PHYS 110A - HW #5 Solutions by David Pace Any Referenced Equations Are from Griffiths
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Electrostatics Vs Magnetostatics Electrostatics Magnetostatics
Electrostatics vs Magnetostatics Electrostatics Magnetostatics Stationary charges ⇒ Constant Electric Field Steady currents ⇒ Constant Magnetic Field Coulomb’s Law Biot-Savart’s Law 1 ̂ ̂ 4 4 (Inverse Square Law) (Inverse Square Law) Electric field is the negative gradient of the Magnetic field is the curl of magnetic vector electric scalar potential. potential. 1 ′ ′ ′ ′ 4 |′| 4 |′| Electric Scalar Potential Magnetic Vector Potential Three Poisson’s equations for solving Poisson’s equation for solving electric scalar magnetic vector potential potential. Discrete 2 Physical Dipole ′′′ Continuous Magnetic Dipole Moment Electric Dipole Moment 1 1 1 3 ∙̂̂ 3 ∙̂̂ 4 4 Electric field cause by an electric dipole Magnetic field cause by a magnetic dipole Torque on an electric dipole Torque on a magnetic dipole ∙ ∙ Electric force on an electric dipole Magnetic force on a magnetic dipole ∙ ∙ Electric Potential Energy Magnetic Potential Energy of an electric dipole of a magnetic dipole Electric Dipole Moment per unit volume Magnetic Dipole Moment per unit volume (Polarisation) (Magnetisation) ∙ Volume Bound Charge Density Volume Bound Current Density ∙ Surface Bound Charge Density Surface Bound Current Density Volume Charge Density Volume Current Density Net , Free , Bound Net , Free , Bound Volume Charge Volume Current Net , Free , Bound Net ,Free , Bound 1 = Electric field = Magnetic field = Electric Displacement = Auxiliary -
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. -
Review of Electrostatics and Magenetostatics
Review of electrostatics and magenetostatics January 12, 2016 1 Electrostatics 1.1 Coulomb’s law and the electric field Starting from Coulomb’s law for the force produced by a charge Q at the origin on a charge q at x, qQ F (x) = 2 x^ 4π0 jxj where x^ is a unit vector pointing from Q toward q. We may generalize this to let the source charge Q be at an arbitrary postion x0 by writing the distance between the charges as jx − x0j and the unit vector from Qto q as x − x0 jx − x0j Then Coulomb’s law becomes qQ x − x0 x − x0 F (x) = 2 0 4π0 jx − xij jx − x j Define the electric field as the force per unit charge at any given position, F (x) E (x) ≡ q Q x − x0 = 3 4π0 jx − x0j We think of the electric field as existing at each point in space, so that any charge q placed at x experiences a force qE (x). Since Coulomb’s law is linear in the charges, the electric field for multiple charges is just the sum of the fields from each, n X qi x − xi E (x) = 4π 3 i=1 0 jx − xij Knowing the electric field is equivalent to knowing Coulomb’s law. To formulate the equivalent of Coulomb’s law for a continuous distribution of charge, we introduce the charge density, ρ (x). We can define this as the total charge per unit volume for a volume centered at the position x, in the limit as the volume becomes “small”. -
Section 14: Dielectric Properties of Insulators
Physics 927 E.Y.Tsymbal Section 14: Dielectric properties of insulators The central quantity in the physics of dielectrics is the polarization of the material P. The polarization P is defined as the dipole moment p per unit volume. The dipole moment of a system of charges is given by = p qiri (1) i where ri is the position vector of charge qi. The value of the sum is independent of the choice of the origin of system, provided that the system in neutral. The simplest case of an electric dipole is a system consisting of a positive and negative charge, so that the dipole moment is equal to qa, where a is a vector connecting the two charges (from negative to positive). The electric field produces by the dipole moment at distances much larger than a is given by (in CGS units) 3(p⋅r)r − r 2p E(r) = (2) r5 According to electrostatics the electric field E is related to a scalar potential as follows E = −∇ϕ , (3) which gives the potential p ⋅r ϕ(r) = (4) r3 When solving electrostatics problem in many cases it is more convenient to work with the potential rather than the field. A dielectric acquires a polarization due to an applied electric field. This polarization is the consequence of redistribution of charges inside the dielectric. From the macroscopic point of view the dielectric can be considered as a material with no net charges in the interior of the material and induced negative and positive charges on the left and right surfaces of the dielectric. -
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. -
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. -
Magnetic Fields
