Electrostatics and Magnetostatics: Charge Densities, Coulomb's Law, and Gauss's Law, Exams of Physics

Download Electrostatics and Magnetostatics: Charge Densities, Coulomb's Law, and Gauss's Law and more Exams Physics in PDF only on Docsity! Section 1: Main results of Electrostatics and Magnetostatics Electrostatics Charge density The most fundamental quantity of electrostatics is electric charge. Charge comes in two varieties, which are called “positive” and “negative”, because their effects tend to cancel. Charge is conserved: it can not be created or destroyed. Charge is quantized. The fact is that electric charges come only in discrete lumps – integers multiples of the basic unit charge e. However, the fundamental unit of charge is so tiny that in macroscopic applications this charge quantization can be ignored. Continuous distribution of charge is described in terms of the charge density. The charge density ( )ρ r is the local charge per unit volume, so that the total charge in volume V is . (1.1) 3( ) ( ) V V V Q dq dV d rρ ρ= = =∫ ∫ ∫r r Fig.1.1 d3r O r ρ(r) V If charge q is localized in a point of space, say r0, this charge is called a point charge. In this case the charge density is given in terms of the delta function: . (1.2) 3 0( ) ( )qρ δ= −r r r For N point charges qi located at positions ri (i = 1…N) the charge density is given by . (1.3) 3 1 ( ) ( ) N i i qρ δ = = −∑r r ir In addition to volume charge density we will also be dealing with surface and line charge densities. The surface charge density ( )σ r is given the local surface charge per unit surface, such that the charge in a small surface area da at point r is then ( )dq daσ= r . Similarly, the line charge density ( )λ r is given by the local line charge per unit length dl, such that the charge in a small element of length dl at point r is then ( )dq dlλ= r . Coulomb’s law All of electrostatics originates from the quantitative statement of Coulomb’s law concerning the force acting between charged bodies at rest with respect to each other. Coulomb, in a series of experiments, showed experimentally that the force between two small charged bodies separated in air a distance large compared to their dimensions • varies directly as the magnitude of each charge, • varies inversely as the square of the distance between them, • is directed along the line joining the charges, and • is attractive for the oppositely charged bodies and repulsive if the bodies have the same type of charge. 1 Therefore, according to Coulomb’s law the force acting on charge q located at point r due to charge q1 located at point r1 is 1 3 0 1 ( ) 4 qq πε 1−= − r rF r r r , (1.4) where ε0 is the permittivity of free space. Although the thing that eventually gets measured is a force, it is useful to introduce a concept of an electric field. The electric field is defined as the force exerted on a unit charge located at position vector r. It is a vector function of position, denoted by E. F = qE, (1.5) where F is the force, E the electric field, and q the charge. According to Coulomb the electric field at the point r due to point charges qi located at positions ri (i=1…N) is given by 3 10 1( ) 4 N i i i i q πε = − = − ∑ r rE r r r , (1.6) which is the superposition of field produced by the individual point charge. The electric field can be visualized in terms of lines starting on positive charges and terminating on negative charges. Fig.1.2 r d3r' O r' r- r'ρ(r') E(r) If there is continuous distribution of the charge it can be described by a charge density ( )ρ ′r as is shown schematically in Fig.1.2. In this case the sum in Eq.