Transcript Chapter 26
Last time… Fields, forces, work, and potential + + Electric forces and work Electric potential energy and electric potential Tues. Oct. 9, 2007 Physics 208 Lecture 11 1 30 Exam 1 percentage MEAN=79% 25 # SCORES Exam 1 20 15 10 5 0 20 30 40 50 60 70 80 90 100 PERCENTAGE SCORE Average = 79% Letter grades indicate how you should interpret this percentage: Average is at the B/BC border. Tues. Oct. 9, 2007 Physics 208 Lecture 11 2 Last time… Fields, forces, work, and potential + + Electric forces and work Electric potential energy and electric potential Tues. Oct. 9, 2007 Physics 208 Lecture 11 3 Electric field from potential Said before that dW Fext ds FCoulomb ds dV E ds 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. 9, 2007 dV dV dV E V , , dx dy dz Physics 208 Lecture 11 4 Quick Quiz Suppose the electric potential is constant everywhere. What is the electric field? A) Positive B) Negative C) Increasing D) Decreasing E) Zero Tues. Oct. 9, 2007 Physics 208 Lecture 11 5 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. 9, 2007 Physics 208 Lecture 11 6 Equipotential lines Lines of constant potential In 3D, surfaces of constant potential Tues. Oct. 9, 2007 Physics 208 Lecture 11 7 Quick Quiz How does the electric potential outside a uniform infinite sheet of positive charge vary with distance from the sheet? A. Is constant B. Increasing as (distance)1 C. Decreasing as (distance)1 D. Increasing as (distance)2 E. Decreasing as (distance)2 Tues. Oct. 9, 2007 Physics 208 Lecture 11 8 V VB VA E d d Electric Potential - Uniform Field V E ds VB VA B B A A B E ds A E ds E ds E x B x A E ||ds E cnst B A x Tues. Oct. 9, 2007 Constant E-field corresponds to linearly decreasing (in direction of E) potential Particle gains kinetic energy equal to the potential energy lost Physics 208 Lecture 11 9 Check of simple cases Previous quick quiz: uniform potential corresponds to zero electric field E V constant 0 Constant electric field corresponds to linear potential E V Ex Ex, Ex, Ex Exˆ y z x Tues. Oct. 9, 2007 Physics 208 Lecture 11 10 Complicated check: point charge E V E points opposite to direction of steepest slope Magnitude proportional to local slope Tues. Oct. 9, 2007 Physics 208 Lecture 11 y +Q x 11 Potential of spherical conductor Charge resides on surface, so this is like the spherical charge shell. Found E = keQ / R2 in the radial direction. What is the electric potential of the conductor? V R V R E ds difficult path E ds easy path R Integral along some path, from point on surface to inf. Tues. Oct. 9, 2007 Easy because is same direction as E, E ds E dr Edr Physics 208 Lecture 11 12 Quick quiz V R V R Q E dr k 2 dr r R Q Q k k rR R So conducting sphere of radius R carrying charge Q is at a potential kQ /R Two conducting spheres of diff radii connected by long conducting wire. What is approximatelytrue of Q1, Q2? R1 Q1 Q2 R2 A) Q2>Q1 B) Q2<Q1 C) Q2=Q1 Tues. Oct. 9, 2007 Physics 208 Lecture 11 13 Connected spheres Since both must be at the same potential, kQ1 kQ2 Q1 R1 R1 R2 Q2 R2 Surface charge densities? Q 1 R2 2 4R 2 R1 Electric field? Since E , 2o Tues. Oct. 9, 2007 Charge proportional to radius Surface charge density proportional to 1/R Local E-field proportional to 1/R (1/radius of curvature) Physics 208 Lecture 11 14 Varying E-fields on conductor Expect larger electric fields near the small end. Electric field approximately proportional to 1/(local radius of curvature). Large electric fields at sharp points, just like square (done numerically previously) Fields can be so strong that air ionized and ions accelerated. Tues. Oct. 9, 2007 Physics 208 Lecture 11 15 Potential and charge Have shown that a conductor has an electric potential, and that potential depends on its charge For a charged conducting sphere: + + + + + + + + + + + Tues. Oct. 9, 2007 Q k V R V k Q R R Electric potential proportional to total charge Physics 208 Lecture 11 16 Quick Quiz Consider this conducting object. When it has total charge Qo, its electric potential is Vo. When it has charge 2Qo, its electric potential A. is Vo B. is 2Vo C. is 4Vo D. depends on shape Tues. Oct. 9, 2007 Physics 208 Lecture 11 17 Capacitance Electric potential of any conducting object proportional to its total charge. 1 V Q C C = capacitance Large capacitance: need lots of charge to change potential Small capacitance: small charge can change potential. Tues. Oct. 9, 2007 Physics 208 Lecture 11 18 Capacitors Where did the charge come from? Usually transferred from another conducting object, leaving opposite charge behind A capacitor consists of two conductors Conductors generically called ‘plates’ Charge transferred between plates Plates carry equal and opposite charges Potential difference between plates proportional to charge transferred Q Tues. Oct. 9, 2007 Physics 208 Lecture 11 19 Definition of Capacitance Same as for single conductor 1 V Q C but V = potential difference between plates Q = charge transferred between plates The SIunit of capacitance is the farad (F) 1 Farad = 1 Coulomb / Volt This is a very large unit: typically use mF = 10-6 F, nF = 10-9 F, pF = 10-12 F Tues. Oct. 9, 2007 Physics 208 Lecture 11 20 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. 9, 2007 Physics 208 Lecture 11 21 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. 9, 2007 Physics 208 Lecture 11 22 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. 9, 2007 Physics 208 Lecture 11 23 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. 