DOCSLIB.ORG

Example 1.1: Calculate the Number of Copper Atoms Present in a Cylinder That Has a Diameter and a Height Both Equal to 1 Μm

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Example 1.1: Calculate the Number of Copper Atoms Present in a Cylinder That Has a Diameter and a Height Both Equal to 1 Μm Example 1.1: Calculate the number of copper atoms present in a cylinder that has a diameter and a height both equal to 1 µm. The mass density of copper is 8.93 x 103 kg/m3 and its atomic mass is 63.55 g/mol. Solution: a) The volume of the copper cylinder is given by: 3.14 x (0.5 x 10-6)2 x 1 x 10-6 m3 = 0.78 x 10-18 m3; b) Its mass is: 8.93 x 103 kg/m3 x 0.78 x 10-18 m3 = 7.96 x 10-15 kg = 7.96 x 10-12 g; c) The number of copper atom present is: 7.96 x 10-12 g x 6.02 x 23 atoms/mol /63.55 g/mol = 6.64 x 1010 atoms. (This problem shows the relatively large number of copper atoms present in a small volume.) Example 1.2: a) Give the electronic configurations of the sodium and chlorine atoms (expressed in terms of the s-orbits and p-orbits); and b) repeat this for their ions. Solution: a) The electronic configuration of the atoms can be obtained from a standard chemistry handbook. They are: Na: 1s2, 2s2, 2p6 and 3s1 and Cl: 1s2, 2s2, 2p6, 3s2, 3p5. (In this configuration, the superscript indicates the number of electrons present in the energy orbits.) Note that both Na and Cl are monovalent. b) In the ion forms: Na+ : 1s2, 2s2 and 2p6 and Cl- : 1s2, 2 s2, 2p6, 3s2, 3p6. In the ions, both the s- and p-orbits are completely filled. This type of electronic configuration (similar to that of the inert gas) will make the ionic solid relatively stable. Example 1.5: Compute the attractive force between a Na+ ion and a Cl- ion at T = 300K. Solution: From Table 1.1 (300K), we find that the ion separation = (0.098 + 0.181) nm = 0.279 nm. 9 Using Eq.(1) in the text, the attractive force is given by: Fa = - 9 x 10 Vm/C x (1.6 x 10-19 C)2/(0.279 x 10-9 m)2 = - 2.96 x 10-9 N. The negative sign indicates an attractive force. Example 1.7: Assuming a 1-dimensional solid, the electrostatic attraction between the neighboring ions produces a potential energy given by: 2 EPE(a) = - 1.75 q /(4πεoa), where a is the ion separation. a) Determine ao , the equilibrium separation, if EPE(ao) = - 0.662 eV; and, b) determine the 4 constant A if the potential due to repulsion has the form: Er(a) = A/a . Solution: 2 a) Making use of the equation provided: ao = - 1.75 q /(4πεoEPE(ao)) = 1.75 x (1.6 X 10-19C)2/(4 x 3.14 x 8.85 x 10-12 F/m x 0.662 x 1.6 x 10-19 V) = 3.8 x 10-9 m. 2 4 b) The totall energy is given by: E’pe(a) = - 1.75 q /(4πεoa) + A/a . 2 2 At equilibrium, dE’PE(a)/da = 0 and this leads to: A = 1.75 q /(4πεoao ) x 5 -19 2 -12 -9 2 ao /4 = 1.75 x (1.6 X 10 ) /(4 x 3.14 x 8.85 x 10 F/m x (3.8 x 10 ) ) x (3.8 x 10-9)5/4 = 5.52 x 10-54 J.m4. (This problem evaluates the equilibrium bond length and bond energy for a 1-dimensional ionic solid. The factor 1.75 actually accounts for the effect of the more distant ions). 2 Example 1.9: Calculate the minimum radius ratio for two different types of ions present in a solid if the coordination number is 8. Solution: Consider a simple cube with dimension ao. If the smaller ion (with radius r) is placed at the center and ions of the opposite type (with radius R’) placed in the corners, automatically we have a structure with a coordination number of 8. In addition, for a close-packed structure, we have: The body diagonal of the cube = √3 x ao = 2 x (r + R’), and ao = 2R’. Eliminating ao from these equations gives: 1 + r/R’ = √3 = 1.732, or r/R’ = 0.732. (Note that if r/R’ is smaller, it will be no longer possible to put in the 8 neighboring ions and yet maintain the close-packed structure.) 