Welcome Back to Physics 1308 Magnetic Fields Sir Joseph John Thomson 18 December 1856 – 30 August 1940 Physics 1308: General Physics II - Professor Jodi Cooley Announcements • Assignments for Tuesday, October 30th: - Reading: Chapter 29.1 - 29.3 - Watch Videos: - https://youtu.be/5Dyfr9QQOkE — Lecture 17 - The Biot-Savart Law - https://youtu.be/0hDdcXrrn94 — Lecture 17 - The Solenoid • Homework 9 Assigned - due before class on Tuesday, October 30th. Physics 1308: General Physics II - Professor Jodi Cooley Physics 1308: General Physics II - Professor Jodi Cooley Review Question 1 Consider the two rectangular areas shown with a point P located at the midpoint between the two areas. The rectangular area on the left contains a bar magnet with the south pole near point P. The rectangle on the right is initially empty. How will the magnetic field at P change, if at all, when a second bar magnet is placed on the right rectangle with its north pole near point P? A) The direction of the magnetic field will not change, but its magnitude will decrease. B) The direction of the magnetic field will not change, but its magnitude will increase. C) The magnetic field at P will be zero tesla. D) The direction of the magnetic field will change and its magnitude will increase. E) The direction of the magnetic field will change and its magnitude will decrease. Physics 1308: General Physics II - Professor Jodi Cooley Review Question 2 An electron traveling due east in a region that contains only a magnetic field experiences a vertically downward force, toward the surface of the earth. -
Equivalence of Current–Carrying Coils and Magnets; Magnetic Dipoles; - Law of Attraction and Repulsion, Definition of the Ampere
GEOPHYSICS (08/430/0012) THE EARTH'S MAGNETIC FIELD OUTLINE Magnetism Magnetic forces: - equivalence of current–carrying coils and magnets; magnetic dipoles; - law of attraction and repulsion, definition of the ampere. Magnetic fields: - magnetic fields from electrical currents and magnets; magnetic induction B and lines of magnetic induction. The geomagnetic field The magnetic elements: (N, E, V) vector components; declination (azimuth) and inclination (dip). The external field: diurnal variations, ionospheric currents, magnetic storms, sunspot activity. The internal field: the dipole and non–dipole fields, secular variations, the geocentric axial dipole hypothesis, geomagnetic reversals, seabed magnetic anomalies, The dynamo model Reasons against an origin in the crust or mantle and reasons suggesting an origin in the fluid outer core. Magnetohydrodynamic dynamo models: motion and eddy currents in the fluid core, mechanical analogues. Background reading: Fowler §3.1 & 7.9.2, Lowrie §5.2 & 5.4 GEOPHYSICS (08/430/0012) MAGNETIC FORCES Magnetic forces are forces associated with the motion of electric charges, either as electric currents in conductors or, in the case of magnetic materials, as the orbital and spin motions of electrons in atoms. Although the concept of a magnetic pole is sometimes useful, it is diácult to relate precisely to observation; for example, all attempts to find a magnetic monopole have failed, and the model of permanent magnets as magnetic dipoles with north and south poles is not particularly accurate. Consequently moving charges are normally regarded as fundamental in magnetism. Basic observations 1. Permanent magnets A magnet attracts iron and steel, the attraction being most marked close to its ends. -
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 -
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. -
Class 35: Magnetic Moments and Intrinsic Spin
Class 35: Magnetic moments and intrinsic spin A small current loop produces a dipole magnetic field. The dipole moment, m, of the current loop is a vector with direction perpendicular to the plane of the loop and magnitude equal to the product of the area of the loop, A, and the current, I, i.e. m = AI . A magnetic dipole placed in a magnetic field, B, τ experiences a torque =m × B , which tends to align an initially stationary N dipole with the field, as shown in the figure on the right, where the dipole is represented as a bar magnetic with North and South poles. The work done by the torque in turning through a small angle dθ (> 0) is θ dW==−τθ dmBsin θθ d = mB d ( cos θ ) . (35.1) S Because the work done is equal to the change in kinetic energy, by applying conservation of mechanical energy, we see that the potential energy of a dipole in a magnetic field is V =−mBcosθ =−⋅ mB , (35.2) where the zero point is taken to be when the direction of the dipole is orthogonal to the magnetic field. The minimum potential occurs when the dipole is aligned with the magnetic field. An electron of mass me moving in a circle of radius r with speed v is equivalent to current loop where the current is the electron charge divided by the period of the motion. The magnetic moment has magnitude π r2 e1 1 e m = =rve = L , (35.3) 2π r v 2 2 m e where L is the magnitude of the angular momentum. -
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.