Application of Vector Calculus to Electrostatics

Topics: 􀂄 􀂄 􀂄 􀂄 8.022 (E&M) – Lecture 4 More applications of vector calculus to electrostatics: Laplacian: Poisson and Laplace equation Curl: concept and applications to electrostatics Introduction to conductors 2 Last time… 􀂄 Electric potential: 􀂄 􀂄 􀂄 􀂄 􀂄 􀂄 i it ( ) -r r E φ φ ∞ = − = ∇ ∫ 􀁇 􀁊􀁇 􀁇 􀁇 􀁇 i 2 1 () 2 8 E U ρφ π = = ∫ ∫ 4 E πρ ∇ = 􀁇 i G. Sciolla – MIT 8.022 – Lecture 4 Work done to move a unit charge from infinity to the point P(x,y,z) It’s a scalar! Energy associated with an electric field: Work done to assemble system of charges is stored in E Gauss’s law in differential form: Easy way to go from E to charge d stribution that created w ith E ds Volume Entire with space charges r dV dV 13 Laplacian operator 2 f f ∇ i 2 2 2 2 2 2 2 2 2 2 2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) f f f f x y z x y z x y z x y z f f f f f x y z x y z ∂ ∂ ∂ ∂ ∂ ∂ ∇ ∇ = + + + + ∂ ∂ ∂ ∂ ∂ ∂ ⎛ ⎞ ∂ ∂ ∂ ∂ ∂ ∂ = + + = + + ⎜ ⎟ ∂ ∂ ∂ ∂ ∂ ∂ ⎝ ⎠ i i G. Sciolla – MIT 8.022 – Lecture 4 What if we combine gradient and divergence? Let’s calculate the div grad f (Q: difference wrt grad div f ?) Laplacian Operator ≡∇ ∇ ≡∇ 4 Interpretation of Laplacian φ 2+y2 2 2 2 2 2 2 2 (2 2) 4 f f x y z a a ⎛ ⎞ ∂ ∂ ∂ ∇ = + + = ⎜ ⎟ ∂ ∂ ∂ ⎝ ⎠ = + = gives the G. Sciolla – MIT 8.022 – Lecture 4 Given a 2d function (x,y)=a(x )/4 calculate the Laplacian As the second derivative, the Laplacian curvature of the function 2Poisson equation Let’s apply the concept of Laplacian to electrostatics. 􀂄 Rewrite Gauss’s law in terms of the potential 􀁇⎧∇i = 4 πρ⎪⎨􀁇⎩ i( 2⎪∇i =∇ −∇ φ) = −∇ φ 2 → ∇ φ= − 4 πρPoisson Equation G. Sciolla – MIT 8.022 – Lecture 4 5 EE Laplace equation and Earnshaw’s Theorem 􀂄 What happens to Poisson’s equation in vacuum? 2 2 ∇ = − 4 πρ⇒ ∇ =0 Laplace Equation φ φ 􀂄 What does this teach us? In a region where φ satisfies Laplace’s equation, then its curvature must be 0 everywhere in the region 􀃆 The potential has no local maxima or minima in that region 􀂄 Important consequence for physics: Earnshaw’s Theorem: It is impossible to hold a charge in stable equilibrium with electrostatic fields (no minima) G. Sciolla – MIT 8.022 – Lecture 4 6 3 G. Sciolla – MIT 8.022 – Lecture 4 C 7 Is the equilibrium stable? No! (does the question sound familiar?) G. Sciolla – MIT 8.022 – Lecture 4 Application of Earnshaw’s Theorem 8 charges on a cube and one free in the middle. 