676_PartUniversity Physics Solution

676_PartUniversity - Gausss Law 22-27 EXECUTE(a At this point the field of the left-hand sphere is zero and the field of the right-hand sphere is

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Gauss’s Law 22-27 EXECUTE: (a) At this point the field of the left-hand sphere is zero and the field of the right-hand sphere is toward the center of that sphere, in the + x -direction. This point is outside the right-hand sphere, a distance 2 rR = from its center. 2 0 1 ˆ 44 Q π R + E =i G P . (b) This point is inside the left-hand sphere, at /2 = , and is outside the right-hand sphere, at 3 /2 = . Both fields are in the + x -direction. 32 2 2 2 00 0 1( / 2 ) 1 4 1 1 7 ˆˆ ˆ 4( 3 2 ) 4 2 9 4 1 8 QR Q Q Q Q π RR π π R ⎡⎤ ++ ⎢⎥ ⎣⎦ E = i = i G PP P . (c) This point is outside both spheres, at a distance = from their centers. Both fields are in the + x -direction. 22 2 1 42 QQ Q π π R + E = i G . (d) This point is outside both spheres, a distance 3 = from the center of the left-hand sphere and a distance = from the center of the right-hand sphere. The field of the left-hand sphere is in the + x -direction and the field of the right-hand sphere is in the -direction x . 2 0 11 1 8 ˆ 3 ) 49 4 9 Q Q Q π π π R −− E = i = i G P . EVALUATE: At all points on the x -axis the net field is parallel to the x- axis. 22.65. IDENTIFY: Let dQ be the electron charge contained within a spherical shell of radius r and thickness dr . Integrate r from 0 to r to find the electron charge within a sphere of radius r . Apply Gauss’s law to a sphere of radius r to find the electric field () E r . SET UP: The volume of the spherical shell is 2 4 dV r dr π = . EXECUTE: (a) / 2 33 0 () r ar a r Qr Q ρ dV Q e r dr Q r e dr π aa =− ∫∫ . 0 2/ 2 0 4 ( ) (2 2 2) [2( / ) 2( / ) 1]. r ra r Qe Q e α r r Q e a α − − = + + Note if , ( ) 0 rQ r →∞ ; the total net charge of the atom is zero. (b) The electric field is radially outward. Gauss’s law gives 2 0 (4 ) Er = P and 0 2 2 (2( ) 2( ) 1) kQe a r a r =+ + . (c) The graph of E versus r is sketched in Figure 22.65. What is plotted is the scaled E , equal to pt charge / EE , versus scaled r , equal to 0 / . pt charge 2 kQ E r = is the field of a point charge. EVALUATE: As 0 r , the field approaches that of the positive point charge (the proton). For increasing r the electric field falls to zero more rapidly than the 2 1/ r dependence for a point charge. Figure 22.65 22.66. IDENTIFY: The charge in a spherical shell of radius r and thickness dr is 2 ()4 dQ r r dr ρπ = . Apply Gauss’s law. SET UP: Use a Gaussian surface that is a sphere of radius r . Let i Q be the charge in the region / 2 and let 0 Q be the charge in the region where / 2 R .
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22-28 Chapter 22 EXECUTE: (a) The total charge is 0 i QQ Q = + , where 33 4( 2 ) 36 i π R απ R Q α == and 44 3 23 0 /2 (8 ) (1 6 ) 1 1 4( 2) ( / ) 8 342 4 R R RR απ R Q πα rrR d r απ R ⎛⎞ −− =− = = ⎜⎟ ⎝⎠ . Therefore, 3 15 24 απ R Q = and 3
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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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676_PartUniversity - Gausss Law 22-27 EXECUTE(a At this point the field of the left-hand sphere is zero and the field of the right-hand sphere is

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