DOCSLIB.ORG

Solution # 3 Instructors: AKJ & SC Solution 3.1: Finding Potential, Given a Charge Distribution (Griffiths 3Rd Ed

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Solution # 3 Instructors: AKJ & SC Solution 3.1: Finding Potential, Given a Charge Distribution (Griffiths 3Rd Ed Department of Physics IIT Kanpur, Semester II, 2017-18 PHY103A: Physics II Solution # 3 Instructors: AKJ & SC Solution 3.1: Finding potential, given a charge distribution (Griffiths 3rd ed. Prob 2.26) z h b x h r’ r a y x FIG. 1: The potential V (a) at point a is given by (see Fig. 1) p 1 Z 2h σ2πx V (a) = dr 4π0 0 r p we have x = r= 2. Therefore we get, p 2πσ Z 2h 1 2πσ 1 p σh V (a) = p dr = p ( 2h) = 4π0 0 2 4π0 2 20 The potential V (b) at point b is given by (see Fig. 1) p 1 Z 2h σ2πx V (b) = 0 dr 4π0 0 r p p p we have x = r= 2 and r0 = h2 + r2 − 2hr. Therefore we get, p p ! 1 Z 2h σ2πx 2πσ 1 Z 2h r V (b) = dr = p dr 0 p p 4π0 0 r 4π0 2 0 h2 + r2 − 2hr p σ q p h q p p 2h 2 2 2 2 = p h + r − 2hr + p ln 2 h + r − 2hr + 2r − 2h 2 20 2 0 σ h p p h p = p h + p ln(2h + 2 2h − 2h) − h − p ln(2h − 2h) 2 20 2 2 p ! p ! σ h h p p i σh 2 + 2 σh 2 + 2 = p p ln(2h + 2h) − ln(2h − 2h) = ln p = ln p 2 20 2 40 2 − 2 40 2 − 2 p ! σh (2 + 2)2 σh p = ln = ln 1 + 2 40 2 20 Thus we get the required potential difference to be σh h p i V (a) − V (b) = 1 − ln 1 + 2 20 1 Solution 3.2: Finding field and charge density, given an electric potential, (Griffiths 3rd ed. Prob 2.46) e−λr (a) The electric potential is V (r) = A . Therefore, the electric field E(r) can be written as r @ e−λr r(−λ)e−λr − e−λr E = −rV = −A ^r = −A ^r @r r r2 ^r = Ae−λr(1 + λr) r2 (b) The corresponding charge density ρ(r) can be calculated by using the differential form of Gauss's Law ρ = 0r · E. Using the product rule for divergence, r · (fA) = f(r · A) + A · (rf), we obtain ^r ^r ρ = r · E = Ae−λr(1 + λr)r · + A · r e−λr(1 + λr) 0 0 r2 0 r2 Next we use the properties of the Dirac-delta function and the formula for gradient in spherical coordinates to get ^r Ae−λr(1 + λr)r · = Ae−λr(1 + λr)4πδ3(r) = A4πδ3(r) 0 r2 0 0 ^r ^r @ A · r e−λr(1 + λr) = A · e−λr(1 + λr)^r 0 r2 0 r2 @r ^r = A · −λe−λr(1 + λr) + e−λrλ^r 0 r2 ^r = A · −λ2re−λr^r 0 r2 λ2 = − A e−λr 0 r Therefore, we get the charge density ρ(r) as λ2 ρ = A 4πδ3(r) − e−λr 0 r (c) The total charge Q can now be calculated to be Z Q = ρdτ Z Z 1 −λr 3 2 e 2 = 0A4π δ (r)dτ − 0Aλ 4πr dr 0 r Z 1 2 −λr = 0A4π − 0Aλ 4π re dr 0 1 = A4π − Aλ24π 0 0 λ2 = 0 Therefore the total charge is zero. 2 Solution 3.3: Verifying Poisson's Equation (Griffiths 3rd ed. Prob 2.29) z ½(r’) r = r-r’ r’ r y x FIG. 2: The potential V (r) at r due to the localized charge distribution is 1 Z ρ(r0) V (r) = dτ 0; 4π0 r We note that the charge distribution has been represented in the (r0; θ0; φ0) coordinates. We take the Laplacian of the potential in (r; θ; φ) coordinates. Therefore, we get 1 Z ρ(r0) r2V (r) = r2 dτ 0 4π0 r 1 Z ρ(r0) = r2 dτ 0 4π0 r 1 Z 1 = ρ(r0) r2 dτ 0 4π0 r 1 Z 1 = ρ(r0) r · r dτ 0 4π0 r Z r 1 0 −^ 0 = ρ(r ) r · 2 dτ 4π0 r 1 Z = −ρ(r0)4πδ3(r)dτ 0 4π0 1 Z = −ρ(r0)4πδ3(r − r0)dτ 0 4π0 1 = − ρ(r) 0 Thus, we see that the given potential satisfies the Poisson's equation. 