From last time. Today: More on electric potential and connection to E-field How to calculate E-field from V Capacitors and Capacitance

Transcription

1 From last time More on electric potential and connection to Efield How to calculate Efield from V Capacitors and Capacitance Today: More on Capacitors and Capacitance Energy stored in Capacitors Current Resistance Ohm s Law

2 How do we charge a capacitor? Battery has fixed electric potential difference across its terminals Conducting plates connected to battery terminals produce a charge separation on plates Electrons move from negative battery terminal to right plate and from left plate to positive battery terminal This charge motion requires work. The work is supplied by the battery. Between plates there is an Efield and a potential difference is estableshed ΔVplates = ΔVbattery Infinite plates: we normally mean to neglect border effets (uniform field)

3 Capacitance Q E E Eleft Eright d E Q E E E A=surface area E left E right = σ /2ε o σ /2ε o = σ /ε o σ=q/a surface charge density Work of the electric force to move q from positive to negative plate = W el = ΔU = q(v V ) = q ΔV = σ ε 0 d = Qd ε 0 A = Q C C = ε o A d ΔV = 1 C Q d 0 r E d r s = qed ΔV = Ed This is a geometrical factor ΔV = V V

4 Spherical capacitor Charge Q moved from outer to inner sphere Gauss law says E=kQ/r 2 until second sphere between the inner sphere and the shell Potential difference Along path shown ΔV = b a kq r 2 b = kq 1 r a C = C = Q Q ΔV = ΔV = 1 k 1 a 1 b 1 = kq 1 a 1 b Path to find ΔV Gaussian surface to find E Q

5 Energy stored in capacitors 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 W = q C Q2 dq = 2C The energy stored in any capacitor is: Q 0 For a parallel capacitor: U = 1/2 ε o AdE 2

6 Energy density The energy stored per unit volume is U/(Ad) = 1/2 ε o E 2 This is a fundamental relationship for the local energy stored in an electric field Not restricted to capacitors: Efield can be from any source Interpretation: energy is stored in the field Capacitors are devices to store electric energy and charge (used in radio receivers, filters of power supplies, electronic flashes)

7 Quick Quiz 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? pull q d q pull A) B) C) 1)The charge decreases 2)The capacitance increases 3)The electric field increases 4)The voltage between the plates increases D) E) 5)The energy stored in the capacitor increases The charge cannot decrease because C is isolated C = ε 0 A/d C decreases! E= Q/(ε 0 A) E constant V= Ed V increases U = Q 2 /(2C) U increases

8 Human capacitors: cell membranes 100 µm Lipid bilayers form a capacitor with charge separation at about 78nm outside cell Extracellular fluid Plasma membrane Cytoplasm inside cell

9 Modeling a cell membrane 78 nm K Na Extracellular fluid Plasma membrane Cytoplasm Na K Na Charges are / ions instead of electrons (atoms with different numbers of electrons and protons) Charge motion is through cell membrane (ion channels) rather than through wire ΔV~0.07 V Channels are selective for each kind of ion Across the cell membrane about ΔV=70 mv (resting potential) since more Na outside and gates are closed while K ions diffuse out and in at the same rate There is more positive charge outside the cell Capacitance: ε o A d 100 µm sphere surface area ( F /m)4π m = m ( ) 2 ~ 3x10 4 cm 2 = F = 35pF ~0.1µF/cm 2

10 Depolarization in a nerve cell

11 Cell membrane depolarization 78 nm K Extracellular fluid Plasma membrane Cytoplasm Na K ΔV~0.07 ΔV0.03 V An influx of Na ions thru Na channel can cause a depolarization of the membrane. The potential difference becomes about 0.03 mv. Potential change = about 0.1 V About 100 channels/µm 2 (each has radius of 50 µm) How many ions flow per channel? Charge xfer required ΔQ=CΔV=(35 pf)(0.1v) =(35x10 12 C/V)(0.1V) = 3.5 x Coulombs 3.5 x C/ (1.6x10 19 C/ion) = 2.2 x 10 7 ions flow (100 channels/µm 2 )x4π(50 µm) 2 =3.14x10 6 ion channels Ion flow / channel =(2.2x10 7 ions) / 3.14x10 6 channels ~ 7 ions/channel

