6.641,Electromagnetic Fields,Forces,and Motion Prof.Markus Zahn Lecture 5:Method of Images I.Point Charge Above Ground Plane 1.Potential and Electric Field 0 (x,y,z) Z Induced Surface Charge q d d - 个 Image Charge Φp= q 1 4πe0 ye-y ax 9 z(xI.tyi,e-0T.)Zxiyi(+ 40 z[2+y+-7z[x+y++形 q (-d) =eg (perpendicular to equipotential ground plane) 6.641,Electromagnetic Fields,Forces,and Motion Lecture 5 Prof.Markus Zahn Page 1 of 11
6.641, Electromagnetic Fields, Forces, and Motion Lecture 5 Prof. Markus Zahn Page 1 of 11 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 5: Method of Images I. Point Charge Above Ground Plane 1. Potential and Electric Field ( ) ( ) p 22 22 2 2 0 q1 1 4 x y zd x y zd ⎡ ⎤ ⎢ ⎥ Φ= − π + +− + ++ ⎣ ⎦ ε ___ ppp E iii p xyz p xyz ⎡ ⎤ ∂Φ ∂Φ ∂Φ = −∇Φ = − + + ⎢ ⎥ ∂∂∂ ⎣ ⎦ 0 2 q 4 = πε ( ) __ _ xy z xi yi z d i 2 ⎛ ⎞ ⎜ ⎟ + +− ⎝ ⎠ ( ) 3 2 2 2 2 2 x y zd − ⎡ ⎤ + +− ⎢ ⎥ ⎣ ⎦ ( ) __ _ xy z xi yi z d i 2 ⎛ ⎞ ⎜ ⎟ + ++ ⎝ ⎠ ( ) 3 2 2 2 2 x y zd ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎡ ⎤ + ++ ⎥ ⎢ ⎢ ⎥ ⎣ ⎦ ⎥ ⎣ ⎦ ( ) (− ) = = π ⎡ ⎤ + + ⎣ ⎦ ε _ p z 3 0 2 2 22 q d Ez0 i 2 xyd (perpendicular to equipotential ground plane)
2.Gauss's Law Boundary Condition ∮E.da=jpav 万=1 area ds 土在土寸 i2=-万 E2 Os (Surface charge density) ()dsd(total charge inside plbox) o,=on…[月，-E2] 3.Back to Point Charge Above Ground Plane n=2 E =-qdiz 个 2e[x2+y2+d2]32 E2=0 2 At z=0: -6]6i.242 -qd -gd 2-[x2+y+d72[P+ r2=x2+y2 a-11ww-wzn 6.641,Electromagnetic Fields,Forces,and Motion Lecture 5 Prof.Markus Zahn Page 2 of 11
6.641, Electromagnetic Fields, Forces, and Motion Lecture 5 Prof. Markus Zahn Page 2 of 11 2. Gauss’s Law Boundary Condition 0 S V ε E da dV = ρ ∫ ∫ i v 0 00 s ( ) 11 22 S εεε E da E n E n dS dS = + =σ ∫ i ii v (total charge inside pillbox) s 0 1 2 σ= − nE E ⎡ ⎤ ⎣ ⎦ ε i 3. Back to Point Charge Above Ground Plane _ s 0 0 0z 1 2 z1 3 3 2 2 22 2 22 qd qd nE E i E E 2x y d 2r d − − σ= − = = = = ⎡ ⎤ ⎣ ⎦ π ++ π + ⎡ ⎤ ⎡⎤ ⎣ ⎦ ⎣⎦ ε εε i i = + 2 22 r xy ( ) 2 T ss y x r0 0 qd q z 0 dxdy rdrd 2 +∞ +∞ ∞ π = −∞ = −∞ = φ= − = = σ = σ φ= ∫ ∫ ∫∫ π ( ) 2 π 3 2 22 r 0 rdr r d ∞ = ⎡ ⎤ + ⎣ ⎦ ∫ At z=0:
