1 / 84

Understanding Capacitors: Potential, Electric Field, and Capacitance

This lecture explains the concepts of potential difference, electric field, and capacitance in parallel-plate capacitors and applies them to understand the behavior of human capacitors and cell membranes. It also discusses the relationship between current and resistivity in wires.

katiej
Download Presentation

Understanding Capacitors: Potential, Electric Field, and Capacitance

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. V=Vo Last time… Potential and electric field Capacitors Friday Honors Lecture: Prof. R. Moss, Physiology Detecting the body’s electrical signals Physics 208 Lecture 12

  2. 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 12

  3. 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 12

  4. - + + - - + - + + - + - - + + - - + + - - + - + + - - + - + 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 12

  5. What is the capacitance? -Q +Q This is a geometrical factor d Physics 208 Lecture 12

  6. -q +q pull pull + + + + - - - - d Quick Quiz 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 E) None of these C = e0A/d  C decreases! E= q/(e0A)  E constant V= Ed  V increases Physics 208 Lecture 12

  7. Work done to charge a capacitor • Requires work to transfer charge dq from one plate: • Total work required = sum of incremental work: Physics 208 Lecture 12

  8. Energy stored in a capacitor • Energy stored = work done • Example: parallel plate capacitor Energy density depends only on field strength Physics 208 Lecture 12

  9. + + + + - - - - -q +q d pull pull 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? 1) Increases 2) Decreases 3) Stays the same Physics 208 Lecture 12

  10. 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 12

  11. 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 12

  12. - - - - - - - - - - - - + + + + + + + + + + + + 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 ~ 10x membrane thickness (~ 100 channels / µm2 ) 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 This is an electric current! Physics 208 Lecture 12

  13. 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 Physics 208 Lecture 12

  14. Electric Current • Electric current = I = amount of charge per unit time flowing through a plane perpendicular to charge motion • SI unit: ampere1 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 Physics 208 Lecture 12

  15. 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 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

  16. Current in a wire • Battery produces E-field in wire • Current flows in response to E-field Physics 208 Lecture 12

  17. But experiment says… • Current constant in time • Proportional to voltage • R = resistance (unit Ohm = ) • Also written • J = current density = I / (cross-section area) •  = resistivity = R x (cross-section area) / (length) • Resistivity is independent of shape Physics 208 Lecture 12

  18. x x x x x x x x 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. Before collision After collision Physics 208 Lecture 12

  19. 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. A. R1 < R2 B. R1 = R2 C. R1 > R2 1 2 Current flow ~ uniform. More cross-sectional area means more current flowing -less resistance. Physics 208 Lecture 12

  20. Resistors Circuits Physical layout Schematic layout Physics 208 Lecture 12

  21. Quick Quiz Which bulb is brighter? A B Both the same Current through each must be same Conservation of current (Kirchoff’s current law) Charge that goes in must come out Physics 208 Lecture 12

  22. I2 I1 I3 I1=I2+I3 I1 I3 I2 I1+I2=I3 Current conservation Iin Iout Iout = Iin Physics 208 Lecture 12

  23. Quick Quiz How does brightness of bulb B compare to that of A? B brighter than A B dimmer than A Both the same Battery maintain constant potential difference Extra bulb makes extra resistance -> less current Physics 208 Lecture 12

  24. R R 2R = Resistors in Series • Potentials add • V = DV1 + DV2 = IR1 + IR2 = = I (R1+R2) • The equivalent resistance Req = R1+R2 • I1 = I2 = I 2 resistors in series: R  L Like summing lengths Physics 208 Lecture 12

  25. Quick Quiz What happens to the brightness of the bulb B when the switch is closed? Gets dimmer Gets brighter Stays same Something else Battery is constant voltage,not constant current Physics 208 Lecture 12

  26. Quick Quiz What happens to the brightness of the bulb A when the switch is closed? Gets dimmer Gets brighter Stays same Something else Physics 208 Lecture 12

  27. R R Resistors in Parallel • DV = DV1 = DV2 • I = I1 + I2 (lower resistance path has higher current) • Equivalent Resistance R/2 Add areas Physics 208 Lecture 12

  28. R1=1W V0 R2=10W R12 V0 Quick Quiz Calculate the voltage across each resistor if the battery has potential V0= 22 V. • R12 = R1 + R2 • V12 = V1 + V2 • I12 = I1 = I2 R1 and R2 are in series = 11 W = V0 = 22 Volts = V12/R12 = 2 Amps • V1 = I1R1 • V2 = I2R2 = 2 x 1 = 2 Volts = 2 x 10 = 20 Volts Check: V1 + V2 = V12 Physics 208 Lecture 12

  29. Quick Quiz What happens to the current through resistor 2 when the switch is closed? • Increases • Remains Same • Decreases Switch open: Ibattery = e/R2 = I2 • Follow Up:What happens to the current through the battery? • Increases • Remains Same • Decreases Switch closed: Ibattery = e/Req = e(1/R2+1/R3) Ibattery = I2 + I3 Physics 208 Lecture 12

  30. 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 12

  31. 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 12

  32. +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 12

  33. 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 12

  34. Current Density • Alternative expression of Ohm’s law • J = current density, = conductivity • Independent of sample geometry • Local relation between J and E Physics 208 Lecture 12

  35. What is the drift velocity? The volume occupied by a mol V=m/r = 7.09 cm3 Copper wire: A = 3.31 x 10-6 m2 and r = 8.95 g/cm3 Copper molar mass m = 63.5 g/mol 1 mol contains NA = 6.02 x 1023 atoms Hence for I = 10 A 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 Density of electrons n = NA/V = 8.49 x 1028 m-3 Physics 208 Lecture 12

  36. 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: ampere1 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 Physics 208 Lecture 12

  37. 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 Physics 208 Lecture 12

  38. 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: Physics 208 Lecture 12

  39. Resistance • The potential difference maintained across the wire of length L creates an electric field DV = EL • Hence J = sE = s DV/L = I/A And • The constant of proportionality is called the resistance of the conductor and SI units of resistance are ohms () • 1  = 1 V / A • Resistance in a circuit arises due to collisions between the electrons carrying the current with fixed atoms Physics 208 Lecture 12

  40. Quick Quiz 1 Two cylindrical resistors are made from the same material. They are of equal length but one has twice the diameter of the other. • R1 > R2 • R1 = R2 • R1 < R2 1 2 Smaller diameter  not as easy for carriers to flow through Physics 208 Lecture 12

  41. 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 DU=QDV • 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 Physics 208 Lecture 12

  42. - - - - - - + + + + + + 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 12

  43. - - + + 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 12

  44. 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 12

  45. - - - - - - + + + + + + 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 12

  46. - - - - - - - - - - - - - + + + + + + + + + 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 12

  47. 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 12

  48. Physics 208 Lecture 12

  49. 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 12

  50. 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 12

More Related