3 Solution 3.4: Electrostatic energy of two spherical shells (Griffiths 3rd ed. Prob 2.34) 1 q (a) The electric field due to the two shells is given by E(r) = 2 ^r, for (a < r < b), and E(r) = 0, otherwise. 4π0 r The energy of this configuration is: Z 2 Z b 2 2 0 2 0 q 1 2 q 1 1 W = E dτ = 2 4πr dr = − 2 2 4π0 a r 8π0 a b (b) Let's us first calculate the energy of the individual shells. The electric filed due to the shell of radius a is 1 q Ea(r) = 2 ^r, for (r > a), and Ea(r) = 0, otherwise. The electric filed due to the shell of radius b is 4π0 r 1 q Eb(r) = − 2 ^r, for (r > b), and Eb(r) = 0, otherwise. Therefore, the energy of the first spherical shell is: 4π0 r Z 2 Z 1 2 2 0 2 0 q 1 2 q Wa = Eadτ = 2 4πr dr = 2 2 4π0 a r 8π0a Similarly, the energy of the second spherical shell is Z 2 Z 1 2 2 0 2 0 q 1 2 q Wb = Eb dτ = 2 4πr dr = 2 2 4π0 b r 8π0b The interaction energy of this system is therefore: Z q2 q2 Wint = 0 Ea · Ebdτ = W − Wa − Wb = − − 8π0b 8π0b q2 = − 4π0b Solution 3.5: Electrostatic Force (a) z (b) z Q R da Q R dτ θ θ FIG. 3: 1 Q (a) The electric filed due to the metal sphere of radius R is given by E(r) = 2 ^r, for (r ≥ R), and E(r) = 0, 4π0 r otherwise. From the symmetry of the problem, it is clear that the total electrostatic force on northern hemisphere will be in the z direction. Now, the electrostatic force per unit area in the z-direction at the area 4 element da, as shown in Fig. 3(a), is: E(r) Q 1 1 Q Q2 fz = σEother · ^z = σ · ^z = 2 2 cos θ = 2 4 cos θ 2 4πR 2 4π0 r 32π 0R Therefore, the total repulsive force on the northern hemisphere is Z Z π=2 Z 2π 2 Q 2 Fz = fzda = 2 4 cos θ R sin θdθdφ θ=0 φ=0 32π 0R Q2 Z π=2 = 2 2 2π cos θ sin θdθ 32π 0R θ=0 Q2 1 = 2 2 2π 32π 0R 2 Q2 = 2 32π0R 1 Qr (b) The electric filed inside a uniformly charged sphere of radius R and charge Q is given by E(r) = 3 ^r. 4π0 R From the symmetry of the problem, it is clear that the total electrostatic force on the northern hemisphere will be in the z direction. Now, the electrostatic force per unit volume in the z-direction on the volume element dτ, as shown in Fig. 3(b), is: 3Q 1 Qr 3Q2 fz = ρE(r) · ^z = 3 3 cos θ = 2 6 r cos θ 4πR 4π0 R 16π 0R Therefore, the electrostatic force on the northern hemisphere is Z Z R Z π=2 Z 2π 2 3Q 2 Fz = fzdτ = 2 6 r cos θ r sin θdθdφ 0 θ=0 φ=0 16π 0R 2 Z R Z π=2 Z 2π 3Q 3 = 2 6 r cos θ sin θdθdφ 16π 0R 0 θ=0 φ=0 2 Z R Z π=2 Z 2π 3Q 3 = 2 6 r dr cos θ sin θdθ dφ 16π 0R 0 θ=0 φ=0 3Q2 R4 1 = 2 6 × × × 2π 16π 0R 4 2 3Q2 = 2 64π0R Solution 3.6: Capacitance of coaxial metal cylinders (Griffiths 3rd ed. Prob 2.39) Suppose that for a length L, the charge on the inner cylinder is Q and the charge on the outer cylinder is −Q. Using Q 1 the Gaussian surface as shown in Fig. 4, it can be shown that the field in between the cylinders is E(s) = ^s. 2π0L s The potential difference between the cylinders is therefore, Z b Q Z b 1 Q b V (b) − V (a) = − E · dl = − ds = − ln : a 2π0L a s 2π0L a Q b We see that a is at a higher potential. So, we take the potential difference as V = V (a) − V (b) = ln . The 2π0L a 5 capacitance C of this configuration is therefore given by Q 2π C = = 0 V b ln a b -Q s a Q L FIG. 4: 6.