3 Example 1.18: a) Determine the number of lattice points present in a fcc unit cell, and b) draw a primitive cell within the unit cell. Solution: a) There are 4 lattice points in the unit cell. b) The primitive cell is shown in Fig.E1.6. Example 1.21: For a cubic crystal structure, draw a) the [1 1 1], [1 3 3 ], and [2 3 6] lattice directions, and b) the (1 1 1), (2,3,4), and (3 2 1) lattice planes. Solution: a) See Fig. E1.10a. b) See Fig.E1-10b. For the (1 1 1) plane, the lattice interceptions should be: 1, 1, 1. For the (2 3 4) plane, they should be: 1, 2/3, 1/2. For the (3 2 1) plane, they should be: 1, 3/2, 3. Example 1.22: a) Determine which of the planes in the fcc crystal structure have the highest density of atoms, and b) evaluate its value for copper. The lattice constant of copper is 0.362 nm. Solution: a) The highest density planes for the fcc crystal structure are the {1 1 1} planes. Note that they also have the shortest interplane distance. b) There is a total of 2 atoms in the (1 1 1) plane in the unit cell. Its area 2 2 2 = 1/2 x √(3/2) x (√2 x ao) = √3/2 x ao = 0.866 ao . Therefore, the density of the copper atoms in the (1 1 1) plane is 2/(0.866 x (0.362 nm)2) = 17.62 atoms/nm2. 4 Example 1.24: Calculate the atomic density of α-Fe which has bcc crystal structure. Note that metals have a close-packed structure. Solution: For the bcc crystal structure, the ratio of the lattice constant to the atomic radius is 4/√3. Since rFe = 0.124 nm, ao = 4/√3 x 0.124 nm = 0.286 nm. 3 Since there are 2 atoms per unit cell, the atomic density of α-Fe is 2/ao = 2/(0.286 x 10-9)3 /m3 = 8.55 x 1028 /m3. Example 1.25: Fig. E1.13 shows the unit cell of a hcp structure. Given that c/ao is 1.63, calculate the APF assuming that all of the atoms have the same radius. Solution: o 3 The volume of the unit cell = ao x ao/sin(60 ) x 1.63 x ao = 1.41 ao . Since ao = 2r for a close-packed structure, the volume = 1.41 x (2r)3 = 11.3 r3. Since there are 2 atoms per unit cell, the APF = 2 x 4π/3 x r3/(11.3 r3) = 0.74. Example 1.27: Calculate the planar density of atoms in the (1 1 1) plane of germanium. Solution: Since the lattice constant of germanium is 0.563 nm, the length of the face diagonal is √2 x 0.563 nm = 0.796 nm. The area of the triangle in the (1 1 1 ) plane within the unit cell is 0.796/2 x √3 x 0.796/2 = 0.274 nm2. Since there are 2 atoms in this area, the planar density is 2/(0.274 x 10-18) /m2 = 7.29 x 1018 /m2. 5 Example 1.32: Assume that Kd = 3 and ∆Ed = 2.4 eV, compute the ratio Nd/Nsite at a) T = 300K, and b) T = 1000K. Solution: a) Based on Eq.(4) in the notes, Nd/Nsite = Kd exp(- ∆Ed/(kT)) = 3 x exp(- 2.4 x 1.6 x 10-19 J/(1.38 x 10-23 J/K x 300K) = 1.4 x 10-40. -19 -23 b) Similarly, at 1000K, Nd/Nsite = 3 x exp(- 2.4 x 1.6 x 10 J/(1.38 x 10 J/K x 1000K) = 2.4 x 10-12. Example 1.34: In an IC fabrication process, a 100 nm thickness boron- doped surface layer has to be formed on silicon. The required boron density within the surface layer must be no less than 1 x 1024 /m3. If the boron density at the upper silicon surface is kept constant at 1 x 1025 /m3, how long does it take for the surface layer to be formed if a) T = 750oC, and b) 1100oC. Fig.E1.16 provides a plot of the diffusivity of boron as a function of temperature. Assume the original boron density in silicon is negligible. Solution: o -21 2 o From Fig.E1.16, the diffusivity Di (750 C) = 1 x 10 m /s and Di (1100 C) = 1 -17 2 x 10 m /s. From Eq.(7): co = 0, ci(x)/cis = 1 - erf(x/(2√(Dit))). o - a) ci(x)/cis = 0.1 at x =100 nm and T = 750 C. This gives 0.1 = 1 - erf(1 x 10 7/(2 x √(1 x 10-21 x t))), or t = 1.86 x 106 s (516 hrs). b) Similarly, at 1100oC, 0.1 = 1 - erf(1 x 10-7/(2 x √(1 x 10-17 x t))), or t = 186 s (0.052 hrs). 6 Example 1.36: In a diffraction experiment involving a polycrystalline sample (see Fig.E1.17), the (1 1 1) diffraction ring measured on a photographic plate is 0.01 m from the center.
