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676_PartUniversity Physics Solution

# 676_PartUniversity Physics Solution - Gausss Law 22-27...

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Gauss’s Law 22-27 E XECUTE : (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 r R = from its center. 2 0 1 ˆ 4 4 Q π R + E = i G P . (b) This point is inside the left-hand sphere, at /2 r R = , and is outside the right-hand sphere, at 3 /2 r R = . Both fields are in the + x -direction. 3 2 2 2 2 0 0 0 1 ( /2) 1 4 1 17 ˆ ˆ ˆ 4 (3 2) 4 2 9 4 18 Q R Q Q Q Q π R R π R R π R + + E = i = i = i G P P P . (c) This point is outside both spheres, at a distance r R = from their centers. Both fields are in the + x -direction. 2 2 2 0 0 1 ˆ ˆ 4 2 Q Q Q π R R π R + E = i = i G P P . (d) This point is outside both spheres, a distance 3 r R = from the center of the left-hand sphere and a distance r R = 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 2 2 2 2 0 0 0 1 1 1 8 ˆ ˆ ˆ 4 (3 ) 4 9 4 9 Q Q Q Q Q π R R π R R π R E = i = i = i G P P P . E VALUATE : At all points on the x -axis the net field is parallel to the x- axis. 22.65. I DENTIFY : 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 . S ET U P : The volume of the spherical shell is 2 4 dV r dr π = . E XECUTE : (a) 0 0 / 2 / 2 2 2 3 3 0 0 0 4 4 ( ) r a r a r Q Q Q r Q ρ dV Q e r dr Q r e dr π a a π = = = . 0 2 / 2 2 2 0 0 3 3 0 4 ( ) (2 2 2) [2( / ) 2( / ) 1]. r r a r Qe Q r Q e α r r Qe r a r a a α α α α = = + + Note if , ( ) 0 r Q r → ∞ ; the total net charge of the atom is zero. (b) The electric field is radially outward. Gauss’s law gives 2 0 ( ) (4 ) Q r E r π = P and 0 2 / 2 0 0 2 (2( ) 2( ) 1) r a kQe E r 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 / E E , versus scaled r , equal to 0 / r a . pt charge 2 kQ E r = is the field of a point charge. E VALUATE : 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. I DENTIFY : The charge in a spherical shell of radius r and thickness dr is 2 ( )4 dQ r r dr ρ π = . Apply Gauss’s law.

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