Load more

Recommended publications
  • Chapter 2 Introduction to Electrostatics
    Chapter 2 Introduction to electrostatics 2.1 Coulomb and Gauss’ Laws We will restrict our discussion to the case of static electric and magnetic fields in a homogeneous, isotropic medium. In this case the electric field satisfies the two equations, Eq. 1.59a with a time independent charge density and Eq. 1.77 with a time independent magnetic flux density, D (r)= ρ (r) , (1.59a) ∇ · 0 E (r)=0. (1.77) ∇ × Because we are working with static fields in a homogeneous, isotropic medium the constituent equation is D (r)=εE (r) . (1.78) Note : D is sometimes written : (1.78b) D = ²oE + P .... SI units D = E +4πP in Gaussian units in these cases ε = [1+4πP/E] Gaussian The solution of Eq. 1.59 is 1 ρ0 (r0)(r r0) 3 D (r)= − d r0 + D0 (r) , SI units (1.79) 4π r r 3 ZZZ | − 0| with D0 (r)=0 ∇ · If we are seeking the contribution of the charge density, ρ0 (r) , to the electric displacement vector then D0 (r)=0. The given charge density generates the electric field 1 ρ0 (r0)(r r0) 3 E (r)= − d r0 SI units (1.80) 4πε r r 3 ZZZ | − 0| 18 Section 2.2 The electric or scalar potential 2.2 TheelectricorscalarpotentialFaraday’s law with static fields, Eq. 1.77, is automatically satisfied by any electric field E(r) which is given by E (r)= φ (r) (1.81) −∇ The function φ (r) is the scalar potential for the electric field. It is also possible to obtain the difference in the values of the scalar potential at two points by integrating the tangent component of the electric field along any path connecting the two points E (r) d` = φ (r) d` (1.82) − path · path ∇ · ra rb ra rb Z → Z → ∂φ(r) ∂φ(r) ∂φ(r) = dx + dy + dz path ∂x ∂y ∂z ra rb Z → · ¸ = dφ (r)=φ (rb) φ (ra) path − ra rb Z → The result obtained in Eq.
    [Show full text]
  • 6.007 Lecture 5: Electrostatics (Gauss's Law and Boundary
    Electrostatics (Free Space With Charges & Conductors) Reading - Shen and Kong – Ch. 9 Outline Maxwell’s Equations (In Free Space) Gauss’ Law & Faraday’s Law Applications of Gauss’ Law Electrostatic Boundary Conditions Electrostatic Energy Storage 1 Maxwell’s Equations (in Free Space with Electric Charges present) DIFFERENTIAL FORM INTEGRAL FORM E-Gauss: Faraday: H-Gauss: Ampere: Static arise when , and Maxwell’s Equations split into decoupled electrostatic and magnetostatic eqns. Electro-quasistatic and magneto-quasitatic systems arise when one (but not both) time derivative becomes important. Note that the Differential and Integral forms of Maxwell’s Equations are related through ’ ’ Stoke s Theorem and2 Gauss Theorem Charges and Currents Charge conservation and KCL for ideal nodes There can be a nonzero charge density in the absence of a current density . There can be a nonzero current density in the absence of a charge density . 3 Gauss’ Law Flux of through closed surface S = net charge inside V 4 Point Charge Example Apply Gauss’ Law in integral form making use of symmetry to find • Assume that the image charge is uniformly distributed at . Why is this important ? • Symmetry 5 Gauss’ Law Tells Us … … the electric charge can reside only on the surface of the conductor. [If charge was present inside a conductor, we can draw a Gaussian surface around that charge and the electric field in vicinity of that charge would be non-zero ! A non-zero field implies current flow through the conductor, which will transport the charge to the surface.] … there is no charge at all on the inner surface of a hollow conductor.
