MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, Erich Ippen, and David Staelin, 6.013/ESD.013J Electromagnetics and Applications, Fall 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons AttributionNoncommercial-Share Alike. 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 6.013, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn September 20, 2005 Lecture 4: The Scalar Electric Potential and the Coulomb Superposition Integral I. Quasistatics Electroquasistatics (EQS) Magnetquasistatics (MQS) 0 ∇×E = − ∂ µ0 H ≈ 0 ∂t ( ) ∇×E = − ∂ µ0 H ∂t ( ) 0 ∂ ε0 E ∇×H = J+ ∂t ( ) ( ) ∇ i ε0 E = ρ ∇×H = J+ ∇iJ+ ∂ ∂t ( ε E) ( ) ∇ i µ0H = 0 0 ∂ρ =0 ∂t ∇iJ = 0 ( ) ∇ i ε0 E = ρ II. Irrotational EQS Electric Field 1. Conservative Electric Field 0 d ∫ E i ds = − dt ∫ µ C 0 H i da ≈ 0 S 6.013, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 4 Page 1 of 6 ∫ b E i ds = C a b ∫ E i ds + ∫ I II a E i ds = 0 ⇒ b ∫ E i ds a I b = ∫ E i ds a II Electromotive Force (EMF) EMF between 2 points (a, b) independent of path E field is conservative () r ref ( ) ∫ E i ds Φ r − Φ r ref = r Scalar electric potential b r ref b a a r ref ∫ E i ds = ∫ E i ds + ∫ E i ds = Φ ( a) − Φ (r ref ) + Φ ( r ) − Φ (b ) = Φ ( a ) − Φ ( b ) ref 2. The Electric Scalar Potential _ _ _ r = x ix + y iy + z iz _ _ _ ∆r = ∆x i x + ∆y i y + ∆z i z ∆n = ∆r cos θ 6.013, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 4 Page 2 of 6 ∆Φ = Φ ( x + ∆x, y + ∆y, z + ∆z ) − Φ ( x, y, z ) = Φ ( x, y, z ) + = ∂Φ ∂Φ ∂Φ ∆x + ∆y + ∆z − Φ ( x, y, z ) ∂x ∂y ∂z ∂Φ ∂Φ ∂Φ ∆x + ∆y + ∆z ∂x ∂y ∂z ⎡ ∂Φ _ ∂Φ _ ∂Φ _ ⎤ =⎢ ix + iy + i i ∆r ∂y ∂z z ⎥⎦ ⎣ ∂x grad Φ = ∇Φ _ ∇ = ix _ _ ∂ ∂ ∂ + iy + iz ∂x ∂y ∂z _ grad Φ = ∇Φ = i x r +∆r ∫ () ∂Φ _ ∂Φ _ ∂Φ + iy + iz ∂x ∂y ∂z ( ) E i ds = Φ r − Φ r + ∆r = −∆Φ = −∇Φ i ∆r = E i ∆r r E = −∇Φ ∆Φ = ∆Φ ∆Φ n i ∆r = ∇Φ i ∆r ∆r cos θ = ∆n ∆n ∇Φ = ∆Φ ∂Φ n= n ∆n ∂n The gradient is in the direction perpendicular to the equipotential surfaces. III. Vector Identity ∇×E = 0 E = −∇Φ ∇ × ( ∇Φ ) = 0 6.013, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 4 Page 3 of 6 IV. Sample Problem Φ ( x, y ) = V0 xy a2 (Equipotential lines hyperbolas: xy=constant) ⎡ ∂Φ _ ∂Φ _ ⎤ E = −∇Φ = − ⎢ ix+ i ⎥ ∂y y ⎦ ⎣ ∂x = _ ⎞ −V0 ⎛ _ y ix+ x iy ⎟ 2 ⎜ a ⎝ ⎠ Electric Field Lines [lines tangent to electric field] dy Ey x = = ⇒ ydy = xdx dx Ex y y2 x2 = +C 2 2 y2 − x2 = y20 − x20 [lines pass through point ( x0 , y0 ) ] (hyperbolas orthogonal to xy) Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 6.013, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 4 Page 4 of 6 V. Poisson’s Equation ∇ i E = ∇ i ( −∇Φ ) = ρ ε0 ⇒ ∇ 2 Φ = − ρ ε 0 _ _ ⎡_ ∂ ∂ ∂ ⎤ ⎡ ∂Φ _ ∂Φ _ ∂Φ _ ⎤ ∇2 Φ = ∇ i ( ∇Φ ) = ⎢ i x + iy + iz ix+ iy+ i ⎥ ⎥i⎢ ∂x ∂y ∂z ⎦ ⎣ ∂x ∂y ∂z z ⎦ ⎣ = ∂2 Φ ∂x2 + ∂2 Φ ∂y2 + ∂2 Φ ∂z2 VI. Coulomb Superposition Integral 1. Point Charge Er = − ∂Φ q q = ⇒Φ= +C 2 ∂r 4πε0r 4πε0r Take reference Φ (r → ∞ ) = 0 ⇒ C = 0 Φ= q 4πε0r 2. Superposition of Charges 6.013, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 4 Page 5 of 6 ⎡ ⎤ ⎢ ⎥ q2 dq1 dq2 1 ⎢ q1 + + ... + + ...⎥ dΦ T ( P ) = ⎥ 4πε0 ⎢ r − r r − r2 1 r − r' r − r' ⎢ ⎥ 1 2 ⎣ ⎦ Φ T (P ) = 1 4πε0 ⎡ ⎢ N qn ⎢ dq + ⎢ ⎢ n =1 r − r n r − r' all line, surface, and ⎢ volume ch arg es ⎣ ∑ ⎡ N qn 1 ⎢ ⎢ = + 4πε0 ⎢n =1 r − r n ⎢⎣ ∑ ∫ ∫ L λ ⎛⎜ r ' ⎞⎟ dl' ⎝ ⎠ + ' r −r ∫ S ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ σs ⎛⎜ r ' ⎞⎟ da' ⎝ ⎠ + ' r −r ∫ V ⎤ ρ ⎛⎜ r ' ⎞⎟ dV ' ⎥ ⎝ ⎠ ⎥ ⎥ ' r −r ⎥⎦ Short-hand notation () ∫ Φ r = V ρ ⎛⎜ r ' ⎞⎟ dV ' ⎝ ⎠ 4πε0 r − r ' 6.013, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 4 Page 6 of 6