9, 2007 Physics 208 Lecture 11 + + + + + + + + + + + + + + + +Q Etotal d - -Q 24 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. 9, 2007 Physics 208 Lecture 11 25 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. 9, 2007 Physics 208 Lecture 11 26 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. 9, 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 11 2 3.5 1011 F 35pF ~0.1µF/cm2 27 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 ~ 10xmembrane thickness (~ 100 channels / µm2 ) + - + - + - + - + - + Cytoplasm How many ions flow through each channel? 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 (100 channels/µm2)x4(50 µm)2=3.14x106 ion channels Ion flow / channel =(2.6x107 ions) / 3.14x106 channels ~ 7 ions/channel Tues. Oct. 9, 2007 Physics 208 Lecture 11 28 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. 9, 2007 Physics 208 Lecture 11 29 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. 9, 2007 + E Physics 208 Lecture 11 30 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. 9, 2007 Physics 208 Lecture 11 Eind E0 31 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. 9, 2007 Physics 208 Lecture 11 32 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. 9, 2007 Physics 208 Lecture 11 33 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. 9, 2007 Physics 208 Lecture 11 + + + + + + + + + 34 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. 9, 2007 U = 1/2 oAdE2 Physics 208 Lecture 11 35 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. 9, 2007 Physics 208 Lecture 11 36 -q Quick Quiz 1 pull - + - + - A parallel plate capacitor given a charge q. The plates are then pulled a small distance further apart. Which of the following apply to the situation after the plates have been moved? - d + +q pull + 1)The charge decreases 2)The capacitance increases 3)The electric field increases C = 0A/d C decreases! E= Q/(0A) E constant 4)The voltage between the plates increases V= Ed V increases 5)The energy stored in the capacitor increases U= QV / 2 Q constant, V increased U increases Tues. Oct. 9, 2007 Physics 208 Lecture 11 37 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. 9, 2007 Physics 208 Lecture 11 15 V Ceq 10 V 38 Capacitors in Series Same Charge: Q1 = Q2 = Qeq Share Voltage:V1+V2=Veq Add d: +Q + + + ++ -Q +Q -Q Tues. Oct. 9, 2007 --++ + + ++ - + -+ + + + 0A C d 1 1 1 Ceq C1 C2 C1 +Q + Ceq C2 -Q Physics 208 Lecture 11 - 39 Tues. Oct. 9, 2007 Physics 208 Lecture 11 40 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: t = p x E of magnitude:t= Fa sin pE sin The potential energy is = work done by the torque to rotate dipole: dW = t d U = - pE cos = - p · E When the dipole is aligned to the field it is minimum U = -pE equilibrium! Tues. Oct. 9, 2007 Physics 208 Lecture 11 41 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. 9, 2007 Physics 208 Lecture 11 42 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. 9, 2007 Physics 208 Lecture 11 43 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. 9, 2007 -A - E+ E+ E+ E- E- E- d 44 2 0=1/(4ke)=8.85x10-12 C2/Nm Physics 208 Lecture 11 Spherical capacitor Capacitance of Spherical Capacitor: Capacitance of Cylindrical Capacitor: Tues. Oct. 9, 2007 Physics 208 Lecture 11 45 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. 9, 2007 Physics 208 Lecture 11 46 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. 9, 2007 Physics 208 Lecture 11 47 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. 9, 2007 Physics 208 Lecture 11 48 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. 9, 2007 Physics 208 Lecture 11 49 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 = /t = (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. 9, 2007 Physics 208 Lecture 11 50 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. 9, 2007 Physics 208 Lecture 11 51 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. 9, 2007 Physics 208 Lecture 11 53 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. 9, 2007 Physics 208 Lecture 11 54 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. 9, 2007 Physics 208 Lecture 11 55 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. 9, 2007 Physics 208 Lecture 11 56 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. 9, 2007 Physics 208 Lecture 11 E d 0 E 57 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. 9, 2007 Physics 208 Lecture 11 58 Quick Quiz 3 Suppose the electric potential is constant everywhere. What is the electric field? A) Positive B) Negative C) Zero Tues. Oct. 9, 2007 Physics 208 Lecture 11 59 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. 9, 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 11 60 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. 9, 2007 dV dV dV E V , , dx dy dz Physics 208 Lecture 11 61 Equipotential lines Lines of constant potential In 3D, surfaces of constant potential Tues. Oct. 9, 2007 Physics 208 Lecture 11 62 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. 9, 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 11 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. 9, 2007 Physics 208 Lecture 11 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. 9, 2007 Physics 208 Lecture 11 E d 0 E 65 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. 9, 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 11 66 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. 9, 2007 + + + + + + + + Q o A V d Physics 208 Lecture 11 + + + + + + + + + + + + + + + +Q - Eleft plate + Eright plate = Etotal - - - d -Q 67