8 The circulation 􀂄 F 􀂄 1 and C2 􀂄 1 and C2 C F ds Γ = ∫ 􀁇 􀁇 i 􀁶 C2 C1 C’ C’ 1 2 1 2 1 2 ' ' ' ' C C C C C C C F ds F ds s F ds F ds − − Γ = = + = − + + = + ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 i i i 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 i i i i 􀁇 􀁇 􀁇 􀁇 i i 􀁶 􀁶 􀁶 􀁶 􀁶 􀁶 􀁶 􀁶 􀁶 1 2 Consider the line integral of a vector function over a closed path C: Let’s now cut C into 2 smaller loops: C Let’s write the circulation C in terms of the integral on C Circulation C C C C F ds F ds F d F ds F ds ⇒ Γ = Γ +Γ 49 The curl of F 􀂄 If we repeat the procedure N times: 􀂄 F as circulation of F in the limit A􀃆0 􀂄 the 1 i i i = = Γ = Γ ∑ C 0 ˆ C A F ds F n A → ≡ ∫ 􀁇 􀁇 i 􀁊􀁇 i 􀁶 G. Sciolla – MIT 8.022 – Lecture 4 Define the curl of per unit area where A is the area inside C The curl is a vector normal to the surface A with direction given by “right hand rule” LargeN lim curl Stokes Theorem F over a closed line C and the surface integral of the curl losed by C 1 1 1 1 L 1 1 1 ˆ A ˆ ˆ curl curl ( ) i i i i C i i C i i i i i C i i i A i i i i i i F ds F ds A A F ds F n A dA A A n = = = = = = = = = Γ = = → → → Γ = = = ∫ ∑ ∑ ∑ ∫ ∫ ∑ ∫ ∑ ∑ 􀁇 􀁇 i 􀁇 􀁇 i 􀁇 􀁇 i 􀁇 i 􀁇 􀁇 􀁇 􀁇 i i i 􀁶 􀁶 􀁶 A = A C C F F dA F ds ds ⎧ → ⎪ ⎨ ⎪ Γ = ⎩ ⇒ ∑ ∫ ∫ ∫ ∫ 􀁇 􀁇 􀁇 i 􀁇 􀁇 i 􀁇 􀁇 􀁇 i i 􀁶 􀁶 C A G. Sciolla – MIT 8.022 – Lecture 4 10 NB: Stokes relates the line integral of a function of the function over the area enc LargeN LargeN LargeN LargeN LargeN LargeN In the limit 0: curl and curl Fn F A FA Γ = argeN curl (definition of circulation) curl Stokes Theorem F dA 5􀂄 􀂄 􀂄 field is zero. = C A F ds ∫ ∫ 􀁇 􀁇 􀁇 􀁇 i i 􀁶 = 0 C F ds ∫ 􀁇 􀁇 i 􀁶 A curl = 0 E dA ⇒ ∫ 􀁇 􀁇 i curl E ⇒ 􀁇 G. Sciolla – MIT 8.022 – Lecture 4 11 Application of Stoke’s Theorem Stoke’s theorem: The Electrostatics Force is conservative: The curl of an electrostatic curl F dA for any surface A = 0 Curl in cartesian coordinates (1) 􀂄 F 􀂄 F ( , , ) ~ ( , , ) 2 2 ( , , ) ~ ( , , ) 2 2 ( , , )( ) ~ ( , , ) 2 2 ( , , )( ) ~ ( , , ) 2 b y y y a c z z z b d y y y c a z z d F z z F d y z F y y F d z y F z z F d y F y z y F d z z ∂ ⎡ ⎤ ∆ ∆ = − ∆ − ∆ ⎢ ⎥ ∂ ⎣ ⎦ ⎡ ⎤ ∂ ∆ ∆ = + ∆ + ∆ ⎢ ⎥ ∂ ⎣ ⎦ ∂ ⎡ ⎤ ∆ ∆ = + − + ∆ ⎢ ⎥ ∂ ⎣ ⎦ ∆ = − − − ∫ ∫ ∫ ∫ 􀁇 􀁇 i 􀁇 􀁇 i 􀁇 􀁇 i 􀁇 􀁇 i 2 z y z z y F F F ds y z ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎡ ⎤ ∆ ⎪ ⎢ ⎥ ∂ ⎣ ⎦ ⎩ ∂ ⎛ ⎞ ∂ ⇒ = − ⎜ ⎟ ∂ ∂ ⎝ ⎠ ∫ 􀁇 􀁇 i 􀁶 P x y z ∆ y ∆ z a b c d G. Sciolla – MIT 8.022 – Lecture 4 12 Consider infinitesimal rectangle in yz plane centered at P=(x,y,z) in a vector filed Calculate circulation of around the square: s F x y z y F x y z s F x y z z F x y z s F x y z x y z s F x y F x y z − ∆ − ∆ squareYZ Adding the 4 compone nts: F y y z ∂ ∆ ∆ ∆ 6G. Sciolla – MIT 8.022 – Lecture 4 14 Curl in cartesian coordinates (2) 􀂄 Combining this result with definition of curl: 􀂄 􀃆 0 0 0 ˆ ) curl C A square y z x x y y z F ds F n F ds F F A F x y y z F F F ds y z → ∆ → ∆ → ⎧ ⎪ ≡ ∂ ⎛ ⎞ ∂ ⎪ ⇒ = = − ⎨ ⎜ ⎟ ∆ ∆ ∂ ∂ ∂ ⎛ ⎞ ∂ ⎝ ⎠ ⎪ = − ⎜ ⎟ ⎪ ∂ ∂ ⎝ ⎠ ⎩ ∫ ∫ ∫ 􀁇 􀁇 i 􀁊􀁇 􀁇 􀁇 i i 􀁇 􀁇 􀁇 i 􀁶 􀁶 􀁶 ˆ ˆ ˆ ˆ ˆ ˆ curl y y x x z z x y z x y z F F F F F F F x y z F y z z x x y x y z F F F ∂ ∂ ⎛ ⎞ ⎛ ⎞ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎛ ⎞ = − + − + − ≡ × ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 􀁇 􀁇 l: easy to calculate! 