u=r2+d2 du 2rdr rdr 37 「d=u为=-1 3/ [r2+ 2+ 92=o)=4 =-q -q2 f。-4co2d -q2 -16mEod2 II.Point Charge and Sphere 1.Grounded Sphere Conducting sphere at zero potential inducing chorgc b outside sphere (a) Inducing charge inside sphere g-8 g D D ) Figure 2-27 (a)The field due to a point charge q.a distance D outside a conducting sphere of radius R.can be found by placing a single image charge -gR/D at a distance b=RD from the center of the sphere.(b)The same relations hold true if the charge q is inside the sphere but now the image charge is outside the sphere,since D<R. Courtesy of Krieger Publishing.Used with permission. 6.641,Electromagnetic Fields,Forces,and Motion Lecture 5 Prof.Markus Zahn Page 3 of 11
6.641, Electromagnetic Fields, Forces, and Motion Lecture 5 Prof. Markus Zahn Page 3 of 11 2 2 u r d du 2rdr =+⇒ = 1 2 3 3 2 2 2 22 2 rdr du 1 u 2u r d r d − = =− =− ⎡ ⎤ + + ⎣ ⎦ ∫ ∫ T ( ) 2 2 0 qd qz0 q r d ∞ + = = =− + ( ) _ _ 2 2 q z 2 2 0 0 q q f i 4 2d 16 d − − = = πε πε II. Point Charge and Sphere 1. Grounded Sphere Courtesy of Krieger Publishing. Used with permission
s-[2+D2-2rDcos,s[b+r2-2rbcoso r=R=0÷g-(g-( q2s2=q2s2q2R2+D2-2RD cos0=q2b2+R2-2Rbcos0 q2(R2+D2)=q2(b2+R2） g心Z0os-70as雨一g-8 BR2)-8R2-2-6g0小R2-0 o-0-)-0 6、R2 D D2→q'=-qR/D force on sphere qq' -q"R/D -q2RD 4πe(D-b)2 4π80 R2 4(D2-R27 2.Isolated Sphere [Put additional Image Charge +q'=+qR/D at center] (zero charge) Φ(r=R)=,g'=q 4xEoR4xED 6.641,Electromagnetic Fields,Forces,and Motion Lecture 5 Prof.Markus Zahn Page 4 of 11
6.641, Electromagnetic Fields, Forces, and Motion Lecture 5 Prof. Markus Zahn Page 4 of 11 0 1 q q' 4 s s' ⎛ ⎞ Φ= + ⎜ ⎟ πε ⎝ ⎠ = + − θ = +− θ ⎡ ⎤⎡ ⎤ ⎣ ⎦⎣ ⎦ 1 1 2 2 2 22 2 s r D 2rD cos , s' b r 2rb cos ( ) 2 2 q q' q q' rR 0 s s' s s' − ⎛⎞ ⎛ ⎞ Φ= =⇒ = ⇒ = ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ 2 2 22 2 2 2 2 2 2 q s ' q' s q' R D 2RD cos q b R 2Rb cos = ⇒ + − θ= + − θ ⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ ( ) ( ) 2 2 2 22 2 q' R D q b R += + 2 +q' 2 RD cos θ 2 = +q 2 Rb cos θ 2 2 q' b q D ⇒ = ( ) 2 b R 22 22 2 2 RD bR bb DR 0 D D ⎛ ⎞ + =+⇒− ++ = ⎜ ⎟ ⎝ ⎠ ( ) 2 R bDb 0 D ⎛ ⎞ − ⎜ ⎟ − = ⎝ ⎠ 2 R b D = 2 22 2 2 b R q' q q q' qR D D D = = ⇒ =− force on sphere ( ) ( ) 2 2 x 22 2 2 2 2 0 0 0 qq' q R D q RD f 4 Db R 4 DR 4 D D − − == = π − ⎛ ⎞ π − π − ⎜ ⎟ ⎝ ⎠ ε ε ε 2. Isolated Sphere [Put additional Image Charge +q' qR D = + at center] (zero charge) ( ) 0 0 q' q r R 4R4D Φ= = = π π ε ε
force on sphere q g' qq'D2-(D-b)-q'R[2bD-b2] fx= '7 4e0 1(D-b2 D2 4xoD2 (D-b)2 4p0-g】 -q'RD2 o R2[ -q2R3 4xEoD (D2-R2)2 D 48D2(D2-R2 [2D2-R2] qR D -qR/D ←b→ D 6.641,Electromagnetic Fields,Forces,and Motion Lecture 5 Prof.Markus Zahn Page 5 of 11