Load more

Recommended publications
  • Chapter 4 SINGLE PARTICLE MOTIONS
    Chapter 4 SINGLE PARTICLE MOTIONS 4.1 Introduction We wish now to consider the effects of magnetic fields on plasma behaviour. Especially in high temperature plasma, where collisions are rare, it is important to study the single particle motions as governed by the Lorentz force in order to understand particle confinement. Unfortunately, only for the simplest geometries can exact solutions for the force equation be obtained. For example, in a constant and uniform magnetic field we find that a charged particle spirals in a helix about the line of force. This helix, however, defines a fundamental time unit – the cyclotron frequency ωc and a fundamental distance scale – the Larmor radius rL. For inhomogeneous and time varying fields whose length L and time ω scales are large compared with ωc and rL it is often possible to expand the orbit equations in rL/L and ω/ωc. In this “drift”, guiding centre or “adiabatic” approximation, the motion is decomposed into the local helical gyration together with an equation of motion for the instantaneous centre of this gyration (the guiding centre). It is found that certain adiabatic invariants of the motion greatly facilitate understanding of the motion in complex spatio-temporal fields. We commence this chapter with an analysis of particle motions in uniform and time-invariant fields. This is followed by an analysis of time-varying electric and magnetic fields and finally inhomogeneous fields. 4.2 Constant and Uniform Fields The equation of motion is the Lorentz equation dv F = m = q(E + v×B) (4.1) dt 88 4.2.1 Electric field only In this case the particle velocity increases linearly with time (i.e.
    [Show full text]
  • Particle Motion
    Physics of fusion power Lecture 5: particle motion Gyro motion The Lorentz force leads to a gyration of the particles around the magnetic field We will write the motion as The Lorentz force leads to a gyration of the charged particles Parallel and rapid gyro-motion around the field line Typical values For 10 keV and B = 5T. The Larmor radius of the Deuterium ions is around 4 mm for the electrons around 0.07 mm Note that the alpha particles have an energy of 3.5 MeV and consequently a Larmor radius of 5.4 cm Typical values of the cyclotron frequency are 80 MHz for Hydrogen and 130 GHz for the electrons Often the frequency is much larger than that of the physics processes of interest. One can average over time One can not however neglect the finite Larmor radius since it lead to specific effects (although it is small) Additional Force F Consider now a finite additional force F For the parallel motion this leads to a trivial acceleration Perpendicular motion: The equation above is a linear ordinary differential equation for the velocity. The gyro-motion is the homogeneous solution. The inhomogeneous solution Drift velocity Inhomogeneous solution Solution of the equation Physical picture of the drift The force accelerates the particle leading to a higher velocity The higher velocity however means a larger Larmor radius The circular orbit no longer closes on itself A drift results. Physics picture behind the drift velocity FxB Electric field Using the formula And the force due to the electric field One directly obtains the so-called ExB velocity Note this drift is independent of the charge as well as the mass of the particles Electric field that depends on time If the electric field depends on time, an additional drift appears Polarization drift.
