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Class Examples for Chapter 23

# Class Examples for Chapter 23 - Class Examples for Chapter...

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1 Class Examples for Chapter 23 8. (a) The total surface area bounding the bathroom is       2 2 2.5 3.0 2 3.0 2.0 2 2.0 2.5 37 m . A The absolute value of the total electric flux, with the assumptions stated in the problem, is 2 3 2 | | | | | | (600 N/C)(37 m ) 22 10 N m /C. E A E A   By Gauss’ law, we conclude that the enclosed charge (in absolute value) is 7 enc 0 | | | | 2.0 10 C. q   Therefore, with volume V = 15 m 3 , and recognizing that we are dealing with negative charges, the charge density is 7 83 enc 3 || 2.0 10 C 1.3 10 C/m . 15 m q V (b) We find (| q enc |/ e )/ V = (2.0 10 –7 C/1.6 10 –19 C)/15 m 3 = 8.2 10 10 excess electrons per cubic meter. 14. Equation 23-6 (Gauss’ law) gives   q enc . (a) Thus, the value 52 2.0 10 N m /C   for small r leads to 12 2 2 5 2 6 6 central 0 (8.85 10 C /N m )(2.0 10 N m /C) 1.77 10 C 1.8 10 C q   (b) The next value that takes is 4.0 10 N m /C    , which implies that 6 enc 3.54 10 C. q  But we have already accounted for some of that charge in part (a), so the result for part (b) is q A = q enc q central = – 5.3 10 6 C. (c) Finally, the large r value for is 6.0 10 N m /C   , which implies that 6 total enc 5.31 10 C. q  Considering what we have already found, then the result is total enc central 8.9 . A q q q C 

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2 17. (a) The charge on the surface of the sphere is the product of the surface charge density and the surface area of the sphere (which is 2 4, r where r is the radius). Thus,   2 2 6 2 5 m 4 4 8.1 10 C/m 3.7 10 C.
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Class Examples for Chapter 23 - Class Examples for Chapter...

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