12 Charge motion: current Cell membrane capacitor: Found ~7 ions flow through ion each ion channel to depolarize membrane Occurs in ~ 1 ms = sec. This is an electric current I: amount of charge per unit time flowing through a plane perpendicular to charge motion Units: 1 Coulomb / sec = 1 Ampere Average current: Instantaneous value: Conventionally we assume: current in direction of charge particles though electrons in reality flow in a wire

13 Current in a conductor wire Battery produces Efield in wire Current flows in response to Efield Average current: E n = number of electrons/volume electrons travel distance L = vd Δt in time Δt # of electrons: n x volume = nal =na vd Δt I av = ΔQ/ Δt = nqal v d /L Current density J = I/A (A/m 2 ) J = nqv d (direction of charge carriers) vd =10 4 m/s

14 Ohm s law When an electric field is applied free electrons experience an acceleration (opposite to E): a = F/m = qe/m Calling t the mean time between 2 collisions electronatoms electrons achieve a drift velocity: v d = at = qe m t r j = nq r v d = nq2 t r E m Ohm s Law: for many materials, the current density is proportional to the electric field producing the current J = σ E σ = conductivity of the conductor E

15 Ohm s Law Ohm s Law: ρ = resistivity J = σ E or E = ρ J E = ΔV L = ρ I A Define: R = ρl/a Resistance in ohms (Ω) Ohm s law becomes: ΔV = R I ohmic device: relationship current and voltage is linear

16 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. R 1 < R 2 B. R 1 = R 2 R = ρ L A C. R 1 > R Current flow ~ uniform. More crosssectional area means more current flowing less resistance. 16

17 Dependence on Temperature L ρ = ρ 0 α(τ Τ 0 ) ρ of a conductor varies approximately linearly with T ρ o is the resistivity at T o = 20 C ρ 0 TT 0 α x (ρρ 0 ) T 0 T α = temperature coefficient in SI units of o C 1 The higher T the greater atomic vibrations that increases collision probability Similarly: R = R o [1 α(t T o )]

18 But other materials Semiconductors: α is negative ρ decreases for increasing T Superconductors ρ = ρ 0 1 αδt ( ) Mercury Below a certain temperature, T C = critical temperature ρ is zero Once a current is set up in a superconductor, it persists without any applied voltage since R = 0

19 Resistivity Resistivity: ρ = 1 / σ SI units of Ω. m The resistance depends resistivity and geometry: on conductor Resistors control the current level in circuits Resistors can be composite or wirewound

20 Resistors in Series I 1 = I 2 = I ΔV = ΔV 1 ΔV 2 R eq = R 1 R 2 R R = 2R 2 resistors in series: R L Like summing lengths R = ρ L A I1 I2 ΔV 1 ΔV 2 Neglect resistance of wires in circuits respect to R resistance ΔV = ΔV 1 ΔV 2 = R 1 I 1 R 2 I 2 = (R 1 R 2 )I = R eq I

21 Resistors in parallel ΔV = ΔV 1 = ΔV 2 I = I 1 I 2 R A A R I = I 1 I 2 = ΔV 1 ΔV 2 = 1 1 R 1 R 2 R 1 R 2 ΔV = ΔV R eq 2A Add areas R = ρ L A

22 Current Conservation: 1 st Kirchoff s law I 2 I in I 1 I 1 =I 2 I 3 I 3 I 1 I 3 I out I out = I in I 2 I 1 I 2 =I 3 Junction Rule: Σ I in = Σ I out A statement of Conservation of Charge 22

23 Electrical power As a charge ΔQ moves from a to b, the electric potential energy of the system increases by ΔQΔV and the chemical energy in the battery decreases by this same amount As the charge moves through the resistor (c to d), the system loses this electric potential energy during collisions of the electrons with the atoms of the resistor This energy is transformed into internal energy in the resistor (vibrational motion of the atoms in the resistor) Power in units of Watts = J/s Power = rate of energy loss P = ΔU Δt = ΔQ Δt ΔV P = IΔV = RI 2 = ΔV 2 23 R

24 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 In A: In B: I A = ε R P A = RI A I B = ε 2R P B = ε2 4R 2 = ε2 R 24

25 Dielectrics in capacitors 78 nm K Extracellular fluid Plasma membrane Cytoplasm Na Membrane is not really empty It has molecules inside that respond to electric field. The molecules in the membrane can be polarized E = E0/ κ and V = V0/ κ κ>1 E 0 E ind The capacitance increases C = κ C0 In cells lipid belayer as κ=10 so 70 ions flow/ channel