    [Show full text]
  • A Problem-Solving Approach – Chapter 2: the Electric Field
    chapter 2 the electric field 50 The Ekelric Field The ancient Greeks observed that when the fossil resin amber was rubbed, small light-weight objects were attracted. Yet, upon contact with the amber, they were then repelled. No further significant advances in the understanding of this mysterious phenomenon were made until the eighteenth century when more quantitative electrification experiments showed that these effects were due to electric charges, the source of all effects we will study in this text. 2·1 ELECTRIC CHARGE 2·1·1 Charginl by Contact We now know that all matter is held together by the aurae· tive force between equal numbers of negatively charged elec· trons and positively charged protons. The early researchers in the 1700s discovered the existence of these two species of charges by performing experiments like those in Figures 2·1 to 2·4. When a glass rod is rubbed by a dry doth, as in Figure 2-1, some of the electrons in the glass are rubbed off onto the doth. The doth then becomes negatively charged because it now has more electrons than protons. The glass rod becomes • • • • • • • • • • • • • • • • • • • • • , • • , ~ ,., ,» Figure 2·1 A glass rod rubbed with a dry doth loses some of iu electrons to the doth. The glau rod then has a net positive charge while the doth has acquired an equal amount of negative charge. The total charge in the system remains zero. £kctric Charge 51 positively charged as it has lost electrons leaving behind a surplus number of protons. If the positively charged glass rod is brought near a metal ball that is free to move as in Figure 2-2a, the electrons in the ball nt~ar the rod are attracted to the surface leaving uncovered positive charge on the other side of the ball.
    [Show full text]
  • Chapter 5 Capacitance and Dielectrics
    Chapter 5 Capacitance and Dielectrics 5.1 Introduction...........................................................................................................5-3 5.2 Calculation of Capacitance ...................................................................................5-4 Example 5.1: Parallel-Plate Capacitor ....................................................................5-4 Interactive Simulation 5.1: Parallel-Plate Capacitor ...........................................5-6 Example 5.2: Cylindrical Capacitor........................................................................5-6 Example 5.3: Spherical Capacitor...........................................................................5-8 5.3 Capacitors in Electric Circuits ..............................................................................5-9 5.3.1 Parallel Connection......................................................................................5-10 5.3.2 Series Connection ........................................................................................5-11 Example 5.4: Equivalent Capacitance ..................................................................5-12 5.4 Storing Energy in a Capacitor.............................................................................5-13 5.4.1 Energy Density of the Electric Field............................................................5-14 Interactive Simulation 5.2: Charge Placed between Capacitor Plates..............5-14 Example 5.5: Electric Energy Density of Dry Air................................................5-15
    [Show full text]
  • Chapter 4. Electric Fields in Matter 4.4 Linear Dielectrics 4.4.1 Susceptibility, Permittivity, Dielectric Constant
    Chapter 4. Electric Fields in Matter 4.4 Linear Dielectrics 4.4.1 Susceptibility, Permittivity, Dielectric Constant PE 0 e In linear dielectrics, P is proportional to E, provided E is not too strong. e : Electric susceptibility (It would be a tensor in general cases) Note that E is the total field from free charges and the polarization itself. If, for instance, we put a piece of dielectric into an external field E0, we cannot compute P directly from PE 0 e ; EE 0 E0 produces P, P will produce its own field, this in turn modifies P, which ... Breaking where? To calculate P, the simplest approach is to begin with the displacement D, at least in those cases where D can be deduced directly from the free charge distribution. DEP001 e E DE 0 1 e : Permittivity re1 : Relative permittivity 0 (Dielectric constant) Susceptibility, Permittivity, Dielectric Constant Problem 4.41 In a linear dielectric, the polarization is proportional to the field: P = 0e E. If the material consists of atoms (or nonpolar molecules), the induced dipole moment of each one is likewise proportional to the field p = E. What is the relation between atomic polarizability and susceptibility e? Note that, the atomic polarizability was defined for an isolated atom subject to an external field coming from somewhere else, Eelse p = Eelse For N atoms in unit volume, the polarization can be set There is another electric field, Eself , produced by the polarization P: Therefore, the total field is The total field E finally produce the polarization P: or Clausius-Mossotti formula Susceptibility, Permittivity, Dielectric Constant Example 4.5 A metal sphere of radius a carries a charge Q.
