671_PartUniversity Physics Solution

# 671_PartUniversity Physics Solution - 22-22 Chapter 22...

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22-22 Chapter 22 EXECUTE: (a) We could place two charges Q + on either side of the charge q + , as shown in Figure 22.56. (b) In order for the charge to be stable, the electric field in a neighborhood around it must always point back to the equilibrium position. (c) If q is moved to infinity and we require there to be an electric field always pointing in to the region where q had been, we could draw a small Gaussian surface there. We would find that we need a negative flux into the surface. That is, there has to be a negative charge in that region. However, there is none, and so we cannot get such a stable equilibrium. (d) For a negative charge to be in stable equilibrium, we need the electric field to always point away from the charge position. The argument in (c) carries through again, this time implying that a positive charge must be in the space where the negative charge was if stable equilibrium is to be attained. EVALUATE: If q is displaced to the left or right in Figure 22.56, the net force is directed back toward the equilibrium position. But if q is displaced slightly in a direction perpendicular to the line connecting the two charges Q , then the net force on q is directed away from the equilibrium position and the equilibrium is not stable for such a displacement. Figure 22.56 22.57. () ( ) 3 00 1 / for where 3 / . rr R r R Q R ρρ ρ π =− = ( ) 0 for R = (a) IDENTIFY: The charge density varies with r inside the spherical volume. Divide the volume up into thin concentric shells, of radius r and thickness dr . Find the charge dq in each shell and integrate to find the total charge. SET UP: The thin shell is sketched in Figure 22.57a. EXECUTE: The volume of such a shell is 2 4 dV r dr = The charge contained within the shell is ( ) ( ) 2 0 41 / dq r dV r r R dr ρπ == Figure 22.57a The total charge Q in the charge distribution is obtained by integrating dq over all such shells into which the sphere can be subdivided: () () 22 3 / 4 / RR Qd q r r R d r rrR d r πρ = ∫∫ () ( ) 34 3 4 33 3 0 0 4 4 4 /12 4 3 / /12 , R QR Q R R Q ⎡⎤ = = = = ⎜⎟ ⎢⎥ ⎣⎦ as was to be shown. (b) IDENTIFY: Apply Gauss’s law to a spherical surface of radius r , where r > R . SET UP: The Gaussian surface is shown in Figure 22.57b. EXECUTE: encl 0 E Q Φ= P 2 0 4 Q Er = P Figure 22.57b 2 0 ; 4 Q E r = P same as for point charge of charge Q . (c) IDENTIFY: Apply Gauss’s law to a spherical surface of radius r , where r < R : SET UP: The Gaussian surface is shown in Figure 22.57c. EXECUTE: encl 0 E Q P ( ) 2 4 E E r Figure 22.57c

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Gauss’s Law 22-23 The calculate the enclosed charge encl Q use the same technique as in part (a), except integrate dq out to r rather than R. (We want the charge that is inside radius r .) 3 22 encl 0 0 00 41 4 rr Qr d r r d r RR πρ ′′ ⎛⎞ =− = ⎜⎟ ⎝⎠ ∫∫ 34 3 encl 0 0 0 0 1 44 4 r r R ⎡⎤⎛⎞ = ⎢⎥ ⎣⎦⎝⎠ 33 0e n c l 3 31
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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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671_PartUniversity Physics Solution - 22-22 Chapter 22...

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