Electricity and Magnetism--Electric field and Gauss's Law

Q q5 qi q3q2 q1 qN ˆi r Topics: 􀂄 􀂄 Electric field: concept and problems 􀂄 Feedback: 􀂄 􀂄 􀂄 Math revi􀂄 l8.022 (E&M) -Lecture 2 Energy stored in a system of charges Gauss’s law and its applications Thanks for the feedback! Scared by Pset 0? Almost all of the math used in the course is in it… ew: too fast? Will review new concepts again before using them Pace ofectures: too fast? We have a lot to cover but… please remind me! 2 Last time… 􀂄 Coulomb’s law: 􀂄 1 2 2 2 21 ˆ | | qqF r r = 􀁇 2 1 ˆ | | iN i Q i i i qQF r r = = = ∑􀁇 2V ρˆr |r|QF = ∫ 􀁇 Q dq r V ˆir September 8, 2004 8.022 – Lecture 1 Superposition principle: 21 dV Q 13 Energy associated with FCoulomb r1r2 q Q How much work do I 1 to r2? x y September 8, 2004 8.022 – Lecture 1 have to do to move q from r4 Work done to move charges 􀂄 How much work do I 1 to r2? 􀂄 􀂄 􀂄 lr1r2 q Q 2 ˆ.I I QqrW F F F r = •∫ 􀁇 􀁇􀁇 2 1 2 21 2 1 ˆ ˆ( ) r I r Qqrr dr r r r → = • = −∫ ∫ 􀁇 􀁇 i ds􀁇 September 8, 2004 8.022 – Lecture 1 have to do to move q from rAssuming radial path: Does this result depend on the path chosen? No! You can decompose any path in segments //to the radial direction and segments |_ to it. Since the component on the |_ is nul the result does not change. where Coulomb ds =− =− Qq Qq W r F ds =− 2September 8, 2004 8.022 – Lecture 1 5 Corollaries 􀂄 1 2 is the 􀂄 x y P1 P2 3 2 112 1 2 3 W F F ds F ds = • = • = • ∫ ∫ ∫ 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 11 0W F= • =∫ 􀁇 􀁇 􀁶 The work performed to move a charge between Pand Psame independently of the path chosen The work to move a charge on a close path is zero: In other words: the electrostatic force is conservative! Path Path Path ds Any ds This will allow us to introduce the concept of potential (next week) 6 Energy of a system of charges x y 1 ( ) P WQ ∞ = • =∫ 􀁇 􀁇 13 2 312 13 23 qqqqW W W W r r r++ + + + = + + = + + q1 q2 q3 12 12 12 I qqW F r+ = • =∫ 􀁇 􀁇 1 1 1 2 jNiN ij i j ijji qqU r == = = ≠ = ∑∑ September 8, 2004 8.022 – Lecture 1 How much work does it take to assemble a certain configuration of charges? 0 no other charges: F=0 F ds 12 3 1 2 1 3 2 3 12 q q ds Energy stored by N charges: 37 􀂄 1 charge in the Universe 􀂄 2 charges in the Universe 􀀯 􀃆 The electric field 2ˆ ||q qQF r r = 􀁇 Q ESF 􀁇 q 2ˆ || qF QE r q r = = 􀁇 􀁇 September 8, 2004 8.022 – Lecture 1 Q: what is the best way of describing the effect of charges? But: the force F depends on the test charge q… define a quantity that describes the effect of the charge Q on the surroundings: Electric Field Units: dynes/e.s.u 8 􀂄 􀂄 Properties: 􀂄 Fi􀂄 ll lElectric field lines Sink + -Demo September 8, 2004 8.022 – Lecture 1 Visualize the direction and strength of the Electric Field: Direction: //to E, pointing towards – and away from + Magnitude: the denser the lines, the stronger the field. eld lines never cross (if so, that’s where E=0) They are orthogona to equipotential surfaces (wil see thisater). Faucet 4109 Problem: Calculate the electric field created by a uniformly charged ri􀂄 􀂄 Electric field of a ring of charge 3 2 2 2 Qz ˆ ( ) E z R z = + 􀁇 R z P September 8, 2004 8.022 – Lecture 1 ng on its axis Special case: center of the ring General case: any point P on the axis Answers: • Center of the ring: E=0 by symmetry • General case: Problem: Find the electric field created by a iiElectric field of disk of charge R z P Trick: And we know Ering… r September 8, 2004 8.022 – Lecture 