    [Show full text]
  • Basic Physics of Magnetoplasmas-I: Single Particle Drift Montions
    310/1749-5 ICTP-COST-CAWSES-INAF-INFN International Advanced School on Space Weather 2-19 May 2006 _____________________________________________________________________ Basic Physics of Magnetoplasmas-I: Single Particle Drift Montions Vladimir CADEZ Astronomical Observatory Volgina 7 11160 Belgrade SERBIA AND MONTENEGRO ___________________________________________________________________________ These lecture notes are intended only for distribution to participants LECTURE 1 SINGLE PARTICLE DRIFT MOTIONS Vladimir M. Cade·z· Astronomical Observatory Belgrade Volgina 7, 11160 Belgrade, Serbia&Montenegro Email: [email protected] Gaseous plasma is a mixture of moving particles of di®erent species ® having mass m® and charge q®. UsuallyP but not necessarily always, such a plasma is globally electro-neutral, i.e. ® q® = 0. In astrophysical plasmas, we often have mixtures of two species ® = e; p or electron-proton plasmas as protons are the ionized atoms of Hydrogen, the most abundant element in the universe. Another plasma constituent of astrophysical signi¯cance are dust particles of various sizes and charges. For example, the electron-dust and electron-proton-dust plasmas (® = e; d and ® = e; p; d resp.) are now frequently studied in scienti¯c literature. Some astrophysical con¯gura- tions allow for more exotic mixtures like electron-positron plasmas (® = e¡; e+) which are steadily gaining interest among theoretical astrophysicists. In what follows, we shall primarily deal with the electron-proton, electro- neutral plasmas in magnetic ¯eld con¯gurations typical of many solar-terrestrial phenomena. To understand the physics of plasma processes in detail it is necessary to apply complex mathematical treatments of kinetic theory of ionized gases. For practical reasons, numerous approximations are introduced to the full kinetic approach which results in simpli¯ed and more applicable plasma theories.
    [Show full text]
  • The Electron Drift Velocity in the HARP TPC
    HARP Collaboration HARP Memo 08-103 8 October 2008 http://cern.ch/harp-cdp/driftvelocity.pdf The electron drift velocity in the HARP TPC V. Ammosov, A. Bolshakova, I. Boyko, G. Chelkov, D. Dedovitch, F. Dydak, A. Elagin, M. Gostkin, A. Guskov, V. Koreshev, Z. Kroumchtein, Yu. Nefedov, K. Nikolaev, J. Wotschack, A. Zhemchugov Abstract Apart from the electric field strength, the drift velocity depends on gas pressure and temperature, on the gas composition and on possible gas impurities. We show that the relevant gas temperature, while uniform across the TPC volume, depends significantly on the temperature and flow rate of fresh gas. We present the correction algorithms for changes of pressure and temperature, and the precise measurement of the electron drift velocity in the HARP TPC. Contents 1 Introduction 2 2 Theoretical expectation on drift velocity variations 2 3 The TPC gas 3 4 Drift velocity variation with gas temperature 4 5 Drift velocity variation with gas pressure 6 6 The algorithm for temperature and pressure correction 7 7 Drift velocity variation with gas composition and impurities 9 8 Small-radius and large-radius tomographies 12 9 Comparison with ‘Official’ HARP's analysis 15 1 1 Introduction In a TPC, the longitudinal z coordinate is measured via the drift time. Therefore, the drift velocity of electrons in the TPC gas is a centrally important quantity for the reconstruction of charged-particle tracks. Since we are not concerned here with the drift velocity of ions, we will henceforth refer to the drift velocity of electrons simply as `drift velocity'. In this paper, we discuss first the expected dependencies of the drift velocity on temperature, pressure and gas impurities.