6.641, Electromagnetic Fields, Forces, and Motion Lecture 5 Prof. Markus Zahn Page 5 of 11 force on sphere ( ) ( ) ( ) 2 2 2 2 x 2 2 2 2 2 2 0 0 3 0 qq' D D b q R 2bD b q q' q' f 4 D b D 4 DDb R 4 DD D ⎡ ⎤ ⎡ ⎤ − − − − ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ ⎣⎦ = − ⎢ ⎥ = = π ⎢ ⎥ − π− ⎛ ⎞ ⎣ ⎦ π − ⎜ ⎟ ⎝ ⎠ ε ε ε ( ) ( ) 22 2 2 2 3 2 2 x 2 2 32 2 32 2 0 0 q RD R R q R f 2D 2D R D D 4 DD R 4 DD R − − ⎡ ⎤ = − = − ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ π − ε ε ⎢ ⎥ ⎣ ⎦ π −
III.Demonstration 4.7.1-Charge Induced in Ground Plane by Overhead Conductor ⊙ Amp X Area A a-R Figure 4.7.2 Charge induced on ground plane by overhead conductor is measured Y10 by probe.Distribution shown is predicted 4 321 0--2-3-4 by(4.7.7. Courtesy of Hermann A.Haus and James R.Melcher.Used with permission. Φ=U R ·-入 a=12-R2 入 fe-srp.n0-xr I 2RE0 [a*x2+y2 4e0 (a+x2+y2 C'= 入 280 Φ(x=1-R,y=0) 元Ina-1+R Φ(x=1-R,y=0)=U -R2+i 2xEoa+1-R In R 6.641,Electromagnetic Fields,Forces,and Motion Lecture 5 Prof.Markus Zahn Page 6 of 11
6.641, Electromagnetic Fields, Forces, and Motion Lecture 5 Prof. Markus Zahn Page 6 of 11 III. Demonstration 4.7.1 – Charge Induced in Ground Plane by Overhead Conductor Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. ( ) ( ) ( ) ( ) 1 2 2 2 2 2 1 2 2 0 0 2 2 2 ax y ax y ln ln 2 4 ax y ax y ⎡ ⎤ − + ⎡ ⎤ −λ −λ ⎢ ⎥ − + ⎣ ⎦ Φ= = ⎢ ⎥ π π ⎢ ⎥ ⎡ ⎤ + + + + ⎣ ⎦ ⎢ ⎥ ⎣ ⎦ ε ε ( ) ( ) λ λ π = = = Φ =− = = Φ =− = −λ − + ⎡ ⎤ − + ⎢ ⎥ π +− ⎢ ⎥ ⎣ ⎦ ε ε 0 2 2 0 2 C ' , x l R, y 0 U x l R, y 0 alR ln lR l 2 alR ln R
=EoE,(x0) 、o迹 2fe-r]-fa. 元「-2(a-x)2(a+x) =4标a-x+ya++yl0 -λ.a (a2+y2） Total Charge per unit length on ground plane is: 入k-0)=y-v tan-1 y =-λ 6架紧y列 -aA d-aAC'dU take U=Uo cos ot C'Aa Vo =-isRs=-- π(a2+y2） Uosint 6.641,Electromagnetic Fields,Forces,and Motion Lecture 5 Prof.Markus Zahn Page 7 of 11
6.641, Electromagnetic Fields, Forces, and Motion Lecture 5 Prof. Markus Zahn Page 7 of 11 s 0x 0 ( ) x 0 Ex0 x = ∂Φ σ = = =− ∂ ε ε + ε = 0 λ π ε 4 0 () () ⎡ ⎡ ⎤⎡ ⎤⎤ −+− ++ ⎢ ⎢ ⎥⎢ ⎥⎥ ⎣ ⎣ ⎦⎣ ⎦⎦ d 2 2 2 2 ln a x y ln a x y dx ( ) ( ) ( ) ( ) 2 2 2 2 x 0 2a x 2a x 4 ax y ax y = ⎡ ⎤ λ −− + ⎢ ⎥ − −+ ++ ⎣ ⎦ = π ( ) 2 2 a a y −λ = π + Total Charge per unit length on ground plane is: ( ) ( ) T s 2 2 y a a x 0 dy dy a y ∞ ∞ = −∞ −∞ −λ −λ λ == σ = = π + ∫ ∫ 1 π a 1 y tan a ∞ − π −∞ = −λ ( ) ( ) s s 22 22 dq aA d aAC dU d i A dt dt dt dt ay ay σ − λ− =≈ = = π+ π+ take U U cos t 0 = ω ' ( ) =− =− ω ω π + 0 ss 0 2 2 C ' Aa v iR U sin t a y