    [Show full text]
  • Physics 2102 Lecture 2
    Physics 2102 Jonathan Dowling PPhhyyssicicss 22110022 LLeeccttuurree 22 Charles-Augustin de Coulomb EElleeccttrriicc FFiieellddss (1736-1806) January 17, 07 Version: 1/17/07 WWhhaatt aarree wwee ggooiinngg ttoo lleeaarrnn?? AA rrooaadd mmaapp • Electric charge Electric force on other electric charges Electric field, and electric potential • Moving electric charges : current • Electronic circuit components: batteries, resistors, capacitors • Electric currents Magnetic field Magnetic force on moving charges • Time-varying magnetic field Electric Field • More circuit components: inductors. • Electromagnetic waves light waves • Geometrical Optics (light rays). • Physical optics (light waves) CoulombCoulomb’’ss lawlaw +q1 F12 F21 !q2 r12 For charges in a k | q || q | VACUUM | F | 1 2 12 = 2 2 N m r k = 8.99 !109 12 C 2 Often, we write k as: 2 1 !12 C k = with #0 = 8.85"10 2 4$#0 N m EEleleccttrricic FFieieldldss • Electric field E at some point in space is defined as the force experienced by an imaginary point charge of +1 C, divided by Electric field of a point charge 1 C. • Note that E is a VECTOR. +1 C • Since E is the force per unit q charge, it is measured in units of E N/C. • We measure the electric field R using very small “test charges”, and dividing the measured force k | q | by the magnitude of the charge. | E |= R2 SSuuppeerrppoossititioionn • Question: How do we figure out the field due to several point charges? • Answer: consider one charge at a time, calculate the field (a vector!) produced by each charge, and then add all the vectors! (“superposition”) • Useful to look out for SYMMETRY to simplify calculations! Example Total electric field +q -2q • 4 charges are placed at the corners of a square as shown.
    [Show full text]
  • Ee334lect37summaryelectroma
    EE334 Electromagnetic Theory I Todd Kaiser Maxwell’s Equations: Maxwell’s equations were developed on experimental evidence and have been found to govern all classical electromagnetic phenomena. They can be written in differential or integral form. r r r Gauss'sLaw ∇ ⋅ D = ρ D ⋅ dS = ρ dv = Q ∫∫ enclosed SV r r r Nomagneticmonopoles ∇ ⋅ B = 0 ∫ B ⋅ dS = 0 S r r ∂B r r ∂ r r Faraday'sLaw ∇× E = − E ⋅ dl = − B ⋅ dS ∫∫S ∂t C ∂t r r r ∂D r r r r ∂ r r Modified Ampere'sLaw ∇× H = J + H ⋅ dl = J ⋅ dS + D ⋅ dS ∫ ∫∫SS ∂t C ∂t where: E = Electric Field Intensity (V/m) D = Electric Flux Density (C/m2) H = Magnetic Field Intensity (A/m) B = Magnetic Flux Density (T) J = Electric Current Density (A/m2) ρ = Electric Charge Density (C/m3) The Continuity Equation for current is consistent with Maxwell’s Equations and the conservation of charge. It can be used to derive Kirchhoff’s Current Law: r ∂ρ ∂ρ r ∇ ⋅ J + = 0 if = 0 ∇ ⋅ J = 0 implies KCL ∂t ∂t Constitutive Relationships: The field intensities and flux densities are related by using the constitutive equations. In general, the permittivity (ε) and the permeability (µ) are tensors (different values in different directions) and are functions of the material. In simple materials they are scalars. r r r r D = ε E ⇒ D = ε rε 0 E r r r r B = µ H ⇒ B = µ r µ0 H where: εr = Relative permittivity ε0 = Vacuum permittivity µr = Relative permeability µ0 = Vacuum permeability Boundary Conditions: At abrupt interfaces between different materials the following conditions hold: r r r r nˆ × (E1 − E2 )= 0 nˆ ⋅(D1 − D2 )= ρ S r r r r r nˆ × ()H1 − H 2 = J S nˆ ⋅ ()B1 − B2 = 0 where: n is the normal vector from region-2 to region-1 Js is the surface current density (A/m) 2 ρs is the surface charge density (C/m ) 1 Electrostatic Fields: When there are no time dependent fields, electric and magnetic fields can exist as independent fields.
