HW3solutions - Physics 213 HW #3 Solutions Spring 2008...

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Physics 213 HW #3 – Solutions Spring 2008 22.31. IDENTIFY: Apply Gauss’s law and conservation of charge. SET UP: E=0 in a conducting material. EXECUTE: (a) Because E=0 inside metal, a Gaussian bubble enclosing the cavity and charge but still within the metal solid must have zero electric flux. Gauss’s law gives zero net charge enclosed in the bubble, requiring + Q on inner surface. (b) The outside of the sphere is grounded, so no excess charge. (c) Consider a Gaussian sphere with the – Q charge at its center and radius less than the inner radius of the metal. This sphere encloses net charge – Q so there is an electric field flux through it; thus there is electric field in the cavity. (d) In an electrostatic situation E = 0 inside a conductor (the metal solid). A Gaussian sphere with the ± Q charge at its center and radius greater than the outer radius of the metal encloses zero net charge (the ± Q charge and the + Q on the inner surface of the metal) so there is no flux through it and E = 0 outside the metal. (e) No, E = 0 there. Yes, the charge has been shielded by the grounded conductor. There is nothing like positive and negative mass (the gravity force is always attractive), so this cannot be done for gravity. EVALUATE: Field lines within the cavity terminate on the charges induced on the inner surface. 22.38. IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a cylinder of radius r , length l and that has the line of charge along its axis. The charge on a length l of the line of charge or of the tube is q = ± l . EXECUTE: (a) (i) For r < a , Gauss’s law gives E (2 ± rl ) = Q encl 0 = ² l 0 and E = 2 ± 0 r . (ii) For a < r < b, the electric field is zero because these points are within the conducting material. (iii) For r > b , Gauss’s law gives E ± rl ) = Q encl 0 = 2 l 0 and E = ± 0 r . (b) (i) The Gaussian cylinder with radius r , for a < r < b , must enclose zero net charge, so the charge per unit length on the inner surface is ± . (ii) Since the net charge per length for the tube is + and there is ± on the inner surface, the charge per unit length on the outer surface must be + 2 . EVALUATE: For r > b the electric field is due only to the charge on the outer surface of the tube.
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This note was uploaded on 03/31/2008 for the course PHYS 2213 taught by Professor Perelstein,m during the Spring '07 term at Cornell University (Engineering School).

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HW3solutions - Physics 213 HW #3 Solutions Spring 2008...

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