IV.Point Electric Dipole 1.Potential 「+ z=d/2 q r 0 (d/2)cose 2=-d/2 (d/2)cose o-品是- 2r- c-v9 Note:Φ(z=0)=0 6.641,Electromagnetic Fields,Forces,and Motion Lecture 5 Prof.Markus Zahn Page 8 of 11
6.641, Electromagnetic Fields, Forces, and Motion Lecture 5 Prof. Markus Zahn Page 8 of 11 IV. Point Electric Dipole 1. Potential 0 q 11 4 rr + − ⎡ ⎤ Φ= − ⎢ ⎥ π ⎣ ⎦ ε 2 2 2 d r xy z 2 + ⎛ ⎞ = + +− ⎜ ⎟ ⎝ ⎠ 2 2 2 d r xy z 2 − ⎛ ⎞ = + ++ ⎜ ⎟ ⎝ ⎠ Note: Φ= = ( ) z0 0
2.Point Electric Dipole (r>>d) d cos *r+号co50 -9 p.gd Courtesy of Krieger Publishing.Used with permission. gr-9cos0rl1-号cos9 Lr+cos0r1+是cos0] 9 1 91+cos0- d cose 1- 1+ d 2 c0S0 4πor'2r ≈qdcos6 4πeor2 !imp=qd(dipole moment)-Φ≈pcos0 d→0 q→0 4πeor2 E=-VΦ=- ar 4πEr 2cos+sinei 6.641,Electromagnetic Fields,Forces,and Motion Lecture 5 Prof.Markus Zahn Page 9 of 11
6.641, Electromagnetic Fields, Forces, and Motion Lecture 5 Prof. Markus Zahn Page 9 of 11 2. Point Electric Dipole (r>>d) Courtesy of Krieger Publishing. Used with permission. d d r r cos r 1 cos 2 2r + ⎡ ⎤ ≈ − θ≈ − θ ⎢ ⎥ ⎣ ⎦ d d r r cos r 1 cos 2 2r − ⎡ ⎤ ≈ + θ≈ + θ ⎢ ⎥ ⎣ ⎦ 0 0 q 1 1 qd d 1 cos 1 cos 4 r 4 r 2r 2r d d 1 cos 1 cos 2r 2r ⎡ ⎤ ⎢ ⎥ ⎡ ⎛ ⎞⎤ Φ ≈ − ≈ + θ− − θ ⎢ ⎜ ⎟⎥ π π ⎢ ⎥ − θ+ θ ⎣ ⎝ ⎠⎦ ⎣ ⎦ ε ε 2 0 qd cos 4 r θ ≈ πε 2 d 0 q 0 p cos lim p qd (dipole moment) → 4 r →∞ θ = ⇒Φ≈ πε __ _ r 1 1 E ii i r r r sin θ φ ⎡ ⎤ ∂Φ ∂Φ ∂Φ = −∇Φ = − + + ⎢ ⎥ ∂ ∂θ θ ∂φ ⎣ ⎦ θ ⎡ ⎤ = θ ⎢ + θ ⎥ πε ⎣ ⎦ _ _ 3 r 0 p 2 cos i sin i 4 r
3.Field Lines:dr==2cose=2cote rde E sine dr=2cotede=Inr=2In(sine)+C r=ro sin2 0 o=r(9=2 2 中=0 4 (a) q 1 4nEod 6 Figure 44.2 (a)Croso section of equipotentials and lines of electric field intensity for the two charges of Figure 4.4.1.(b)Limit in which pair of charges form a dipole at the origin. Courtesy of Hermann A.Haus and James R.Melcher.Used with permission. 6.641,Electromagnetic Fields,Forces,and Motion Lecture 5 Prof.Markus Zahn Page 10 of 11
6.641, Electromagnetic Fields, Forces, and Motion Lecture 5 Prof. Markus Zahn Page 10 of 11 3. Field Lines: dr 2 cos Er 2 cot rd E sin θ θ == = θ θ θ ( ) dr 2 cot d lnr 2ln sin C r = θ θ⇒ = θ + 2 0 r r sin = θ r r 0 ( 2) = θ= π Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 0 q 1 4 d = πε