    [Show full text]
  • Exam 1 Solutions
    PHY2054 Spring 2009 Prof. Pradeep Kumar Prof. Paul Avery Prof. Yoonseok Lee Feb. 4, 2009 Exam 1 Solutions 1. A proton (+e) originally has a speed of v = 2.0 ×105 m/s as it goes through a plate as shown in the figure. It shoots through the tiny +++ +++ holes in the two plates across which 16 V | 100 V | 2 V of electric potential is applied. Find the speed as it leaves the second plate. V − − − − − − Answer: 1.92 × 105 m/s | 1.44 × 105 m/s | 1.99 × 105 m/s Solution: The proton is slowed down by the electric field between v the two plates, so the velocity will be reduced. Conservation of en- 11222 ergy yields 22mvp ipf−Δ eV = mv , so vveVmf =−Δip2/. 2. Five identical charges of +1μC are located on the 5 vertices of a regular hexagon y of 1 m side leaving one vertex empty. A −1μC | −2μC | +2μC point charge is placed at the center of the hexagon. Find the magnitude and direction of the net x Coulomb force on this charge. Answer: 9.0 × 10−3 N −x-direction | 1.8 × 10−2 N −x-direction | 1.8 × 10−2 N +x-direction Solution: Only the charge on the far left contributes to the net force on the charge in the center because its contribution is not cancelled by an opposing charge. If Q represents the charge at each hexagon vertex, q is the charge at the center and d is the length of a hexagon side (note that the distance from each hexagon point to the center is also d), the force on q is kQq/ d 2 along the +x direction if Qq is positive and along the −x direction if Qq is negative.
    [Show full text]
  • Physics 1E3 Test Monday, Feb. 8
    Physics 1E3 Test Monday, Feb. 8 Time: 7:00 to 8:20 Locations: see “My Grades” on WebCT. Students with conflicts: early session 5:30 to 6:50. (See “My Grades” on webCT). Unresolved conflicts: e-mail your instructor yesterday. (Conductors NOT in equilibrium; E ≠ 0) Text sections 27.1, 27.2 Current and current density Ohm’s Law, resistivity, and resistance Practice problems: Chapter 27, problems 5, 7, 11, 13 1 CURRENT I is the charge per unit time flowing along a wire: if charge dQ flows past in time dt Units: 1 ampere (A) = 1 C/s Direction: by convention, the direction of movement of positive charge + + + - - - + + + - - - I I + + + + + + A + + + + + + L = vd Δt Charge passing through the shaded circle in time Δt : Q = (number of charges/volume) x (charge on each one) x volume Q = n ·q ·(AL) = nqAvd Δt Current: I = Q/Δt = nqAvd Δt /Δt So, v I = nqAv d = average (“drift”) d velocity of each charge 2 Units: Amps/m2 (Note that the “current through a surface” is the flux of the current density through that surface.) So, J = nqvd (a vector equation) In normal conductors, J is caused by an electric field in the conductor—which is not in equilibrium. Quiz Mobile positively-charged sodium ions in a salt solution carry an electrical current when a battery is connected. There are also some negatively-charged chloride ions in the solution. The presence of mobile chloride ions A) causes the net current to be even larger B) causes the net current to be smaller C) causes the net current to be zero D) has no effect on the net current 3 The mobile charges in most metals are electrons, with about one or two electrons per atom being free to move.
    [Show full text]
  • Number of Electrons/M3 • Drift Velocity • Example: a Copper Wire (2Mm Dia
    From: https://ece.uwaterloo.ca/~dwharder/Analogy/Resistors/ Current = I = dQ/dt, unit is C/Sec which is called Ampere, A w/o proof: I=nqAvd Example: A copper wire (2mm dia) carries 10A current. If 3 =8.9x10 kg/m3, amu = 63.5 gr/mole and there is one electron/atom, calculate: •Number of electrons/m3 •Drift velocity PHYS42-9-24-2015 Page 1 •Drift velocity This is the AVERAGE speed of electrons in the wire. Why is it so low? example: An electron is orbiting in a Circle of radius R. What is the current associated with this motion? Electron is moving with speed of V. PHYS42-9-24-2015 Page 2 Symbol for resistance is Symbol for Battery is Symbol for Capacitor is This means VARIABLE PHYS42-9-24-2015 Page 3 Conductivity Unit of Conductivity? Example: A 8mm diameter plastic wire (L=1m) is coated with gold to a thickness of 1 micron. If for gold is 2.44x10-8 ohms-m, what is the resistance of wire end to end? A wire is drawn to 3 times of it's original Length. what is the ratio of new resistance to the old one? PHYS42-9-24-2015 Page 4 R=R0(1+α(T)) or =0(1+α(T)) NOTE: R0 and 0 are not resistance and resistivity at 0 degrees. They are the constant at the temperature given when α is specified wrong way: PHYS42-9-24-2015 Page 5 Right Way: Current Density Use of Resistance for "strain. Gauge" This section will not be in your test.
