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Chapter 23: Gauss’s Law Gauss’s Law E dA Qenc / 0 We got the same flux through our boxes and sphere no matter what size or shape the closed surfaces are as long as they fully enclose the charge. S closed surface = constant = qenclosed o • S is any closed surface. • Qenc is the net charge enclosed within S. • dA is an element of area on the surface of S. • dA is in the direction of the outward normal. Check Quiz, August 30, 2013 Find the flux through the Gaussian surface. A. q1 q3 q2 q5 o q2=q3=-2 nC C. 0 D. q4=q5=4 nC 6nC o q3=-2 nC q4=q5=4 nC qenc = -2nC + 4nC + 4 nC 7 nC B. o q1 q3 q6 6nC o E. none of the above q1 = +1 nC D. Gauss’s law says count only the charges inside the surface! 1nC q6 q4 Answer 1 nC = 10-9C q4 q2 q5 q6 = -6 nC Gauss Coulomb dA r Given a point charge, draw a concentric sphere and apply Gauss’s Law: Spherical Charge densities d = r2sindrdd qenc E dA 4r 2 E r o S Uniform Charge density o 4 qenc (r )d od r 3 o 3 V V E E (r )rˆ and dA = da r E dA E (r ) 4 r 2 Non-uniform Charge density For example, let o=A/r r Gauss q / 0 E (r ) q /(4 r 0 ) kq / r 2 qenc (r )d V 2 ! V A r 2 sin drdd d A r r r A(2 )(2) rdr 2Ar 2 0 1 Line of Charge E dA2 We know E is radial. Gaussian Surface Line of Charge Solution dA1 dA3 ++++++++++++++++++++++++++ L Area 1: left endcap Area 2: cylinder wall Area 3: right endcap Flux = enclosed charge Line of Charge -- Direct Method Mathematical Box Box Integration Mathematical Cylinder 2 Mathematical Sphere Cylinder Integration Spherical Integration Conductors Definition -- Conductor: A material that allows free motion of charges. Metals are good examples of conductors. A free charge will move until it feels no net force. Remember that F = ma = qE. Therefore charges will move in a conductor until the electric field inside the conductor is ZERO. All free charges on a conductor reside on their SURFACES. Conductors - - Example 1 We put +Q on an arbitrary conductor that is hollow in the middle. How will it distribute itself? Conductors - - Example 1 Solution hollow conductor qouter = +Q qinner +Q qinner Gaussian surface inside the conductor hollow conductor qouter +Q +Q Gaussian surface inside the conductor 0 E field is 0 inside a conductor We know: qinner + qouter = +Q 0 qenclosed = qinner = 0 qouter = +Q 0 3 Conductors - - Example 2 hollow, neutral conductor qouter qinner Conductors - - Example 2 Solution hollow conductor qouter = -3Q +3Q= qinner 0 Gaussian surface inside the conductor -3Q -3Q Gaussian surface inside the conductor Now we put a -3Q charge in the center. How will the charges redistribute themselves? E field is 0 inside a conductor 0 qenclosed = qinner + -3Q = 0 0 qinner = +3Q Where did that +3Q come from? qinner + qouter = 0 qouter = -3Q +3Q Conductors - - Example 3 Now combine the two examples. +Q on the conductor and -3Q inside the cavity. How will the charges distribute themselves? +Q Conductors - - Example 3 Solution +Q qinner = +3Q hollow conductor qouter = -2Q qinner = +3Q hollow conductor qouter = -3Q + Q = -2Q -3Q Charge left behind -3Q Gaussian surface inside the conductor From Example 1: +Q all on qouter From Example 2: +3Q was attracted from qouter leaving -3Q + Q on qouter qouter = -3Q +Q = -2Q Van de Graaff Demonstration Gaussian surface inside the conductor when +3Q was pulled to qinner Charge initially put on conductor For this Example: Inside cavity: qenc = -3Q, Ecavity = 0 Inside conductor: qenc = 0, qinner = +3Q, Econductor = 0 Outside conductor: qenc = -3Q + qinner + qouter = -2Q, Eoutside = 0 +3Q -2Q Fields in Good Conductors Fact: In a steady state the electric field inside a good conductor must be zero. Why? If there were a field, charges would move. Charges will move around until they find the arrangement that makes the electric field zero in the interior. 4 Field at the Surface of a Conductor Charges in Good Conductors Fact: In a steady state, any net charge on a good conductor must be entirely on the surface. Why? If there were a charge in the interior, then by Gauss’s Law there would be a field in the interior, which we know cannot be true. • Construct closed gaussian surface, sides perpendicular to metal surface, face area = A. • Flux through left face is zero because E=0. • Field is perpendicular to surface or charges would move, therefore flux through sides is 0. • So net outward flux is EA. Field at the Surface of a Conductor • Let σ stand for the surface charge density (C/m2) • Then Qenc = σ A. • Now Gauss’s Law gives us Qenc / 0 EA A / 0 E / 0 5