1 disk of charges on the axs of the dsk a disk is the sum of an infinite number of infinitely thin concentric rings. (creative recycling is fair game in physics) 5Electric fi􀃆 􀃆 Integrating on r: 0􀃆R: E of disk of charge (cont.) R z P r 3 2 2 2 20 0 2 2 1 1ˆ ˆ2 ||( ) rR rR r r zE z z R zr z σπ πσ = = = = ⎛ ⎞ = = = −⎜ ⎟ +⎝ ⎠+ ∫ ∫ 􀁇 􀁇 3 2 2 2 2 ˆ ( ) zdE z r z σ π = + 􀁇 3 2 2 2 ˆ() ( ) zQE r z r z = + 􀁇 2dq da rσ πσ= = September 8, 2004 8.022 – Lecture 1 11 eld of a ring of radius r: If charge is uniformly spread: Electric field created by the ring is: dr dE zz rdr ring dr Special case 1: R􀃆infinity For finite R: 􀃆R z P 2 2 1 1ˆ2 || E z R z πσ ⎛ ⎞ = −⎜ ⎟ +⎝ ⎠ 􀁇 Since 􀃆 Conclusion: 2 2 1lim 0 R R z→∞ = + ˆ2E zπσ= 􀁇 􀂄 􀂄 Magnitude: 2πσ September 8, 2004 8.022 – Lecture 1 12 What if Rinfinity? E.g. what if R>>z? zz Electric Field created by an infinite conductive plane: Direction: perpendicular to the plane (+/-z) (constant!) 6September 8, 2004 8.022 – Lecture 1 For finite R: Special case 2: h>>R R z P 2 2 1 1ˆ2 || E zz z R z πσ ⎛ ⎞ = −⎜ ⎟ +⎝ ⎠ 􀁇 􀂄 􀂄 σπ r2 􀃆 2 􀂄 􀂄 1/22 2 2 2 2 2 1 1 1ˆ ˆ2 2 1 1 || 1ˆ ˆ ˆ~2 1 (1 )2 RE zz zz z z zR z R R Qz z z z z z πσ πσ πσ πσ −⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟= − = − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠+⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎛ ⎞⎛ ⎞ ⎛ ⎞− − = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ 􀁇 13 What happens when h>>R? Physicist’s approach: The disk will look like a point charge with Q=E=Q/zMathematician's approach: Calculate from the previous result for z>>R (Taylor expansion): v 􀁇 The concept of flux 􀂄 􀂄 􀂄 􀂄 Define the loop area vector as ˆ ˆ ˆ) ( )x y z x y z≡ + + ≡􀁇 ˆn ˆA An≡ 􀁇 September 8, 2004 8.022 – Lecture 1 14 Consider the flow of water in a river The water velocity is described by Immerse a squared wire loop of area A in the water (surface S) Q: how much water will flow through the loop? E.g.: What is the “flux of the velocity” through the surface S? (,,, , vxyz vx vy vz v v v 7... ... What is the flux of the velocity? l􀂄 􀂄 General case (definition of flux): v v v 2) if ; = . Av AvvA AvvAθ θ ⊥ = 􀁇 􀁇 􀁇 􀁇􀀦 􀁇 􀁇􀁇 􀁇􀀨 i vS vdAΦ=∫􀁇 􀁇􀁇 i ˆn v 􀁇 ˆn v 􀁇 ˆn v 􀁇 θ September 8, 2004 8.022 – Lecture 1 15 It depends on how the oop is oriented w.r.t. the water… Assuming constant velocity and plane loop: 1) if 0; 3) ifcos vA → Φ= →Φ= →Φ= F.A.Q.: what is the direction of dA? 􀂄 􀂄 􀂄 􀂄 􀂄 􀂄 “dA iˆn September 8, 2004 8.022 – Lecture 1 16 Defined unambiguously only for a 3d surface: At any point in space, dA is perpendicular to the surface It points towards the “outside” of the surface Examples: Intuitively: s oriented in such a way that if we have a hose inside the surface the flux through the surface will be positive” 8Flux of Electric Field : 􀂄 􀂄 Interpretation: Represent E using field lines: ΦE iield and/or surface! E S∫􀁇 􀁇􀁇 i E 􀁇 θcos S EdA θΦ= = =∫ 􀁇 􀁇􀁇 􀁇 i i September 8, 2004 8.022 – Lecture 1 17 DefinitionExample: uniform electric field + flat surface Calculate the flux: s proportional to Nfield lines that go through the loop NB: this interpretation is valid for any electric fEdA Φ≡ Φ = A EA E􀂄 􀂄 Calculate Φ1, Φ2,Φ3 􀂄 􀂄 Φ2=0􀂄 | Φ1|=|Φ3􀃆 ΦE through closed (3d) surface S EdA θΦ= =∫ 􀁇􀁇 i 1 2 3totΦ= +Φ +ΦΦ 1 2 3 E 􀁇 ˆE n⊥ 􀁇 dA dA September 8, 2004 8.022 – Lecture 1 18 Consider the total flux of E through a cylinder: Cylinder axis is //to field lines because | but opposite sign since The total flux through the