    [Show full text]
  • Chapter 2 Motion of Charged Particles in Fields
    Chapter 2 Motion of Charged Particles in Fields Plasmas are complicated because motions of electrons and ions are determined by the electric and magnetic fields but also change the fields by the currents they carry. For now we shall ignore the second part of the problem and assume that Fields are Prescribed. Even so, calculating the motion of a charged particle can be quite hard. Equation of motion: dv m = q ( E + v ∧ B ) (2.1) dt charge E­field velocity B­field � �� � Rate of change of momentum � �� � Lorentz Force Have to solve this differential equation, to get position r and velocity (v= r˙) given E(r, t), B(r, t). Approach: Start simple, gradually generalize. 2.1 Uniform B field, E = 0. mv˙ = qv ∧ B (2.2) 2.1.1 Qualitatively in the plane perpendicular to B: Accel. is perp to v so particle moves in a circle whose radius rL is such as to satisfy 2 2 v⊥ mrLΩ = m = |q | v⊥B (2.3) rL Ω is the angular (velocity) frequency 2 2 2 1st equality shows Ω = v⊥/rL (rL = v⊥/Ω) 17 Figure 2.1: Circular orbit in uniform magnetic field. v⊥ 2 Hence second gives m Ω Ω = |q | v⊥B |q | B i.e. Ω = . (2.4) m Particle moves in a circular orbit with |q | B angular velocity Ω = the “Cyclotron Frequency” (2.5) m v⊥ and radius rl = the “Larmor Radius. (2.6) Ω 2.1.2 By Vector Algebra • Particle Energy is constant. proof : take v. Eq. of motion then � � d 1 2 mv.v˙ = mv = qv.(v ∧ B) = 0.
    [Show full text]
  • Ionic Drift Velocity and Its Relation with Current Density
    394 A Textbook of Physical Chemistry – Volume I Ionic Drift Velocity and Its Relation with Current Density In this section, we will try to explain how the ionic movement is quantitatively related to current density flowing through an electrolytic solution under the influence of an applied electric field. To do so, we need to understand the ionic drift velocity and then its relationship with current density. Ionic Drift Velocity When an ion in the electrolytic solution is placed under the externally applied electric field, the electric field will make the ion to accelerate as per Newton’s second law. Now if the ion is in the vacuum, it would show an acceleration until it strikes with the respective electrode. However, it will not happen since a large number of other ions also present in the same electrolytic solution along with the solvent as well. Consequently, the ion is almost bound to collide with other ions or solvent particles in its journey. The ion will stop for some time and then will start to accelerate again. This stop-start phenomenon will impart a discontinuity in the speed and direction of this moving ion. It means that ionic movement is not very much smooth but actually a resistance is offered by the surrounding medium. Therefore, we can say that the application of the external electric field will make the ion move towards the oppositely charged electrode but in a stops-starts and zigzag fashion. An ion starts moving towards the positively-charged electrode only after the application of the electric field. The initial velocity before that can simply be neglected because it arises from random collisions which can be in any random direction.