13 Similar results orienting the rectangles in //(xz) and (xy) planes square lim (curl lim y z ∆ ∆ ≡∇ This is the usable expression for the cur Summary of vector calculus in 􀂄 Gradient: 􀂄 􀂄 Divergence: 􀂄 􀂄 l i 􀂄 Curl: 􀂄 􀂄 , , x y z φ φ ⎛ ⎞ ∂ ∂ ∂ ∇ ≡ ⎜ ⎟ ∂ ∂ ∂ ⎝ ⎠ E φ 􀁇 y x z F F F F x y z ∂ ∂ ∂ ∇ = + + ∂ ∂ ∂ 􀁇 i S V E dA = ∇ ∫ ∫ 􀁇 􀁇 􀁇 i i 4 E πρ ∇ = 􀁇 i Purcell Chapter 2 A = C F ds F dA ∫ ∫ 􀁇 􀁇 􀁇 􀁇 i i 􀁶 F F = ∇ × 􀁇 􀁇 0 E ∇ × = 􀁇 G. Sciolla – MIT 8.022 – Lecture 4 electrostatics (1) In E&M: Gauss’s theorem: In E&M: Gauss’ aw in different al form Stoke’s theorem: In E&M: = − ∇ EdV curl curl 715 Summary of vector calculus in 􀂄 Laplacian: 􀂄 In E&M: 􀂄 􀂄 􀂄 l with electrostati 2 φ φ ∇ i Purcell Chapter 2 2 4 φ πρ 2 0 φ ∇ = G. Sciolla – MIT 8.022 – Lecture 4 electrostatics (2) Poisson Equation: Laplace Equation: Earnshaw’s theorem: impossib e to hold a charge in stable equilibrium c fields (no local minima) ≡ ∇ ∇ ∇ = − Comment: This may look like a lot of math: it is! Time and exercise will help you to learn how to use it in E&M Conductors and Insulators Conductor 􀂄 􀂄 2O Insulator 􀂄 􀂄 Inorganic materials: quartz, glass,… Au Free electrons Na+ ClGG Sciolla – MIT 8.022 – Lecture 4 16 : a material with free electrons Excellent conductors: metals such as Au, Ag, Cu, Al,… OK conductors: ionic solutions such as NaCl in H : a material without free electrons Organic materials: rubber, plastic,… 8. Electric Fields in Conductors (1) 􀂄 􀂄 􀂄 E=0 􀂄 Why? If E 􀃆 i i 􀂄 -17 -16 s (typi i E E + -E ---+ + ---+ + + + G. Sciolla – MIT 8.022 – Lecture 4 17 A conductor is assumed to have an infinite supply of electric charges Pretty good assumption… Inside a conductor, is not 0 charges w ll move from where the potential is higher to where the potential is lower; m gration will stop only when E=0. How long does it take? 10 -10 cal resist vity of metals) Electric Fields in Conductors (2) 􀂄 Electric potential inside a conductor is constant 􀂄 1 and P2 the ∆φ would be: 2 1 0 P P φ ∆ = = ∫ 􀁇 􀁇 i 􀂄 Net charge can only reside on the surface 􀂄 􀃆 􀂄 External field li 􀂄 ∆φ=0 􀂄 􀂄 G. Sciolla – MIT 8.022 – Lecture 4 18 Given 2 points inside the conductor P since E=0 inside the conductor. E ds If net charge inside the conductor Electric Field .ne.0 (Gauss’s law) nes are perpendicular to surface E//component would cause charge flow on the surface until Conductor’s surface is an equipotential Because it’s perpendicular to field lines 9Corollary 1 φ Why? 􀂄 􀂄 If no charge inside the cavity 􀃆 lds 􀃆 φ or minima 􀃆 φ 􀃆 E=0 􀂄 Shielding of external electric fi E=0 G. Sciolla – MIT 8.022 – Lecture 4 19 In a hollow region inside conductor, =const and E=0 if there aren’t any charges in the cavity Surface of conductor is equipotential Laplace ho cavity cannot have max must be constant Consequence: elds: Faraday’s cage Corollary 2 l conductor Why? 