17_InstSolManual_PDF_Part21

17_InstSolManual_PDF_Part21 - (b for the Gaussian surface...

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Solve: (a) and (b) Apply Gauss’s law to a surface that is within the conductor, just outside the cavity, as shown in Figure 17.63. everywhere on the Gaussian surface so for that surface. Gauss’s law then says that for this surface is zero. The conductor is neutral, so if the outer surface has charge the inner surface must have charge To make there must be within the hole. Reflect: The charge in the hole creates the charge separation in the conductor. It pulls to the inner surface and that leaves on the outer surface. 17.64. Set Up: everywhere within the conductor. Any net charge must be on the inner and outer surfaces of the conductor. Solve: Apply Gauss’s law to a surface that is within the conductor, just outside the cavity, as shown in Figure 17.64. everywhere on the Gaussian surface so for that surface. Figure 17.64 (a) for the Gaussian surface means that on the inner surface, so the of net charge is on the outer surface of the conductor.
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Unformatted text preview: (b) for the Gaussian surface means that there must be nC on the inner surface of the conductor. Since the total net charge of the conductor is nC, if there is nC on the inner surface there must be nC on the outer surface. 17.65. Set Up: Like charges repel and unlike charges attract. Charges and and the forces they exert on at the origin are sketched in Figure 17.65a. For the net force on to be zero, and from and must be equal in magnitude and opposite in direction. Figure 17.65 y F 1 F 2 0.20 m 0.30 m q 2 q 3 q 1 x 1 1 2 (a) F 2 F 1 F 1 F 2 F 2 F 1 q 2 q 3 q 3 q 1 q 3 x y 1 1 1 2 1 I II III (b) q 2 q 1 F S 2 F S 1 q 3 q 3 q 2 q 1 F 5 k qq r r 2 . 1 5 1 11 1 16 1 11 Q encl 5 1 16 m C q 5 Q encl 5 Gaussian surface F E 5 E 5 E 5 2 12 m C 1 12 m C 2 12 m C Q encl 5 1 12 m C. 2 12 m C Q encl F E 5 E 5 Electric Charge and Electric Field 17-21...
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