-field=0 outside Another Gaussian surface: - -Q - + + +Q + + + - - - E-field zero inside outer cond. E-field zero outside outer cond. No flux -> no charge on outer surface! Tues. Oct. 11, 2007 Physics 208 Lecture 12 46 Spherical capacitor Charge Q moved from outer to inner sphere Gauss’ law says E=kQ/r2 until second sphere Potential difference V b E ds a Along path shown Gaussian surface to find E b 1 1 kQ 1 V 2 kQ kQ a b ra a r b 1 1 1 Q C k a b V Path to find V Tues. Oct. 11, 2007 Physics 208 Lecture 12 + + + + + + + + + 47 Tues. Oct. 11, 2007 Physics 208 Lecture 12 48 Electric Dipole alignment The electric dipole moment (p) along line joining the charges from –q to +q Magnitude: p = aq The dipole makes an angle with a uniform external field E The forces F=qE produce a net torque: = p x E of magnitude:= Fa sin pE sin The potential energy is = work done by the torque to rotate dipole: dW = d U = - pE cos = - p · E When the dipole is aligned to the field it is minimum U = -pE equilibrium! Tues. Oct. 11, 2007 Physics 208 Lecture 12 49 Polar Molecules Molecules are said to be polarized when a separation exists between the average position of the negative charges and the average position of the positive charges Polar molecules are those in which this condition is always present (e.g. water) Tues. Oct. 11, 2007 Physics 208 Lecture 12 50 How to build Capacitors Roll metallic foil interlaced with thin sheets of paper or Mylar Interwoven metallic plates are immersed in silicon oil Electrolitic capacitors: electrolyte is a solution that conducts electricity by virtue of motion of ions contained in the solution Tues. Oct. 11, 2007 Physics 208 Lecture 12 51 Capacitance of Parallel Plate Capacitor The electric field from a charged plane of charge per unit area = Q/A is E = /20 For 2 planes of opposite charge V E= /0 = Q/(0A) +E A + Tues. Oct. 11, 2007 -A - E+ E+ E+ E- E- E- d 52 2 0=1/(4ke)=8.85x10-12 C2/Nm Physics 208 Lecture 12 Spherical capacitor Capacitance of Spherical Capacitor: Capacitance of Cylindrical Capacitor: Tues. Oct. 11, 2007 Physics 208 Lecture 12 53 Charge, Field, Potential Difference Capacitors are devices to store electric charge and energy They are used in radio receivers, filters in power supplies, electronic flashes Charge Q on plates V =VA – VB = +E0 d + + + + + E=E0 d Charge 2Q on plates VA – VB = +2E0 d + + + + + + + + + - E=2E0 d - Potential difference is proportional to charge: Double Q Double V EQ, VE, QV Tues. Oct. 11, 2007 Physics 208 Lecture 12 54 Human capacitors: cell membranes Membranes contain lipids and proteins Lipid bilayers of cell membranes can be modeled as a conductor with plates made of polar lipid heads separated by a dielectric layer of hydrocarbon tails QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. Due to the ion distribution between the inside and outside of living cells there is a potential difference called resting potential Tues. Oct. 11, 2007 Physics 208 Lecture 12 55 http://www.cytochemistry.net/Cell-biology/membrane.htm Human capacitors: cell membranes The inside of cells is always negative with respect to the outside and the DV ≈ 100 V and 0.1 V Cells (eg. nerve and muscle cells) respond to electrical stimuli with a transient change in the membrane potential (depolarization of the membrane) followed by a restoration of the resting potential. Remember EKG! The Nobel Prize in Chemistry (2003) for fundamental discoveries on how water and ions move through cell membranes. - Peter Agre discovered and characterized the water channel protein - Roderick MacKinnon has elucidated the structural and mechanistic basis for ion channel function. http://nobelprize.org/nobel_prizes/chemistry/laureates/2003/chemadv03.pdf#search=%2 2membrane%20channels%22 Tues. Oct. 11, 2007 Physics 208 Lecture 12 56 Ion channels Membrane channels are protein/sugar/fatty complexes that act as pores designed to transport ions across a biological membrane In neurons and muscle cells they control the generation of electrical signals They exist in a open or closed state when ions can pass through the channel gate or not Voltage-gated channels in nerves and muscles open due to a stimulus detected by a sensor Eg: in muscles there are 50-500 Na channels per mm2 on membrane surface that can be opened by a change in electric potential of membrane for ~1 ms during which about 103 Na+ ions flow into the cell through each channel from the intracellular medium. The gate is selective: K+ ions are 11 times less likely to cross than Na+ Na channel dimension and the interaction with negative O charges in its interior selects Na+ ions Tues. Oct. 11, 2007 Physics 208 Lecture 12 57 How much charge flow? How much charge (monovalent ions) flow through each open channel making a membrane current? Data: Resting potential = 0.1 V Surface charge density: Q0/A = 0.1 mC/cm2 surface density of channels = C = 10 channels/mm2 = 109 channels/cm2 1 mole of a monovalent ion corresponds to the charge F = Faraday Constant = NA e = 6.02 x 1023 x 1.6 x 10-19 ≈ 105 C/mole NA = Avogadro’s number = number of ions in a mole Hence surface charge density = (Q0/A)/F = (10-7 C/cm2)/(105 C/mole) = 1 picomoles/cm2 Current/area =I/A = / = (10-7 C/cm2)/(10-3 s) = 100 mA/cm2 Current/channel = IC = (I/A)/C = = (10-4 A/cm2)/(109 channel/cm2) = = 0.1 pA/channel (10-13 C/s/channel)/(105 C/mole) = 10-18 moles of ions/s in a channel (10-18 moles of ions/s)/(6.02 x 1023 ions/mole)= 6 x 105 ions/s !! Tues. Oct. 11, 2007 Physics 208 Lecture 12 58 Honors lecture this Friday Superconductivity, by yours truly. 12:05 pm, 2241 Chamberlin Everyone welcome! HW 2 due Thursday midnite Lab 2 this week - bring question sheet available on course web site. Tues. Oct. 11, 2007 Physics 208 Lecture 12 59 Capacitance and Dielectrics 1. 2. This lecture: Definition of capacitance Capacitors Combinations of capacitors in circuits Energy stored in capacitors Dielectrics in capacitors and their polarization Cell Membranes From previous lecture: Gauss’ Law and applications Electric field calculations from Potential Charge distribution on conductors Rectangular conductor (40 electrons) Edges are four lines Charge concentrates at corners Equipotential lines closest together at corners Potential changes faster near corners. Electric field larger at corners. Tues. Oct. 11, 2007 Physics 208 Lecture 12 61 Capacitors with Dielectrics Placing a dielectric between the plates increases the capacitance: Dielectric constant (k > 1) C = k C0 Capacitance with dielectric Capacitance without dielectric The dielectric reduces the potential difference V = V0/ k Tues. Oct. 11, 2007 Physics 208 Lecture 12 62 Dielectrics – An Atomic View Molecules in a dielectric can be modeled as dipoles The molecules are randomly oriented in the absence of an electric field When an external electric field is applied, this produces a torque on the molecules The molecules partially align with the electric field (equilibrium) The degree of alignment depends on the magnitude of the field and on the temperature Tues. Oct. 11, 2007 Physics 208 Lecture 12 63 The Electric Field qo dV FCoulomb d F dV Coulomb d qo E is the Electric Field E d It is independent of the test charge, just like the electric potential It is a vector, with a magnitude and direction, When potential arises from other charges, E = Coulomb force per unit charge on a test charge due to interaction with the other charges. We’ll see later that E-fields in electromagnetic waves exist w/o charges! Tues. Oct. 11, 2007 Physics 208 Lecture 12 64 Electric field and potential Said before that dV E d Electric field strength/direction shows how the potential changesin different directions For example, Potential decreases in direction of local E field at rate E Potential increases in direction opposite to local E-field at rate potential constant in direction perpendicular to local E-field Tues. Oct. 11, 2007 Physics 208 Lecture 12 E d 0 E 65 Potential from electric field dV E d Electric field can be V Vo E d d used to find changes V V d o in potential Potential changes E largest in direction of E-field. d V=Vo Smallest (zero) perpendicular to V Vo E d E-field Tues. Oct. 11, 2007 Physics 208 Lecture 12 66 Quick Quiz 3 Suppose the electric potential is constant everywhere. What is the electric field? A) Positive B) Negative C) Zero Tues. Oct. 11, 2007 Physics 208 Lecture 12 67 V VB VA E d d Electric Potential - Uniform Field + V E ds V V B A B B A A B E ds A E ds E ds Ex E ||ds E cnst B A x Tues. Oct. 11, 2007 Constant E-field corresponds to linearly increasing electric potential The particle gains kinetic energy equal to the potential energy lost by the charge-field system Physics 208 Lecture 12 68 Electric field from potential Said before that dV E d Spell out the vectors: for This works dV Ex dx Ey dy Ez dz dV dV dV Ex , Ey , Ez dx dy dz Usually written Tues. Oct. 11, 2007 dV dV dV E V , , dx dy dz Physics 208 Lecture 12 69 Equipotential lines Lines of constant potential In 3D, surfaces of constant potential Tues. Oct. 11, 2007 Physics 208 Lecture 12 70 Electric Field and equipotential lines for + and - point charges The E lines are directed away from the source charge A positive test charge would be repelled away from the positive source charge Tues. Oct. 11, 2007 The E lines are directed toward the source charge A positive test charge would be attracted toward the negative