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Last time… Fields, forces, work, and potential

+. +. Last time… Fields, forces, work, and potential. Electric forces and work. Electric potential energy and electric potential. Exam 1. Average = 79% Letter grades indicate how you should interpret this percentage: Average is at the B/BC border. +. +.

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Last time… Fields, forces, work, and potential

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  1. + + Last time… Fields, forces, work, and potential Electric forces and work Electric potential energy and electric potential Physics 208 Lecture 11

  2. Exam 1 • Average = 79% • Letter grades indicate how you should interpret this percentage: • Average is at the B/BC border. Physics 208 Lecture 11

  3. + + Last time… Fields, forces, work, and potential Electric forces and work Electric potential energy and electric potential Physics 208 Lecture 11

  4. Electric field from potential • Said before that • Spell out the vectors: • This works for Usually written Physics 208 Lecture 11

  5. Quick Quiz Suppose the electric potential is constant everywhere. What is the electric field? Positive Negative Increasing Decreasing Zero Physics 208 Lecture 11

  6. Potential from electric field • Electric field can be used to find changes in potential • Potential changes largest in direction of E-field. • Smallest (zero) perpendicular to E-field V=Vo Physics 208 Lecture 11

  7. Equipotential lines • Lines of constant potential • In 3D, surfaces of constant potential Physics 208 Lecture 11

  8. Quick Quiz How does the electric potential outside a uniform infinite sheet of positive charge vary with distance from the sheet? Is constant Increasing as (distance)1 Decreasing as (distance)1 Increasing as (distance)2 Decreasing as (distance)2 Physics 208 Lecture 11

  9. B A E cnst x Electric Potential - Uniform Field 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

  10. Check of simple cases • Previous quick quiz: uniform potential corresponds to zero electric field • Constant electric field corresponds to linear potential Physics 208 Lecture 11

  11. y +Q x Complicated check: point charge • E points opposite to direction of steepest slope • Magnitude proportional to local slope Physics 208 Lecture 11

  12. difficult path easy path Integral along some path, from point on surface to inf. Easy because is same direction as E, 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? Physics 208 Lecture 11

  13. A) Q2>Q1 B) Q2<Q1 C) Q2=Q1 Q1 Q2 R1 R2 Quick quiz So conducting sphere of radius R carrying charge Q is at a potential Two conducting spheres of diff radii connected by long conducting wire. What is approximately true of Q1, Q2? Physics 208 Lecture 11

  14. Connected spheres Charge proportional to radius Surface charge densities? Surface charge density proportional to 1/R • Since both must be at the same potential, • Electric field? • Since , Local E-field proportional to 1/R (1/radius of curvature) Physics 208 Lecture 11

  15. Varying E-fields on conductor • Expect larger electricfields 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. Physics 208 Lecture 11

  16. Potential and charge • Have shown that a conductor has an electric potential, and that potential depends on its charge • For a charged conducting sphere: + + + + + + + Electric potential proportional to total charge + + + + Physics 208 Lecture 11

  17. 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 is Vo is 2Vo is 4Vo depends on shape Physics 208 Lecture 11

  18. Capacitance • Electric potential of any conducting object proportional to its total charge. • C = capacitance • Large capacitance: need lots of charge to change potential • Small capacitance: small charge can change potential. Physics 208 Lecture 11

  19. 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 platesproportional to charge transferred Q Physics 208 Lecture 11

  20. Definition of Capacitance • Same as for single conductor • but V = potential difference between plates • Q = charge transferred between plates • The SI unit 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 Physics 208 Lecture 11

  21. Parallel plate capacitor +Q -Q outer 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 Physics 208 Lecture 11

  22. How did the charge get transferred? DV • Battery has fixed electric potential difference across its terminals • Conducting plates connected to battery terminals by conducting wires. • DVplates = DVbattery across plates • Electrons move • from negative battery terminal to -Q plate • from +Q plate to positive battery terminal • This charge motion requires work • The battery supplies the work Physics 208 Lecture 11

  23. 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 Physics 208 Lecture 11

  24. - + + - - + - + + - + - - + + - - + + - - + - + + - - + - + What is the potential difference? • Electric field between plates • Uniform electric field Etotal Potential difference = V+-V- = (1/q)x(- work to move + charge from + to minus plate) d -Q +Q Physics 208 Lecture 11

  25. What is the capacitance? -Q +Q This is a geometrical factor d Physics 208 Lecture 11

  26. Extracellular fluid Plasma membrane Cytoplasm Human capacitors • Cell membrane: • ‘Empty space’ separating charged fluids (conductors) • ~ 7 - 8 nm thick • In combination w/fluids, acts as parallel-plate capacitor 100 µm Physics 208 Lecture 11

  27. A- K+ Extracellular fluid 7-8 nm V~0.1 V Plasma membrane - - - - - - Cytoplasm Na+ Cl- + + + + + + Modeling a cell membrane • 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 Ionic charge at surfaces of conducting fluids ~ 3x10-4 cm2 100 µm sphere surface area Capacitance: ~0.1µF/cm2 Physics 208 Lecture 11

