661_PartUniversity Physics Solution

661_PartUniversity Physics Solution - 22-12 Chapter 22 (b)...

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22-12 Chapter 22 (b) (i) The Gaussian cylinder with radius r , for arb < < , must enclose zero net charge, so the charge per unit length on the inner surface is . α (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 rb > the electric field is due to the charge on the outer surface of the tube. Figure 22.38 22.39. (a) IDENTIFY: Use Gauss’s law to calculate E ( r ). (i) SET UP: : ra < Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r , where , < as sketched in Figure 22.39a. EXECUTE: ( ) 2 E E rl π Φ= encl Ql = (the charge on the length l of the line of charge) Figure 22.39a () encl 00 gives 2 E Er l = PP 0 . 2 E r = P The enclosed charge is positive so the direction of E G is radially outward. (ii) : << Points in this region are within the conducting tube, so E = 0. (iii) SET UP: : > Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r , where , > as sketched in Figure 22.39b. EXECUTE: ( ) 2 E E rl Φ = encl = (the charge on length l of the line of charge) l (the charge on length l of the tube) Thus encl 0. Q = Figure 22.39b encl 0 gives 2 0 and 0. E Q l E = = P The graph of E versus r is sketched in Figure 22.39c. Figure 22.39c (b) IDENTIFY: Apply Gauss’s law to cylindrical surfaces that lie just outside the inner and outer surfaces of the tube. We know E so can calculate encl . Q (i) SET UP: inner surface Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r , where .
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Gauss’s Law 22-13 EXECUTE: This surface lies within the conductor of the tube, where E = 0, so 0. E Φ = Then by Gauss’s law encl 0. Q = The surface encloses charge l α on the line of charge so must enclose charge l on the inner surface of the tube. The charge per unit length on the inner surface of the tube is . (ii) outer surface The net charge per unit length on the tube is . We have shown in part (i) that this must all reside on the inner surface, so there is no net charge on the outer surface of the tube. EVALUATE: For r < a the electric field is due only to the line of charge. For r > b the electric field of the tube is the same as for a line of charge along its axis. The fields of the line of charge and of the tube are equal in magnitude and opposite in direction and sum to zero. For r < a the electric field lines originate on the line of charge and terminate on the surface charge on the inner surface of the tube. There is no electric field outside the tube and no surface charge on the outer surface of the tube.
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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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661_PartUniversity Physics Solution - 22-12 Chapter 22 (b)...

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