(1.6) is replaced by an integral: 33 0 0 1 1( ) ( ) 4 4V V dq d rρ πε πε 3 ′ ′− −′ ′= = ′ ′− −∫ ∫ r r r rE r r r r r r . (1.7) Similarly we can write expressions for the field produced by surface and line charge densities. Gauss’s law There is an important integral expression, called Gauss’s law, which relates the flux of electric field across a closed surface and the electric charge inside this surface electric E(r). The flux of electric field through a surface S is the integral of the normal component of the electric field over the surface, S da⋅∫E n , where n is the normal to this surface. The flux over any closed surface is a measure of the total charge inside. Indeed for charges inside the closed surface field lines starting on positive charges should terminate on negative charges inside or pass through the surface. On the other hand, charges outside the surface will contribute noting to the total flux, since its field lines pass in one side and out the other. This is the essence of Gauss’s law. For a continuous charge density ρ(r), Gauss’s law takes the form: 3 0 1 ( ) S V da d r Qρ ε ⋅ =∫ ∫E n r = , (1.8) 2 Poisson and Laplace Equations We see that the behavior of an electrostatic field can be described by the two differential equations: 0 ρ ε ∇ ⋅ =E , (1.23) 0∇× =Ε . (1.24) the latter equation being equivalent to the statement that E is the gradient of a scalar function, the scalar potential Φ: = −∇ΦE . (1.25) Equations (1.23) and (1.25) can be combined into one partial differential equation for the single function Φ(x): 2 0 ρ ε ∇ Φ = − . (1.26) This equation is called the Poisson equation. In regions of space that lack a charge density, the scalar potential satisfies the Laplace equation: 2 0∇ Φ = . (1.27) Dipole potential At a large distance from a localized change distribution the electrostatic potential exhibits a multipole nature. This follows from eq.(1.15) if we assume that ′r r .The first non-vanishing term gives a monopole contribution to the potential: 0 ( ) 4 Q rπε Φ =r , (1.28) where 3( )Q dρ r′ ′= ∫ r , (1.29) is the total charge. If Q = 0, the potential is given by a dipolar term 3 0 1( ) 4 rπε ⋅ Φ = p rr , (1.30) where p is the dipole moment 3( ) d rρ ′ ′ ′= ∫p r r . (1.31) Energy in a slowly varying electric field If the external electric field varies slowly over the region where a charge distribution ( )ρ r is localized one expend the electrostatic energy given by (1.20) (1.32) 3( ) ( )U ρ= Φ∫ r r d r in multipoles. By expending the potential ( )Φ r around the origin of coordinates in a Taylor series we obtain 0 ( ) (0) ( ) ... (0) (0) ... = Φ = Φ +∇Φ ⋅ + = Φ − ⋅ + r r r r E r , (1.33) 5 so that . (1.34) 3 3( ) (0) ( ) (0)U d r d r Qρ ρ⎡ ⎤ ⎡ ⎤≈ Φ − ⋅ = Φ −⎣ ⎦ ⎣ ⎦∫ ∫r r r E ⋅p E The first term is a monopole contribution and the second term is the dipole contribution to the electrostatic energy. Higher order terms are not included. Dielectrics If electric field is applied to a medium made up of large number of atoms or molecules, the charges bound in each molecule will respond to applied field which will results in the redistribution of charges leading to a polarization of the medium. The polarization P(r') is defined as the dipole moment per unit volume. Polarization is a macroscopic quantity. The potential produced by polarization P(r') is ( ) 33 0 ( )1( ) 4 V d r πε ′ ′⋅ − ′Φ = ′−∫ P r r r r r r . (1.35) As was shown in Phys.913, this eq. can be rewritten 3 0 0 1 ( ) 1 ( )( ) 4 4V S d r da πε πε ′ ′ ′∇ ⋅ ⋅′Φ = − + ′− −∫ ∫ P r P r nr r r r r′ . (1.36) As follows from this expression, the polarization of the medium produces an effective charge which can be interpreted as bound charge or polarization charge. There are two contributions to the bound charge – bulk and surface. The bulk polarization charge density is given by ( ) ( )Pρ = −∇ ⋅r P r . (1.37) The surface polarization charge density is ( ) ( )Pσ = ⋅r P r n . (1.38) If the integration in eq.(1.36) is performed over all space the second term can formally be omitted (it is due to the abrupt change of the polarization on the surface of the dielectric). We can, therefore, make a general statement that the presence of the polarization produces an additional polarization charge so that the total charge density becomes total free Pρ ρ ρ ρ= + = −∇ ⋅P . (1.39) The respective electrostatic equations involve a macroscopic electric field which is the average over volume which includes a large number of atoms. The accurate procedure for the macroscopic averaging will be discussed later. Taking into account eq.