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8/30/16 Quiz 2 – 8/25/16 A positive test charge qo is released from rest at a distance r away from a charge of +Q and a distance 2r away from a charge of +2Q. How will the test charge move immediately after being released? qo +Q r A. B. C. D. E. +2Q 2r To the left To the right It doesn’t move at all Moves upward Moves downward Continuous Charge Distributions: Electric Field and Electric Flux Lecture 3 Last Week Discrete Charges • Magnitude of the electric field due to a point charge: ! E= 1 q 4πε 0 r 2 E+ P ! E− • Electric field due to a dipole at point P: E= qd p = 2πε o z 3 2πε o z 3 r+ z r- + d q+ Dipole - q- center 1 8/30/16 Today Continuous Charge Distributions • Two approaches to calculating electric field – Extension of point charge methods – Coulomb’s Law ! kq E = 2 rˆ → r ! k dq dE = 2 rˆ r Focus on this today • Integrate – Gauss’ Law !∫ • Flux (E•A) = enclosed charge S En dA = 1 Q ε 0 inside Text Reference: Chapter 22.1 Many useful examples – 22-1 to 8, etc. Electric Field from Coulomb’s Law • Discrete charges vs. continuous charge distribution • Principles remain the same – Coulomb’s Law + Law of Superposition Only change: + + + + - + - ∑ → + + + ∫ + + + + + + + + - Electric Field from Coulomb’s Law Bunch of Charges ! P + + + - qi+ - - ! E= ri Continuous Charge Distribution ! P r + dq - ! E= + k q 1 ∑ i rˆ 4πε 0 i ri 2 i Summation over discrete charges 1 dq rˆ = r2 0 ∫ 4πε "! ∫ dE ⎧ ρdV volume charge ⎪ dq = ⎨σ dA surface charge ⎪λ dL line charge Charge ⎩ density Integral over continuous charge distribution http://www.falstad.com/vector3de/ Good visualization 2 8/30/16 Charge Densities • How do we represent the charge Q on an extended object? small pieces total charge of charge Q dq Line of charge: λ = charge per unit length [C/m] dq = λ dx Surface of charge: σ = charge per unit area [C/m2] dq = σ dA = σ dx dy (Cartesian coordinates) Volume of Charge: ρ = charge per unit volume [C/m3] dq = ρ dV = ρ dx dy dz (Cartesian coordinates) Continuous Charge Distributions • Terminology used in the text – Source point, xs, where charge ΔQ(xs) is located – Field point, xp, where field ΔE(xp) is evaluated – Vector from xs to xp: r = xp - xs • Superposition – Add up all the ΔE created by all charge elements ΔQ • Limiting case: replace sum by integral over charge distribution "! ! E= 1 dq rˆ = r2 0 ∫ 4πε ∫ dE – Need to rewrite before we are able to evaluate this – Some examples… Charge Distribution Problems 1) Understand the geometry 2) Choose dq 3) Evaluate dE contribution from the infinitesimal charge element 4) Exploit symmetry as appropriate 5) Set up the integral 6) Solve the integral 7) The result! 8) Check limiting cases 3 8/30/16 Field Due to Charge on a Line Calculate the electric field at point P due to a thin rod of length L of uniform positive charge, with density λ=Q/L Steps 1-3 ! E = E x iˆ + E y jˆ • Pick any infinitesimal segment of the rod dxs – Charge of the segment is then dq = λdxs dE x = dEr cosθ dE y = dEr sinθ • Solve first for Ex k dq r2 −x cosθ = s r dEr = dE x = k λ dxs r2 x2 dE x = k λ ∫ cosθ cosθ dxs r2 x1 Field Due to Charge on a Line Calculate the electric field at point P due to a thin rod of length L of uniform positive charge, with density λ=Q/L Steps 4-7 • Change integration variable from xs to θ tanθ = yp sinθ = dxs = −y p x2 ∫ x1 cosθ dxs r2 θ2 = ⇒ xs = −y p cot θ −xs ∫ θ1 yp r ⇒r= yp