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Chapter 2 Introduction to Electrostatics

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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. 1.82 is independent of the path taken between the points ra and rb. It follows that the integral of the tangential component along a closed path is zero, E (r) d` = dφ (r)=0. (1.83) · I I This last result actually follows from the requirement that E (r)=0 and the application of Stoke’s theorem. ∇ × To obtain the scalar potential due to the charge density ρ0 (r) we note that 1 (1.84) ∇ (x x )2 +(y y )2 +(z z )2 − 0 − 0 − 0 q(x x0) i +(y y0) j+(z z0) k = − − − 3/2 − 2 2 2 (x x ) +(y y ) +(z z ) − 0 − 0 − 0 hr r0 i = − . − r r 3 | − 0| Comparing the expression on the right hand side of Eq. 1.84 to the integrand in Eq. 1.80 we find that can write that the scalar potential due to the charge density ρo (r) is 1 ρ (r ) φ (r)= o 0 d3r + φ , (1.85) 4πε r r 0 0 ZZZ | − 0| where φ0 is a constant which fixes the (arbitrary) location of the zero for the scalar potential. Since the observed quantity is the electric force and, therefore, the electric field, only the difference in the values of the scalar potential at any two different points is significant. (See Eq. 1.82) 19 Section 2.3 Surface charges and charge dipoles 2.3 Surface charges and charge dipoles A surface charge is a charge density which is ‘restricted to lie on a surface’. It is characterized by the equation defining the surface, say F (r)=0, (or F (x, y, z)=0)and a surface charge density, σ (r) , with dimensions of Coulombs/m2 in SI units and statcoulombs per square centimeter in gaussian units. Restricting a volume integral to a surface with a delta function in the integrand The charge density associated with a surface charge will be the product of two terms. One term will contain a delta function which restricts the density to the surface, δ (F (r)) f (r). This term should have dimensions of inverse length, 1 ∗ m− in our units. and the other will be the surface charge density, σ (r) . The function f (r) should be determined such that H (r) δ (F (r)) f (r) d3r = H (r) dS (r) (1.86) ∗ ZZZ ZZ F (r)=0 where H (r) is any ‘smooth’ function and the surface integral on the right hand side is restricted to the surface F (r)=0. Before attacking this problem we should note that, in one dimension, ∞ a (x) δ (b (x)) b0 (x) dx = a (x) δ (b (x)) db (x) (1.87) | | Z−∞ Z = a (x0) , with b (x0)=0 Since the delta function will restrict the integral to the surface we need only consider the region close to the surface. In this region let the coordinates system consist of a coordinate axis perpendicular to the surface at each point and two orthogonal axes which are tangent to the surface at each point. The unit vector which is perpendicular to the surface at each point is either n =+ F (r) / F (r) or n = F (r) / F (r) . ∇ |∇ | −∇ |∇ | 3 Ifb we let ξn be the displacementb of a point from the surface at r0 then d r =[dS (r0) n] dξn. In addition if r0 is a point on the surface then · b b b ∂F (r0 + ξn) F (r0)= . (1.88) ∇ ∂ξ · ¸ξ=0 b When performing the volume integration we carry out the integration perpendicular to the surface at each point r0 and then multiply the