(1.39) we can write the divergence of E as follows: [ 0 1 ρ ε ]∇ ⋅ = −∇ ⋅E P . (1.40) It is convenient to define the electric displacement D, 0ε= +D E P , (1.41) Because this field is generated is generated by free charges only. Using the electric displacement the Gauss’s law takes the form ρ∇ ⋅ =D . (1.42) In the integral form it reads as follows: 3( ) S V da d rρ⋅ =∫ ∫D n r . (1.43) 6 This is particularly useful way to represent Gauss’s law because it makes reference only on free charges. Connecting D and E is necessary before a solution for the electrostatic potential or fields can be obtained. For a linear response of the system the displacement D is proportional to E, 0ε ε= + =D E P E , (1.44) where ε is the electric permittivity. If the dielectric is not only isotropic, but also uniform, then ε is independent of position. The Gauss’s law (1.42) can then be written ρ ε ∇ ⋅ =E . (1.45) In this case all problems in that medium are reduced to those with no electric polarization, except that the electric fields produced by given charges are reduced by a factor ε/ε0. The reduction can be understood in terms of a polarization of the atoms that produce fields in opposition to that of the given charge. One immediate consequence is that the capacitance of a capacitor is increased by a factor of ε/ε0 if the empty space between the electrodes is filled with a dielectric with dielectric constant ε/ε0. Boundary conditions For solving electrostatics problems one needs to know boundary conditions for the electric field. Consider a boundary between different media, as is shown in Fig.1.4. The boundary region is assumed to carry idealized surface charge σ. Consider a small pillbox, half in one medium and half in the other, with the normal it to its top pointing from medium 1 into medium 2. According to the Gauss’s law S D da Aσ⋅ =∫ n , (1.46) where the integral is taken over the surface of the pillbox and A is the area of the pillbox lid. In the limit of zero thickness the sides of the pillbox contribute nothing to the flux. The contribution from the top and bottom surfaces to the integral gives ( )2 1A − ⋅D D n , resulting in 2 1D D σ ⊥ ⊥− = , (1.47) where is the component of the electrical displacement perpendicular to the surface. Eq. D⊥ (1.47) tells us that there is a discontinuity of the at the interface which is determined by the surface charge. D⊥ E2 D2 σ E1 D1 l A Fig. 1.4 Schematic diagram of the boundary surface between different media. 7 Fig. 1.6 Using the Stokes’s theorem it can be transformed into 0 C S d μ⋅ = ⋅∫ ∫B l J nda . (1.63) Since the surface integral of the current density is the total current I passing through the closed curve C, Ampere’s law can be written in the form: 0μ⋅ =∫ C dB l I . (1.64) Just as Gauss’s law can be used for calculation of the electric field in highly symmetric situations, so Ampere's law can be employed in analogous circumstances. Vector Potential The basic differential laws of magnetostatics are 0∇ ⋅ =B , (1.65) 0μ∇× =B J . (1.66) According to eq. (1.65) the divergence of B is zero which implies that there are no sources which produce a magnetic field. There exist no magnetic analog to electric charge. According to eq. (1.66) a magnetic field curls around current. Magnetic field lines do not begin or end anywhere – to do so would require a nonzero divergence. They either form closed loops or extend out of infinity. Now the problem is how to solve differential equations (1.65) and (1.66). If the current density is zero in the region of interest, permits the expression of the magnetic field B as the gradient of a magnetic scalar 0∇× =B potential, = −∇ΦMB . Then (1.65) reduces to the Laplace equation for ΦM, and all our techniques for handling electrostatic problems can be brought to bear. A large number of problems fall into this class, as was discussed in Phys. 913. A general method of attack is to exploit equation (1.65). If 0∇ ⋅ =B everywhere, B must be the curl of some vector field A(r), called the vector potential: ( ) ( )= ∇×B r A r . (1.67) We have, in fact, already written B in this form (1.59). Evidently, from (1.59), the general form of A is 30 ( )( ) ( ) 4 d rμ λ π ′ = + ′−∫ J rA r r r r ∇ . (1.68) The added gradient of an arbitrary scalar function Ψ shows that for a given magnetic induction B, the vector potential can be freely transformed according to ( ) ( ) ( )λ→ +∇A r A r r . (1.69) This transformation is called a gauge transformation. We will discuss gauge transformations in detail later. 