  28. - - - - - - - - - - - - + + + + + + + + + + + + Cell membrane depolarization A- K+ Extracellular fluid • Cell membrane can reverse potential by opening ion channels. • Potential change ~ 0.12 V • Ions flow through ion channels Channel spacing ~ 10xmembrane thickness (~ 100 channels / µm2 ) • How many ions flow through each channel? 7-8 nm V~0.1 V V~-0.02 V Plasma membrane Cytoplasm Na+ Cl- Charge xfer required Q=CV=(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)x4π(50 µm)2=3.14x106 ion channels Ion flow / channel =(2.6x107 ions) / 3.14x106 channels ~ 7 ions/channel Physics 208 Lecture 11

  29. - - - - - - + + + + + + Cell membrane as dielectric A- K+ Extracellular fluid • Membrane is not really empty • It has molecules inside that respond to electric field. • The molecules in the membrane can be polarized 7-8 nm Plasma membrane Cytoplasm Na+ Cl- Dielectric: insulating materials can respond to an electric field by generating an opposing field. Physics 208 Lecture 11

  30. - - + + 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 E Physics 208 Lecture 11

  31. 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/ k andV = V0/ k • The capacitance increases C = k C0 Eind E0 Physics 208 Lecture 11

  32. - - - - - - + + + + + + Cell membrane as dielectric A- K+ Extracellular fluid • 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? 7-8 nm Plasma membrane Cytoplasm 70 7 0.7 Na+ Cl- C increases by factor of 10 10 times as much charged needed to reach potential Physics 208 Lecture 11

  33. - - - - - - - - - - - - - + + + + + + + + + Charge distributions -Q +Q • -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: • E-field zero inside outer cond. • E-field zero outside outer cond. • No flux -> no charge on outer surface! Physics 208 Lecture 11

  34. Gaussian surface to find E + + + + + + + + + Path to find V Spherical capacitor Charge Q moved from outer to inner sphere Gauss’ law says E=kQ/r2until second sphere Potential difference Along path shown Physics 208 Lecture 11

  35. 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: U = 1/2 oAdE2 Physics 208 Lecture 11

  36. 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 Physics 208 Lecture 11

  37. -q +q pull pull + + + + - - - - d Quick Quiz 1 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? 1)The charge decreases 2)The capacitance increases 3)The electric field increases 4)The voltage between the plates increases 5)The energy stored in the capacitor increases C = e0A/d  C decreases! E= Q/(e0A)  E constant V= Ed  V increases U= QV / 2 Q constant, V increased U increases Physics 208 Lecture 11

  38. Ceq Capacitors in Parallel • Both ends connected together by wire • Add Areas: Ceq = C1+C2 • Share Charge: Qeq = Q1+Q2 • Same voltage: V1 = V2 = Veq 15 V 15 V 15 V C1 C2 10 V 10 V 10 V Physics 208 Lecture 11

  39. +Q -Q + + - - Ceq +Q -Q Capacitors in Series • Same Charge: Q1 = Q2 = Qeq • Share Voltage:V1+V2=Veq • Add d: + + + +Q + + C1 + + - + + + + C2 -Q - + - + + + Physics 208 Lecture 11

  40. Physics 208 Lecture 11

  41. 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 dq U = - pE cos  = - p·E When the dipole is aligned to the field it is minimum U = -pE equilibrium! Physics 208 Lecture 11

  42. 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) Physics 208 Lecture 11

  43. 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 Physics 208 Lecture 11

  44. Capacitance of Parallel Plate Capacitor The electric field from a charged plane of charge per unit area s = Q/A is E = s/2e0 For 2 planes of opposite charge E= s/e0 = Q/(e0A) DV + E - A A + - d E+ E+ E+ E- E- Physics 208 Lecture 11 E- e0=1/(4pke)=8.85x10-12 C2/Nm2

  45. Spherical capacitor Capacitance of Spherical Capacitor: Capacitance of Cylindrical Capacitor: Physics 208 Lecture 11

  46. - - - - - - - - - + + + + + - - - - - E=2E0 E=E0 d d 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 Charge 2Q on plates VA – VB = +2E0 d V =VA – VB = +E0 d + + + + + + + + + Potential difference is proportional to charge: Double Q  Double V EQ, VE, QV Physics 208 Lecture 11

  47. 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 • Due to the ion distribution between the inside and outside of living cells there is a potential difference called resting potential • http://www.cytochemistry.net/Cell-biology/membrane.htm Physics 208 Lecture 11

  48. 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=%22membrane%20channels%22 Physics 208 Lecture 11

  49. 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 Physics 208 Lecture 11

  50. 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 = sC = 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 s = (Q0/A)/F = (10-7 C/cm2)/(105 C/mole) = 1 picomoles/cm2 Current/area =I/A = s/t = (10-7 C/cm2)/(10-3 s) = 100 mA/cm2 Current/channel = IC = (I/A)/sC = = (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 !! Physics 208 Lecture 11

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