􀂄 l l i +Q -------+ + + + + + + + 4 ( ) ( ) Q Q Q Q Q π ⎧ = ⎪⎨ = + ⎪⎩ ⇒ = − ⇒ = − = ∫ ∫ 􀁇 􀁇 i 􀁇 􀁇 i 􀁶 􀁶 G. Sciolla – MIT 8.022 – Lecture 4 20 A charge +Q in the cavity wil induce a charge +Q on the outside of the App y Gauss’s aw to surface ---ins de the conductor 0 because E=0 inside a conductor Gauss's law conductor is overall neutral inside inside outside inside E dA E dA Q Q 1022 Corollary 3 σ =E /4π Why? 􀂄 ∆E=4πσ 􀂄 Since Einside=0 􀃆 E =4πσ σ +Q -------+ + + + + + + + G. Sciolla – MIT 8.022 – Lecture 4 21 The induced charge density on the surface of a conductor caused by a charge Q inside it is induced surface For surface charge layer, Gauss tells us that surface induced Uniqueness theorem ρ potential φ ion φ i 􀂄 φ1 and φ2 i : 􀂄 φ1 and φ2 φ1 = φ2 =φ 􀂄 φ1and φ2 will i φ3 = φ2 – φ1: 􀂄 φ3 l 􀂄 φ3=0 􀃆 φ3 􀃆 φ1 = φ2 2 2 2 3 2 1 ( ) ( ) r r r r r φ φ φ πρ πρ ∇ = − = 􀁇 􀁇 􀁇 􀁇 􀁇 2 1 2 2 ( ) r r r r φ πρ φ πρ ∇ = ∇ = 􀁇 􀁇 􀁇 􀁇 Why do I care? G. Sciolla – MIT 8.022 – Lecture 4 Given the charge density (x,y,z) in a region and the value of the electrostatic (x,y,z) on the boundaries, there is only one funct (x,y,z) wh ch describes the potential in that region. Prove: Assume there are 2 solutions: ; they w ll satisfy Poisson Both satisfy boundary conditions: on the boundary, Superposition: any combination of be solution, includ ng satisfies Lap ace: no local maxima or minima inside the boundaries On the boundaries =0 everywhere inside region everywhere inside region ( ) 4 ( ) 4 ( ) 0 = ∇ − ∇ ( ) 4 ( ) 4 ( ) A solution is THE solution! 1112 G. Sciolla – MIT 8.022 – Lecture 4 23 Uniqueness theorem: application 1 􀂄 A hollow conductor is charged until its external surface reaches a potential (relative to infinity) φ=φ0. What is the potential inside the cavity? + + ++ + + + + φ0 φ=? Solution φ=φ0 everywhere inside the conductor’s surface, including the cavity. Why? φ=φ0 satisfies boundary conditions and Laplace equation 􀃆 The uniqueness theorem tells me that is THE solution. G. Sciolla – MIT 8.022 – Lecture 4 24 Uniqueness theorem: application 2 􀂄 Two concentric thin conductive spherical shells or radii R1 and R2 carry charges Q1 and Q2 respectively. 􀂄 What is the potential of the outer sphere? (φinfinity=0) 􀂄 What is the potential on the inner sphere? 􀂄 What at r=0? R1 R2 Q1 Q2 1 1 2 2 2 2 1 2 1 2 2 2 2 1 2 2 2 2 1 Because of uniqueness: ( ) R R R RQ Q Q E ds dr r R R Q Q r r R R R φ φφ φ φ − = − =− = − ⇒ = + = ∀ < ∫ ∫ 􀁇 􀁇 i Solution 􀂄 Outer sphere: φ1=(Q1+Q2)/R1 􀂄 Inner sphere 13 G. Sciolla – MIT 8.022 – Lecture 4 25 Next time… 􀂄 More on Conductors in Electrostatics 􀂄 Capacitors 􀂄 NB: All these topics are included in Quiz 1 scheduled for Tue October 5: just 2 weeks from now!!! 􀂄 Reminders: 􀂄 Lab 1 is scheduled for Tomorrow 5-8 pm 􀂄 Pset 2 is due THIS Fri Sep 24