21. (a) Consider a Gaussian surface that is completely within the conductor and surrounds the cavity. Since the electric field is zero everywhere on the surface, the net charge it encloses is zero. The net charge is the sum of the charge q in the cavity a
22. We imagine a cylindrical Gaussian surface A of radius r and unit length concentric q with the metal tube. Then by symmetry E dA = 2rE = enc .
(a) For r < R, qenc = 0, so E = 0. (b) For r > R, qenc = , so E (r ) = / 2 r 0 . With = 2.00 108 C/m and
23. The magnitude of the electric field produced by a uniformly charged infinite line is E = /20r, where is the linear charge density and r is the distance from the line to the point where the field is measured. See Eq. 23-12. Thus, = 2 0 Er = 2 ( 8.85 10
24. We combine Newtons second law (F = ma) with the definition of electric field ( F = qE ) and with Eq. 23-12 (for the field due to a line of charge). In terms of magnitudes, we have (if r = 0.080 m and = 6.0 106 C/m ) ma = eE = e 2o r a= e = 2.1 1017 m/
25. (a) The side surface area A for the drum of diameter D and length h is given by A = Dh . Thus, q = A = Dh = 0 EDh = 8.85 1012 C2 /N m 2 2.3 105 N/C ( 0.12 m )( 0.42 m ) = 3.2 107 C. (b) The new charge is
A Dh 7 q = q = q = 3.2 10 C A Dh