    [Show full text]
  • Electron Charge Density: a Clue from Quantum Chemistry for Quantum Foundations
    Electron Charge Density: A Clue from Quantum Chemistry for Quantum Foundations Charles T. Sebens California Institute of Technology arXiv v.2 June 24, 2021 Forthcoming in Foundations of Physics Abstract Within quantum chemistry, the electron clouds that surround nuclei in atoms and molecules are sometimes treated as clouds of probability and sometimes as clouds of charge. These two roles, tracing back to Schr¨odingerand Born, are in tension with one another but are not incompatible. Schr¨odinger'sidea that the nucleus of an atom is surrounded by a spread-out electron charge density is supported by a variety of evidence from quantum chemistry, including two methods that are used to determine atomic and molecular structure: the Hartree-Fock method and density functional theory. Taking this evidence as a clue to the foundations of quantum physics, Schr¨odinger'selectron charge density can be incorporated into many different interpretations of quantum mechanics (and extensions of such interpretations to quantum field theory). Contents 1 Introduction2 2 Probability Density and Charge Density3 3 Charge Density in Quantum Chemistry9 3.1 The Hartree-Fock Method . 10 arXiv:2105.11988v2 [quant-ph] 24 Jun 2021 3.2 Density Functional Theory . 20 3.3 Further Evidence . 25 4 Charge Density in Quantum Foundations 26 4.1 GRW Theory . 26 4.2 The Many-Worlds Interpretation . 29 4.3 Bohmian Mechanics and Other Particle Interpretations . 31 4.4 Quantum Field Theory . 33 5 Conclusion 35 1 1 Introduction Despite the massive progress that has been made in physics, the composition of the atom remains unsettled. J. J. Thomson [1] famously advocated a \plum pudding" model where electrons are seen as tiny negative charges inside a sphere of uniformly distributed positive charge (like the raisins|once called \plums"|suspended in a plum pudding).
    [Show full text]
  • Charge Density Empty 1/1
    9/30/2004 sec 3_2 charge density empty 1/1 3-2 Charge and Charge Density (pp.61-63) HO: Electric Charge Q: A: HO: Charge Density Q: A: HO: Total Charge Jim Stiles The Univ. of Kansas Dept. of EECS 9/30/2004 Electric Charge 1/2 Electric Charge Most of classical physics can be described in terms of three fundamental units, which define our physical “reality” 1. Mass (e.g., kg) 2. Distance (e.g., meters) 3. Time (e.g., seconds) From these fundamental units, we can define other important physical parameters. For example, energy can always be described in units of kg m2/s2. But, these three fundamental units alone are insufficient for describing all of classic physics—we require one more to completely describe physical reality! This fourth fundamental unit is Coulomb, the unit of electric charge. All electromagnetic phenomena can be attributed to electric charge! Jim Stiles The Univ. of Kansas Dept. of EECS 9/30/2004 Electric Charge 2/2 We shall find that electric charge is somewhat analogous to mass. However, one important difference between mass and charge is that charge can be either positive or negative! Essentially, charge (like mass) is a property of atomic particles. Specifically, we find that: The charge “on” a proton is +1.6 x 10-19 C The charge “on” a neutron is 0.0 C The charge “on” an electron is -1.6 x 10-19 C Charged particles (of all types) can be distributed (unevenly) across a volume, surface, or contour. Jim Stiles The Univ.
