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9/6/16 Continuous Charge Distributions: Electric Field and Electric Flux III Lecture 5 Gauss’ Law: Cylindrical Symmetry Two long, charged concentric cylinders have radii of r1=3.0 cm and r2=6.0 cm. The charge per unit length is 5.0 x 10-6 C/m on the inner cylinder and -7.0 x 10-6 C/m on the outer cylinder. Find the electric field at (a) r = 4.0 cm and (b) r = 8.0 cm, where r is the radial distance from the common axis. r (a) r1 b (b) ! ! ε o "∫ E • dA = qenc ! ε oE "∫ dA = q E= ra r2 (λ + λ )h (λ + λ ) q = 1 2 = 1 2 2ε o π rh 2ε o π rh 2ε o π r 2π h ε oE ∫ ∫ r dθ dz = ε E2π rh = q o E= 0 0 E= E= q λh λ = = 2ε o π rh 2ε o π rh 2ε o π r −2 x10−6 2(8.55x10−12 ) π (0.08) = -4.654 x105 N / C 5 x10−6 = 2.327 x10 6 N / C 2(8.55x10−12 ) π (0.04) Gauss’ Law: Spherical Symmetry A spherically symmetric charge distribution has a charge density given by ρ=a/r where a is a constant. Find the electric field as a function of r. ! ! ε o "∫ E • dA = qenc 2π π ε oEr 2 ∫ d φ ∫ sinθ dθ = ε oE4π r 2 = q 0 0 2 r ε oE4π r = q = ∫ ρ 4π r 0 2 r dr = a4π ∫ 0 r2 4π ar 2 dr = r 2 a E= 2ε o Figure from: http://www.oceanopticsbook.info/view/light_and_radiometry/geometry 1 9/6/16 Thin Sheet of Any Charge Distribution !" !" !" ΔE = (E '+ E disk+ ) !" !" − (E '+ E disk− ) !" !" = E disk+ − E disk− ⎛ σ ⎛ σ ⎞⎞ = ⎜⎜ − ⎜− ⎟⎟⎟ n# ⎝ 2ε 0 ⎝ 2ε 0 ⎠⎠ σ = n# ε0 tiny disk !" !" !" E L = E '+ E disk− Just to left of disk Discontinuity of magnitude: !" !" !" E R = E '+ E disk+ ΔEn = Just to right of disk σ ε0 From Lecture 1 Conductors vs. Insulators • Insulators: material in which electric charges are “frozen” in place • Conductors: material in which electric charges can move around “freely Demo Conductors 5A-12 Gauss' Law: Charge Within a Conductor 2 9/6/16 Charges on a Conductor • Why do charges always move to the surface of a conductor? – Gauss’ Law Charges on a Conductor • Why do charges always move to the surface of a conductor? – Gauss’ Law • Charge redistributes inside conductor • Cancels external field inside • No field across surface => no flux through the surface • No net charge density inside conductor – Charge must reside on surface Electric Field in a Conductor Conducting cylinder • Electric field lines for an oppositely charged metal cylinder and metal plate. • Note that: 1. Electric field lines are perpendicular to the conductors 2. No electric field lines inside the cylinder • Interior shielded from the outside Conducting plate 3 9/6/16 Electric Field in a Conductor • Key points – An external electric field exerts a force on charge carriers which causes them to move – They will continue to move until their electric field negates the external field – The motion stops when the net electric field is zero = Electrostatic equilibrium • Main point to remember – The electric field is always zero inside a conductor • IF it is in a state of electrostatic equilibrium Gauss’ Law: Spherical Symmetry Electric Field inside and outside a shell of uniform charge distribution Demo Conductors: Shielding from an Electrical Field student sensor sparks screened cage Van de Graaff generator 4 9/6/16 Demo Charge Distribution on a Conductor http://cnx.org/content/m42317/latest/Figure_19_07_07a.jpg Demo Charge Distribution on a Conductor Infinite Sheet of Charge Planar E= Qenc Aε 0 A = 2π r 2 σ E= 2ε 0 • An infinite plane sheet of charge creates a CONSTANT electric field – Note: 2x smaller than for conducting plates 5 9/6/16 Conductors: External Electric Field • General case • Choose a cylindrical Gaussian surface with one end inside the conductor • Ignore the curvature – Cross section of the cylinder is chosen to be very small • The electric field must be perpendicular to the surface – Otherwise, there would be perpetual currents flowing along the surface Conductors: External Electric Field σ = surface charge density • Left end cap ! ∫ E dA = 0 because E = 0 n in a conductor • Right end cap ! E || n̂ ⇒ ∫ EndA = EA • Cylinder ∫ EndA = 0! ! because E ⊥ n̂ and E ⋅ n̂ = 0 Conductors: External Electric Field ∴ !∫ En dA = 0 + EA + 0 = EA S qinside = σ A Gauss' law: ⇒ EA = !∫ S En dA = 1 q ε 0 inside 1 σA ε0 E= σ ε0 6 9/6/16 Two Conducting Plates • The electric field near a charged conducting plate is given by - σ1 σ1 σ1 + = 2ε 0 2ε 0 ε 0 E= • Similarly E =− σ1 σ1 σ − =− 1 2ε 0 2ε 0 ε0 • What happens if the two plates are pulled toward each other? Two Conducting Plates • Excess charge on one plate attracts the excess charge on the other plate – Surface charge density • Vanishes on the outer faces - • Doubled on the inner faces • Electric field magnitude – 2x that near an isolated plate between the plates – 0 everywhere else E= 2σ 1 ε0 • (c) not a superposition of (a) and (b) – Charges redistributed Uniform Charge Density: Summary Non-conductor Cylindrical symmetry Planar Spherical symmetry ρr E= 2ε 0 E= E =0 2 R ρ 2ε 0r σ 2ε 0 1 Q E= r 4πε 0 R 3 E= E= Conductor 1 Q 4πε 0 r 2 E= λ 2πε 0r E= σ ε0 E =0 E= 1 Q 4πε 0 r 2 inside outside inside outside 7 9/6/16 Summary of Lectures 3, 4 & 5 • Gauss’ law relates net flux, F, of an electric field through a closed surface to the net charge that is enclosed by the surface ! ! ε oφ = ε o "∫ E • dA = qenc • Takes advantage of certain symmetries (spherical, cylindrical, planar) • Gauss’ Law proves electric fields vanish in conductors while extra charges reside on surface 8 9/6/16 9