sinθ d cot θ = y p csc 2 θ dθ dθ cosθ y p csc 2 θ dθ = y p 2 / sin2 θ 1 yp θ2 ∫ cosθ dθ θ1 Field Due to Charge on a Line Calculate the electric field at point P due to a thin rod of length L of uniform positive charge, with density λ=Q/L Steps 4-7 • Evaluate integral, solve for Ex Ex = kλ yp θ2 kλ θ1 p ∫ cosθ dθ = y (sinθ 2 − sinθ1) ⎛ 1 1⎞ kλ ⎛ y p y p ⎞ ⎜ − ⎟ = kλ ⎜ − ⎟ Ex = y p ⎜⎝ r2 r1 ⎟⎠ ⎝ r2 r1 ⎠ r2>0, r1>0 • Similarly for Ey ⎛ cot θ cot θ ⎞ 2 1 E y = −k λ ⎜ − ⎟ y ≠0 r1 ⎠ p ⎝ r2 E y = 0 yp=0 4 8/30/16 Field Due to Charge on a Ring Calculate the electric field along z-axis due to a circular ring of radius R of uniform positive charge, with density λ Steps 1-3 • Pick any infinitesimal segment of the ring – Total charge of the segment is then dq = λds • The electric field due to this (point) charge dE = 1 dq 4πε 0 r 2 • Symmetry: direction of the field at point P must be along the positive z direction Field Due to Charge on a Ring Step 4 – Exploit symmetry • The z-component of the field: 1 dq cosθ 4πε 0 r 2 dE z = dE ⋅ cosθ = 1 λ ds z ⋅ 4πε 0 z 2 + R 2 z 2 + R 2 = ( )( zλ = ( 4πε 0 z 2 + R 2 ) 3/2 1/2 ) ds Field Due to Charge on a Ring Steps 5-7 • Integrate over the entire ring with λ uniform – z and R fixed E = Ez = = ∫ dE z = 2π R zλ ( 4πε 0 z 2 + R 2 ) 3/2 ∫ ds 0 zλ (2π R) ( 4πε 0 z 2 + R 2 E= ) 3/2 qz ( 2 4πε 0 z + R 2 ) 3/2 q: the total charge of the ring = 2λπR 5 8/30/16 Limiting Cases Step 8 E= qz ( 4πε 0 z 2 + R 2 ) 3/2 • Two limiting cases: – The electric field vanishes as z approaches zero (or at the center of the ring) – The electric field approaches that of a point charge at large distances (i.e., large z) E= qz ( 2 4πε 0 z + R 2 ) 3/2 ≈ 1 q 4πε 0 z 2 Field From Charged Disk Problem: calculate the electric field along zaxis due to a circular disk of radius R of uniform positive charge, with density σ. • Pick any ring element (of infinitesimal width dr) of the disk • The charge of the element is dq=σ dA=σ (2π r dr) • The electric field due to the ring element dE = (σ ⋅ 2π rdr )z ( 4πε 0 z 2 + r 2 ) 3/2 Field From Charged Disk • Integrate over the entire disk R E= 0 = σz ∫ dE = ∫ 4ε σz 4ε 0 z 2 +R 2 ∫ z2 0 2rdr (z 2 +r 2 ) 3/2 = dX σz = ⋅ −2X −1/2 X 3/2 4ε 0 ( σz 4ε 0 ) R ∫ 0 ( d z2 + r 2 (z 2 +r 2 ) ) 3/2 z 2 +R 2 z2 where we have set X=z2+r2 and dX=2rdr z>0 z<0 ⎞ σ ⎛ z E= ⎜⎜1− ⎟⎟ 2 2 2ε 0 ⎝ z +R ⎠ E =− ⎞ σ ⎛ z ⎜⎜1− ⎟⎟ 2 2 2ε 0 ⎝ z +R ⎠ 6 8/30/16 Limiting Cases z > 0 E= • Small distance z (or large disk) ⎞ σ ⎛ z ⎜⎜1− ⎟⎟ 2ε 0 ⎝ z2 + R 2 ⎠ z → 0 (or R → ∞) , E → σ 2ε 0 Ez Discontinuity at z= 0 z Limiting Cases z > 0 E= • Large distance z E= = ⎞ σ ⎛ z ⎜⎜1− ⎟⎟ 2 2 2ε 0 ⎝ z +R ⎠ ⎛ ⎞ ⎟ σ R2 σ ⎜ 1 ⋅ 2 ⎜1− ⎟≈ 2 2ε 0 ⎜ 1+ R 2 ⎟ 2ε 0 2z ⎝ z ⎠ 1 σ (π R 2 ) 1 q = 4πε 0 z 2 4πε 0 z 2 Infinite Sheet of Positive Charge Ex = − σ 2ε 0 Uniform electric fields generated on both sides of sheet + + + + + + + + + + + + + + + y Ex = σ 2ε 0 x Discontinuity at x = 0 7 8/30/16 y 90° Arc of Charge x E Yʹ In this coordinate system you have to deal with both Ex and Ey In this coordinate system you only have to deal with the horizontal component of E E Xʹ uniform charge distribution total charge = Q 90° Arc of Charge (cont’d) y rdθ λ= θ x dE Q 2π r = 2Q πr 4 dq = λ r dθ k dq k λ r cosθ dθ dE x = 2 cosθ = r r2 k λ θ2 Ex = ∫ cosθ dθ r θ1 kλ = (sinθ2 − sinθ1) r For the 900 arc, θ1 = −450 and θ 2 = 450 , so Ex = k λ ⎡ 1 ⎛ 1 ⎞⎤ 2k λ 2k 2Q 2 2kQ − ⎜− = = ⎢ ⎟⎥ = r ⎣ 2 ⎝ 2 ⎠⎦ r r πr πr 2 8