    [Show full text]
  • Ampere's Law Was Still Pretty Easy for Me
    Ampere’s Law Phys 122 Lecture 20 G. Rybka Some good questions from the Prelecture Electric fields do not exist inside conductors. Do magnetic fields exist? Could we spend more time talking about the properties of magnetic fields inside and outside hollow tubes? How does that differ from magnetic fields inside and outside solid cylinders? So, I can't go to your office hours because I have class right after Physics on Friday, what is the best source of getting down the right hand rule down in all of the different cases? This is still holding me back, even though this Ampere's Law was still pretty easy for me. We learned that charges are distributed on the surface of a conductor, and the current is the result of flowing charges. If the charges are all distributed on the surface of a wire, does that mean the current only flows on the surface of the wire? Right-Hand-Rules: Get ready to Click ! ! ! ! ! ! F = qv × B F = IL× B ! ! ! ! τ = r × F τ = µ × B ! ! µ I d s ×rˆ dB = 0 4π r 2 Clicker L1 rail A conducting rail of length L1 rests on the top of the circuit loop as L2 C S shown. It is free to move. A B uniform magnetic field exists in the box of dimension L by L . V 2 3 When switch s is closed, which way R does the rail move (if at all) ? L3 A) Left B) Right C) Rotates clockwise D) Rotates counterclockwise E) Does not move • Current through rail is down • B is pointed toward you • IL x B is to the LEFT Clicker L1 rail A conducting rail of length L1 rests on the top of the circuit loop as L2 C S shown.
    [Show full text]
  • Dielectric Polarization, Bound Charges, and the Electric Displacement Field
    Dielectric Polarization, Bound Charges, and the Electric Displacement Field Any kind of matter is full of positive and negative electric charges. In a dielectric, these charges are bound — they cannot move separately from each other through any macroscopic distance, — so when an electric field is applied there is no net electric current. However, the field does push the positive charges just a tiny bit in the direction of E while the negative charges are pushed in the opposite directions. Consequently, the atoms and the molecules comprising the dielectric acquire tiny electric dipole moments in the direction of E. −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+
    [Show full text]
  • Electromagnetic Fields and Energy
    MIT OpenCourseWare http://ocw.mit.edu Haus, Hermann A., and James R. Melcher. Electromagnetic Fields and Energy. Englewood Cliffs, NJ: Prentice-Hall, 1989. ISBN: 9780132490207. Please use the following citation format: Haus, Hermann A., and James R. Melcher, Electromagnetic Fields and Energy. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed [Date]). License: Creative Commons Attribution-NonCommercial-Share Alike. Also available from Prentice-Hall: Englewood Cliffs, NJ, 1989. ISBN: 9780132490207. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms 9 MAGNETIZATION 9.0 INTRODUCTION The sources of the magnetic fields considered in Chap. 8 were conduction currents associated with the motion of unpaired charge carriers through materials. Typically, the current was in a metal and the carriers were conduction electrons. In this chapter, we recognize that materials provide still other magnetic field sources. These account for the fields of permanent magnets and for the increase in inductance produced in a coil by insertion of a magnetizable material. Magnetization effects are due to the propensity of the atomic constituents of matter to behave as magnetic dipoles. It is natural to think of electrons circulating around a nucleus as comprising a circulating current, and hence giving rise to a magnetic moment similar to that for a current loop, as discussed in Example 8.3.2. More surprising is the magnetic dipole moment found for individual electrons. This moment, associated with the electronic property of spin, is defined as the Bohr magneton e 1 m = ± ¯h (1) e m 2 11 where e/m is the electronic charge­to­mass ratio, 1.76 × 10 coulomb/kg, and 2π¯h −34 2 is Planck’s constant, ¯h = 1.05 × 10 joule­sec so that me has the units A − m .
    [Show full text]