666_PartUniversity Physics Solution

666_PartUniversity Physics Solution - Gausss Law 22-17 (b)...

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Gauss’s Law 22-17 (b) (i) small shell inner surface: Since a Gaussian surface with radius r , for arb < < , must enclose zero net charge, the charge on this surface is zero. (ii) small shell outer surface: 2 q + . (iii) large shell inner surface: Since a Gaussian surface with radius r , for crd < < , must enclose zero net charge, the charge on this surface is 2 q . (iv) large shell outer surface: Since there is 2 q on the inner surface and the total charge on this conductor is 2 q , the charge on this surface is zero. EVALUATE: The outer shell has no effect on the electric field for rc < . For rd > the electric field is due only to the charge on the outer surface of the larger shell. Figure 22.46 22.47. IDENTIFY: Apply Gauss’s law SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the charged shells. EXECUTE: (a) (i) For ,0 , raE <= since charge enclosed is zero. (ii) , arbE < since the points are within the conducting material. (iii) For 2 0 2 1 ,, 4 q brcE π r << = P outward, since charge enclosed is +2 q . (iv) For , crdE = since the points are within the conducting material. (v) For 2 0 2 1 4 q rdE π r >= P inward, since charge enclosed is –2 q . The graph of the radial component of the electric field versus r is sketched in Figure 22.47, where we use the convention that outward field is positive and inward field is negative. (b) (i) small shell inner surface: Since a Gaussian surface with radius r , for a < r < b , must enclose zero net charge, the charge on this surface is zero. (ii) small shell outer surface: +2 q . (iii) large shell inner surface: Since a Gaussian surface with radius r , for c < r < d , must enclose zero net charge, the charge on this surface is –2 q . (iv) large shell outer surface: Since there is –2 q on the inner surface and the total charge on this conductor is –4 q , the charge on this surface is –2 q . EVALUATE: The outer shell has no effect on the electric field for < . For > the electric field is due only to the charge on the outer surface of the larger shell. Figure 22.47 22.48. IDENTIFY: Apply Gauss’s law. SET UP: Use a Gaussian surface that is a sphere of radius r and that is concentric with the sphere and shell. The volume of the insulating shell is () 33 3 42 8 [2 ] VR RR π =− = . EXECUTE: (a) Zero net charge requires that 3 28 3 πρ R Q −= , so 3 28 Q ρ π R 3 . (b) For , 0 rRE since this region is within the conducting sphere. For 2, 0 , rR E since the net charge enclosed by the Gaussian surface with this radius is zero. For 2 R < < , Gauss’s law gives 23 3 00 4 (4 ) ( ) 3 Q E π rr R =+ PP and 22 43 Q ρ E π . Substituting ρ from part (a) gives 0 0 2 . 7 28 QQ r E π r π R P P The net enclosed charge for each r in this range is positive and the electric field is outward.
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22-18 Chapter 22 (c) The graph is sketched in Figure 22.48. We see a discontinuity in going from the conducting sphere to the insulator due to the thin surface charge of the conducting sphere. But we see a smooth transition from the uniform insulator to the surrounding space.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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666_PartUniversity Physics Solution - Gausss Law 22-17 (b)...

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