10 Such transformations on A are possible because (1.67) specifies only the curl of A. The freedom of gauge transformations allows us to make have any convenient functional form we wish. ∇ ⋅A If (1.67) is substituted into the equation (1.66), we find 0μ∇×∇× =A J . (1.70) or ( ) 2 0μ∇ ∇ ⋅ −∇ =A A J . (1.71) If we now exploit the freedom implied by (5.29), we can make the convenient choice of gauge, 0∇ ⋅ =A . (1.72) In this case each rectangular component of the vector potential satisfies the Poisson equation, 2 0μ∇ = −A J . (1.73) From our discussions of electrostatics it is clear that the solution for A in unbounded space is 30 ( )( ) 4 d rμ π ′ ′= ′−∫ J rA r r r . (1.74) It is easy to see by taking directly the divergence of eq. (1.74) that indeed 0∇ ⋅ =A . Magnetic dipole moment At a large distance from a localized current distribution we can consider the asymptotic behavior of the vector potential. We have shown in Phys. 913 that the first non-vanishing contribution is given by the magnetic dipole term 0 3( ) 4 r μ π × = m rA r . (1.75) where m is the magnetic moment: [ ] 31 ( ) 2 d r= ×∫m r J r . (1.76) Forces on a Localized Current Distribution If a localized current distribution is placed in an external magnetic field B(r), it experiences a force according to Ampere’s law. The general expression for the force is given by eq. (1.57) ( ) 3( ) d r= ×∫F J r B r . (1.77) If the external magnetic field varies slowly over the region of current, a Taylor series expansion can be used to find the dominant term in the force. We expand the applied field around some suitably chosen origin within the current distribution ( ) ( ) ( ) 00 ( ) ...′=′ ′= + ⋅∇ +rB r B r B r . (1.78) The force that the field exerts on a localized current distribution is then expanded as follows: ( ) ( )3 00 ( ) ( ) ( ) ...d r d r′=′ ′= − × + × ⋅∇ +∫ ∫ rF B J r J r r B r 3 . (1.79) Now, the first integral in the last line vanishes for a localized steady state current distribution (there can't be any net flow of charge in any direction). The second integral after some transformations (see Phys. 913) gives 11 ( )= ∇ ⋅F m B , (1.80) where the gradient is to be evaluated at the center of the current distribution and m is the magnetic moment (1.76). Notice in particular that there is no force if the applied magnetic induction is uniform. More generally, the force is in the direction of the gradient of the component of B in the direction of m. The potential energy of a permanent magnetic moment in an external magnetic field can be obtained from the expression for the force (1.80). If we interpret the force as the negative gradient of the potential energy U, we find U = − ⋅m B . (1.81) This is well-known result which shows that the dipole tends to orient itself parallel to the field in the position of lowest potential energy. Magnetic Fields in Matter In the presence of matter, atomic electrons give rise to effective atomic currents (or bound currents) which produce the orbital magnetic moment. In addition, electrons have their intrinsic magnetic moments due to electron’s spin. Thus matter has the magnetic dipole moment. The magnetic dipole moment per unit volume is known as the magnetization M. The magnetization produces a magnetic field which can be described through the vector potential. In the magnetized object, each volume element 3d r′ carries a dipole moment (Fig. 1.7), so the 3( )d r′M r ′ total vector potential is given by 30 3 ( ) ( )( ) 4 d rμ π ′ ′× − ′= ′−∫ M r r rA r r r . (1.82) r - r' d3r' r r' O A(r) Fig.1.7 It was shown in Phys.913 course that eq. (1.82) can be transformed to 30 ( ) ( )( ) 4 d r daμ π ⎧ ⎫′ ′ ′ ′∇ × ×⎪ ⎪′ ′= +⎨ ⎬′ ′− −⎪ ⎪⎩ ⎭ ∫ ∫ M r M r nA r r r r r . (1.83) The first term looks like the potential of a volume current density, M = ∇×J M , (1.84) while the second term looks like the potential of a surface current density M = ×K M n , (1.85) where n is the normal unit vector. With these definitions 12