source charge Blue dashed lines are equipotential Physics 208 Lecture 12 71 The Electric Field qo dV FCoulomb d F dV Coulomb d qo E is the Electric Field E d It is independent of the test charge, just like the electric potential It is a vector, with a magnitude and direction, When potential arises from other charges, E = Coulomb force per unit charge on a test charge due to interaction with the other charges. We’ll see later that E-fields in electromagnetic waves exist w/o charges! Tues. Oct. 11, 2007 Physics 208 Lecture 12 72 Electric field and potential Said before that dV E d Electric field strength/direction shows how the potential changesin different directions For example, Potential decreases in direction of local E field at rate E Potential increases in direction opposite to local E-field at rate potential constant in direction perpendicular to local E-field Tues. Oct. 11, 2007 Physics 208 Lecture 12 E d 0 E 73 Electric Field and equipotential lines for + and - point charges The E lines are directed away from the source charge A positive test charge would be repelled away from the positive source charge Tues. Oct. 11, 2007 The E lines are directed toward the source charge A positive test charge would be attracted toward the negative source charge Blue dashed lines are equipotential Physics 208 Lecture 12 74 Qinner Qouter Q = Qinner+Qouter Total E-field = Eleft plate+Eright plate In between plates, Q Eleft plate inner outer 2o 2o 2Ao inner outer Q Eright plate 2o 2o 2Ao Q Etotal Eleft plate Eright plate o A right plate Q Qd V E ds left plate o A dx o A C Tues. Oct. 11, 2007 + + + + + + + + Q o A V d Physics 208 Lecture 12 + + + + + + + + + + + + + + + +Q - Eleft plate + Eright plate = Etotal - - - d -Q 75 Drift Speed Conductor of cross-sectional area A and length x n = # of charge carriers per unit volume nA x = # of charge carriers in this elementary volume total charge = number of carriers x charge of carrier q Q = (nA x)q Average current: Iav = Q/ t = nqvdA where drift speed at which carriers move is vd = x / t Current density 2 Magnitude J = I/A (A/m ) J = nqvd in the direction of carriers Tues. Oct. 11, 2007 Physics 208 Lecture 12 positive charge 76 How small is the drift velocity? Copper wire: A = 3.31 x 10-6 m2 and = 8.95 g/cm3 Copper molar mass m = 63.5 g/mol The volume occupied by a mol V=m/ = 7.09 cm3 1 mol contains NA = 6.02 x 1023 atoms Density of electrons n = NA/V = 8.49 x 1028 m-3 I Hence for I = 10 A vd 2.22 104 m / s nqA Even if the drift velocity is so low the effect of flipping a switch is instantaneous since electrons do not have to travel from the light switch to the light bulb in order for the light to operate but they are already in the light filament Tues. Oct. 11, 2007 Physics 208 Lecture 12 77 Work done and energy stored During the charging of a capacitor, when a charge q is on the plates, the work needed to transfer further dq from one plate to the other is: The total work required to charge the capacitor is The energy stored in any capacitor is: For a parallel capacitor: Tues. Oct. 11, 2007 U = 1/2 oAdE2 Physics 208 Lecture 12 78 Drift Velocity Electric field produces accel a=qE/m velocity v = vo + qEt / m vo= velocity after last collision t = time since last collision Lots of particles: average over all particle velocities vave = (vo)ave + qEtave / m =0 vave = qE / m vd= Drift velocity= qE / m x =ave. time since last collision Tues. Oct. 11, 2007 Physics 208 Lecture 12 x x x x x 79 Drift Speed n = # of charge carriers per unit volume nA x = # of charge carriers in test volume (length x charge in volume = (# carriers) x (charge of carrier q) Q = (nA x)q Cross-sectional area A Average current: Iav = Q/ t = nqvdA x / t = vd vd= Drift velocity= qE / m Tues. Oct. 11, 2007 Physics 208 Lecture 12 80 Resistance I = current = nqvdA 2 vd= Drift velocity= /m qE qE nq I nq AE A m m Constant E-field -> E=V/L (L=sample length) qE nq2 A I nq A V V IR m m L nq2 A 1 R m L Voltage proportional to current! This is Ohm’s law Tues. Oct. 11, 2007 Physics 208 Lecture 12 81 Producing an electric current To make an electric current, must get charges to move. To more charges requires a force Force on a charge can be produced by an electric field. Tues. Oct. 11, 2007 Physics 208 Lecture 12 82 Energy density The energy stored per unit volume is U/(Ad) = 1/2 oE2 This is a fundamental relationship for the local energy stored in an electric field Not restricted to capacitors: E-field can be from any source Interpretation: energy is stored in the field Tues. Oct. 11, 2007 Physics 208 Lecture 12 83 Resistivity Resistivity A R L Independent of sample geometry SI units Ω-m Tues. Oct. 11, 2007 Physics 208 Lecture 12 84