cylinder is zero! cos A E 9Q1: 􀂄 Clue: 􀂄 iΦEl􀂄 Answer: 􀂄 No: all field liConclusion: ΦE through closed empty surface September 8, 2004 8.022 – Lecture 1 19 Is this a coincidence due to shape/orientation of the cylinder? Think about interpretaton of : proportional # of fied lines through the surface… nes that get into the surface have to come out! The electric flux through a closed surface that does not contain charges is zero. Q1: 􀂄 Clue: 􀂄 iΦEiConclusion: ΦE through surface containing Q Sink + -September 8, 2004 8.022 – Lecture 1 20 What if the surface contains charges? Think about interpretaton of : the lines will ether originate in the surface (positive flux) or terminate inside the surface (negative flux) The electric flux through a closed surface that does contain a net charge is non zero. Faucet 10.. Problem: 􀂄 Calculate ΦE +Q sphere of radius R Solution: 􀂄 􀂄 􀃆 Simple example: ΦE +Q 2 2 2 2 2 //ˆ 4 4 S S S E QE r R Q Q Q R QR R R π π = = = = =∫ ∫ ∫ 􀁇􀁇 􀁇 􀁇􀁇 i September 8, 2004 8.022 – Lecture 1 21 for point charge at the center of a of charge at center of sphere everywhere on the sphere Point charge at distance R: dA E dA dA dA Φ = Impossible integral? Use intuition and interpretation of flux! 􀂄 Version 1: 􀂄 1 􀂄 􀃆 ΦS1 =ΦS=4πQ 􀂄 Version 2: 􀂄 ΦE through a generic surface +Q S1 S Conclusion: Φis proportional4 S E dA QπΦ= =∫ 􀁇􀁇 i􀁶 September 8, 2004 8.022 – Lecture 1 22 What if the surface is not spherical S? Consider the sphere SField lines are always continuous Purcell 1.10 or next lecture The electric flux through any closed surface S containing a net charge Q to the charge enclosed: enc Gauss’s law 11Thoughts on Gauss’s law 􀂄 􀂄 􀂄 􀃆 i􀂄 Given E 􀃆 􀂄 􀂄 πQ 􀂄 Is Gauss’s law always useful? 􀂄 4)S QπΦ= =∫ 􀁇􀁇 i􀁶 September 8, 2004 8.022 – Lecture 1 23 Why is Gauss’s law so important? Because it relates the electric field E with its sources Q Given Q distribution find E (ntegral form) find Q (differential form, next week) Is Gauss’s law always true? Yes, no matter what E or what S, the flux is always = 4No, it’s useful only when the problem has symmetries (Gauss's law in integral formencl EdA Electric field of spherical distribution of charges ρ) + R i2 l iiiComment: correct but usually ' ' ' ' 2 2 2 '0 '0 ' 0 '0 ' ' ' rr rr r r rr r r r r dq dV rdE d d r r r ρ ρθ φ = = = = = = = = = = =∫ ∫ ∫ ∫ ∫ ∫ i ix y September 8, 2004 8.022 – Lecture 1 24 Applications of Gauss’s law: Problem: Calculate the electric field (everywhere in space) due to a spherical distribution of positive charges or radius R. (NB: solid sphere with volume charge density Approach #1 (mathematician) • I know the E due to a pont charge dq: dE=dq/r• I know how to integrate • Sove the integralnsde and outsde the sphere (e.g. rR) heavy on math! sin '' d d θθφ Approach #2 (physicist) • Why would I ever solve an ntegrals somebody (Gauss) already did it for me? • Just use Gauss’s theorem… Comment: correct, much much less time consuming! 12Applications of Gauss’s law: Electric field of spherical distribution of charges Physicist’s solution: 1) Outside the sphere (r>R) Apply Gauss on a sphere S1 of radius r: S2􀁇􀁇 Φ=􀁶∫S1 EdA =4πQi enclosed r 􀁇 + S ym m etry: E is constant on S and 􀀦 to dA .􀁇􀁇1 R S1 􀁶∫S1 EdA E 4πr =4πQi = 2 r Q For r>R, sphere looks → E = r2 like a point charge! 2) Inside the sphere (rR for rR) Apply Gauss on a sphere of radius r: Sym m etry: E is constant on S and to dA. sam e as point charge! enclosed encl EdA EdA E σ π 1) Inside the sphere (r