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Transcript
Physics 2113 Jonathan Dowling Flux Capacitor (Schematic) Physics 2113 Lecture: 09 MON 14 SEP Gauss’ Law II Carl Friedrich Gauss 1777 – 1855 Gauss’ Law: General Case • Consider any ARBITRARY CLOSED surface S -- NOTE: this does NOT have to be a “real” physical object! qins S • The TOTAL ELECTRIC FLUX through S is proportional to the TOTAL CHARGE ENCLOSED! • The results of a complicated integral is a very simple formula: it avoids long calculations! (One of Maxwell’s 4 equations!) Gauss’ Law: ICPP "+ " = +q "- " = -q F1 = +2 + 5 + 7 - 3- 4 - 7 F2 = +3+ 5 +10 - 3- 4 - 6 0 + F 3 = +2 + 5 + 8 - 5 - 6 - 7 - 23-3 A Charged Isolated Conductor Learning Objectives • • • 23.17 For a conductor with a cavity that contains a charged object, 23.14 Apply the relationship determine the charge on the cavity between surface charge density wall and on the external surface. σ and the area over which the • 23.18 Explain how Gauss’ law is charge is uniformly spread. used to find the electric field 23.15 Identify that if excess magnitude E near an isolated charge (positive or negative) is conducting surface with a uniform placed on an isolated conductor, surface charge density σ. that charge moves to the surface • 23.19 For a uniformly charged and none is in the interior. conducting surface, apply the 23.16 Identify the value of the relationship between the charge electric field inside an isolated density σ and the electric field conductor. magnitude E at points near the conductor, and identify the direction © 2014 John Wiley &of Sons, All vectors. theInc. field • rights reserved. Properties of Conductors Inside a Conductor in Electrostatic Equilibrium, the Electric Field Is ZERO. Why? Because If the Field Is Not Zero, Then Charges Inside the Conductor Would Be Moving. SO: Charges in a Conductor Redistribute Themselves Wherever They Are Needed to Make the Field Inside the Conductor ZERO. Excess Charges Are Always on the Surface of the Conductors. ICPP: Conducting Sphere • A spherical conducting shell has an excess charge of +10 C. • A point charge of –15 C is located at center of the sphere. • Use Gauss’ Law to calculate the charge on inner and outer surface of spherical shell R2 + + R1 + –15C - (a) Inner: +15 C; outer: 0 (b) Inner: 0; outer: +10 C (c) Inner: +15 C; outer: –5 C - +10 C S1 + + - S2 "+ " = +5C "- " = -5C Hint: E-Field is Zero inside conductor so Gauss’ Law: Conducting Sphere • Inside a conductor, E = 0 under static equilibrium! Otherwise electrons would keep moving! • Construct a Gaussian surface inside the metal as shown. (Does not have to be spherical!) –5 C • Since E = 0 inside the metal, flux through this surface = 0 • Gauss’ Law says total charge enclosed = 0 • Charge on inner surface = +15 C Since TOTAL charge on shell is +10 C, Charge on outer surface = +10 C - 15 C = -5 C! +15C –15C Faraday’s Cage • Given a hollow conductor of arbitrary shape. Suppose an excess charge Q is placed on this conductor. Suppose the conductor is placed in an external electric field. How does the charge distribute itself on outer and inner surfaces? (a) Inner: Q/2; outer: Q/2 (b) Inner: 0; outer: Q (c) Inner: Q; outer: 0 • Choose any arbitrary surface inside the metal • Since E = 0, flux = 0 • Hence total charge enclosed = 0 All charge goes on outer surface! Safe in the Plane!? Inside cavity is “shielded” from all external electric fields! “Faraday Cage effect” Safe in the Plane!? Faraday’s Cage: Electric Field Inside Hollow Conductor is Zero Safe in the Car!? E=0 • Choose any arbitrary surface inside the metal • Since E = 0, flux = 0 • Hence total charge enclosed = 0 All charge goes on outer surface! Inside cavity is “shielded” from all external electric fields! “Faraday Cage effect” Field on Conductor Perpendicular to Surface We know the field inside the conductor is zero, and the excess charges are all on the surface. The charges produce an electric field outside the conductor. On the surface of conductors in electrostatic equilibrium, the electric field is always perpendicular to the surface. Why? Because if not, charges on the surface of the conductors would move with the electric field. Charges in Conductors • Consider a conducting shell, and a negative charge inside the shell. • Charges will be “induced” in the conductor to make the field inside the conductor zero. • Outside the shell, the field is the same as the field produced by a charge at the center! Gauss’ Law: Conducting Plane • Infinite CONDUCTING plane with uniform areal charge density s • E is NORMAL to plane • Construct Gaussian box as shown. • Note that E = 0 inside conductor Applying Gauss' law, we have, As e0 = AE s Solving for the electric field, we get E = e0 Gauss’ Law: Conducting ICPP • Charged conductor of arbitrary shape: no symmetry; non-uniform charge density • What is the electric field near the surface where the local charge density is ? (a) 0 (b) Zero + + + + ++ + (c) 0 Applying Gauss' law, we have, + + + + + E=0 As e0 = AE s Solving for the electric field, we get E = e0 THIS IS A GENERAL RESULT FOR CONDUCTORS! 23-4 Applying Gauss’ Law: Cylindrical Symmetry Learning Objectives • • 23.20 Explain how Gauss’ law is used to derive the electric field magnitude outside a line of charge or a cylindrical surface (such as a plastic rod) with a uniform linear charge density λ. 23.21 Apply the relationship between linear charge density λ on a cylindrical surface and the electric field magnitude E at radial distance r from the central axis. • 23.22 Explain how Gauss’ law can be used to find the electric field magnitude inside a cylindrical non-conducting surface (such as a plastic rod) with a uniform volume charge density ρ. © 2014 John Wiley & Sons, Inc. All rights reserved. 23-4 Applying Gauss’ Law: Cylindrical Symmetry Figure shows a section of an infinitely long cylindrical plastic rod with a uniform charge density λ. The charge distribution and the field have cylindrical symmetry. To find the field at radius r, we enclose a section of the rod with a concentric Gaussian cylinder of radius r and height h. The net flux through the cylinder from Gauss’ Law reduces to yielding © 2014 John Wiley & Sons, Inc. All rights reserved. A Gaussian surface in the form of a closed cylinder surrounds a section of a very long, uniformly charged, cylindrical plastic rod. Gauss’ Law: Cylindrical Symmetry • Charge of q = 10 C is uniformly spread over a line of length L = 1 m. E=? • Use Gauss’ Law to compute magnitude of E at a perpendicular distance of 1 mm from the center of the line. r = 1 mm L=1 m • Approximate as infinitely long line — E radiates outwards. • Choose cylindrical surface of radius r, length L co-axial with line of charge. Line of Charge: = q/L Units: [C/m] Gauss’ Law: Cylindrical Symmetry • Approximate as infinitely long line — E radiates outwards. • Choose cylindrical surface of radius r, length L co-axial with line of charge. F = EA = E2p rL q lL F= = e0 E=? r = 1 mm L=1m l e0 lL l l E= = = 2k 2pe 0 rL 2pe 0 r r Acyl = 2p rL dA || E so cos = 1 Compare to Finite Line Example From Last Week! Ey = k l r L/2 dx ò (r 2 + x 2 )3/2 - L/2 L/2 é ù x = kl r ê 2 2 2 ú ë r x + r û- L/2 = 2k l L • Add the Vectors! • Horrible Integral! • Trig Substitution! r r 4r 2 + L2 If the Line Is Infinitely Long (L >> r) … Blah, Blah, Blah… 2k l L 2kl Ey = = 2 r r L With Gauss’s Law We Got Same Answer With Two Lines of Algebra! r A-Rod