    [Show full text]
  • Steady State
    Electric Current • The amount of charge that flows by per unit time. Q • I t Steady state • A system (e.g. circuit) is in the steady state when the current at each point in the circuit is constant (does not change with time). – In many practical circuits, the steady state is achieved in a short time. • In the steady state, the charge (or current) flowing into any point in the circuit has to equal the charge (or current) flowing out. – Kirchhoff’s Node (or Current) Rule. n = number of charges per unit volume = “charge-number density” (n > 1027 m-3 for a good metal) -5 Figure 25.26 Typically vd ~ 10 m/s ~1 m/day Why do the electrons have a drift velocity? • They feel a force due to an electric field. • But then they should accelerate!: F = ma • Each electron does accelerate for some time but then it collides with something (a nucleus, another electron, etc.). • After the collision, the electron goes off in some random direction, giving it momentarily a zero average velocity. • The drift velocity is the average velocity in the a F qE v time between collisions: d 2 2mm 2 Which of the following statements is false? 1) An electric field is needed to produce an electric current. 2) A potential difference between two points is needed to produce an electric current. 3) For a steady current to flow in a wire, the wire must be part of a closed circuit. 4) The electric field is constant along all parts of the circuit when a steady current is flowing.
    [Show full text]
  • Electrical Current
    Electrical Current So far we only have discussed charges that were not in motion -- electrostatics. Now we turn to the study of charges in motion -- electrodynamics. When charges are in motion, then you have current. Current in a wire In a time ∆t, a total charge ∆Q flows through the wire’s cross sectional area A. We define the current as I = ∆Q/∆t Units: Ampere = Coulomb/sec Drift Velocity: vd r As charges are now in motion, then the electric E ≠ 0 field is not zero inside the conductor. But the charges in motion do not undergo a simple constant acceleration due to this field. The charges are constantly colliding with the atoms of the conducting material. This endless cycle of acceleration and collision results in a net “drift velocity” vd for the charges in motion. r E vd Drift Velocity and Current # of charge carriers let n = (density of charge carriers) unit volume q = charge carried by each charge carrier A = wire cross sectional area vd = drift velocity then Avd = volume of charge carriers through A per second nAvd = number of charge carriers through A per second and finally I = qnAvd is the current in the wire Typical Value of Drift Velocity Let’s say we have a wire 1 m long, and we apply a potential difference of 1 V from one end of the wire to the other. That means the electric field in the wire is 1 V/m. If an electron were freely accelerated from rest in that field, it would achieve an energy of 1 eV at the end of the wire, resulting in a speed of about 600,000 m/s.
    [Show full text]
  • Surface Charge Density
    EECS 117 Lecture 13: Method of Images / Steady Currents Prof. Niknejad University of California, Berkeley University of California, Berkeley EECS 217 Lecture 13 – p. 1/21 Point Charge Near Ground Plane Consider a point charge q sitting a distance d above a conductive ground plane. We'd like to find the electric field E everywhere Clearly E = 0 for z < 0 due to the conductor. Also, since ρ = 0 everywhere except for point (0; 0; d). Thus we can say that φ satisfies Laplace's equation for all points z > 0 except (0; 0; d) where the point charge lives (0; 0;d) q d z>0 E=¡^zEz E=0 University of California, Berkeley EECS 217 Lecture 13 – p. 2/21 Method of Images Consider now a problem we have met before. If we have two charges of equal and opposite magnitude at z = d, then the solution is trivial. How is this problem related? Notice that for this new problem the electric field is normal to the xy plane at z = 0 q(xx^ + yy^ + d^z) −q(xx^ + yy^ − d^z) E(z = 0) = + + 4π(x2 + y2 + d2)3=2 4π(x2 + y2 + d2)3=2 Combining the terms we have a z-directed vector −q2d^z E(z = 0) = 4π(x2 + y2 + d2)3=2 University of California, Berkeley EECS 217 Lecture 13 – p. 3/21 The Image Charge Now consider placing a conductive plane at z = 0. Do the fields need to change? The boundary condition is already satisfied by this field configuration. In other words, this problem satisfies the same boundary conditions as our original problem This negative charge is the “image” charge that actually solves our original problem! Recall that the solution of Poisson's equation is uniquely determined by the boundary conditions.
    [Show full text]