result by the differential surface element at r0,dS(r0) . We have in this way reduced our problem to a set of one dimensional problems, one for each point r0 on the surface. Viewing the problem in this way suggests that the appropriate expression for the function f (r) is f (r0 + ξn)=n 0F (r0 + ξn) (1.89) · ∇ = 0F (r0 + ξn) ∇ b b b It follows then that the volume charge density associated with¯ a surface charge¯ density, σ (r) , on the surface defined by ¯ b ¯ F (r)=0is ρ (r)=σ (r) δ (F (r)) F (r) (1.90) |∇ | 20 Section 2.3 Surface charges and charge dipoles Note: A simpler way to obtain this result is as follows, where δ(F (r))dF =1: R σ (r) dS (r)= σ (r) 0 δ(F(r))dF dS (r) (1.90b) ZZ F (r)=0 ZZ F (r)=0 ·Z ¸ 0 = σ (r) δ(F(r))dr0 F dS (r) ·∇ ZZ F (r)=0 ·Z ¸ 0 F = σ (r) δ(F(r))dr0 ∇ F dS (r) · F |∇ | ZZ F (r)=0 ·Z |∇ | ¸ 0 = σ (r) δ(F (r)) F dr0 nˆs dS (r) |∇ | · ZZ F (r)=0 Z 0 = σ (r) δ(F (r)) F dr0 dS (r) |∇ | · ZZ F (r)=0 Z where the primed integration symbol indicates that the integration must be taken along a direction, dr0 which is along the normal to the surface. 0 = σ (r) δ(F (r)) F dV 0 (1.90c) |∇ | ZZ F (r)=0 Z σ (r) dS (r)= σ (r) δ(F (r)) F dV (1.90d) |∇ | ZZ F (r)=0 ZZZ V contains F (r)=0 Note that the final volume integral must be over a volume which contains all of the surface, Using this form for the charge density allows one to manipulate the variables of integration more freely . 2.3.1 Example: uniformly charged ellipsoidal surface 2 2 2 The system is a uniformly charged ellipsoidal surface (Fig. ??)defined by x +y + z 1=0. The surface charge a2 b2 − density is σ0. Find the potential on the z axis. Solution: From the class notes, Eq. 1.90, the volume charge density is given by x2 + y2 z2 x2 + y2 z2 ρ (r)=σ0δ + 1 ( + 1) (1.91) a2 b2 − |∇ a2 b2 − | µ 2 2 2 ¶ 2 2 2 x + y z x + y z 1/2 = σ0δ + 1 2( + ) a2 b2 − a4 b4 µ ¶ Because of the cylindrical symmetry of the charge distribution it is convenient to work in cylindrical coordinates, x = ξ cos ϕ and y = ξ sin ϕ. In these coordinate the charge density is 21 Section 2.3 Surface charges and charge dipoles .Fig. 3 ξ2 z2 ξ2 z2 ρ (r)=2σ0δ + 1 + a2 b2 − sa4 b4 µ ¶ Taking the zero of potential at infinity, the potential on the z axis due to the surface charge is ξ2 z2 ξ2 z2 δ 2 + 2 1 4 + 4 2σ0 a b a b 3 φ (zˆz)= − d r0 SI units (1.92) 4πε ³ 2 ´ q 2 ZZZ ξ0 +(z z0) − 3 q with d r0 = ξ0dξ0 dϕ0 dz0. The integrand is independent of ϕ0 and the angle integration can be performed giving 2π. The 2 2 2 2 integrand is also only a function of ξ0 . This suggests that we define an η = a− ξ0 . In this case ξ0 dξ0 =0.5 a dη. With these changes the potential is given by 2 2 2 2 4 2 2πσ0a ∞ ∞ δ η + b− z0 1 a− η + b− z0 φ (zˆz)= − dη dz0 (1.93) 4πε 0 2 2 Z−∞ Z ¡ a η +(¢zp z0) − q The delta function allows the η integration to be carried out yielding 1/2 b 2 2 2 2 2 σ0a a− + b− z0 b− a− φ (zˆz)= − dz0 (1.94) 2ε 2 2 2 2 b "(1 b− z0 )+a− (z z0) # Z− − ¡ − ¢ 2 It is convenient to express the z coordinates in units of b, z = bξ0 and z0 = bξ0, then, with β =(b/a) , 1/2 1 2 σ0a β + ξ0 (1 β) φ (ξ bˆz)= dξ0 (1.95) 0 2 − 2 2ε 1 " 1 ξ0 + β ξ ξ0 # Z− − 0 − ¡ 22 ¢ ¡ ¢ Section 2.3 Surface charges and charge dipoles .Fig.
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