73. The electric field is radially outward from the central wire. We want to find its magnitude in the region between the wire and the cylinder as a function of the distance r from the wire. Since the magnitude of the field at the cylinder wall is known,
4. There is no flux through the sides, so we have two inward contributions to the flux, one from the top (of magnitude (34)(3.0)2) and one from the bottom (of magnitude (20)(3.0)2). With inward flux being negative, the result is = 486 Nm2/C. Gauss law the
77. (a) Since E points down and we need an upward electric force (to cancel the downward pull of gravity), then we require the charge of the sphere to be negative. The magnitude of the charge is found by working with the absolute value of Eq. 22-28: |q|=
76. The electric field at a point on the axis of a uniformly charged ring, a distance z from the ring center, is given by qz E= 3/ 2 2 4 0 z + R 2
c
h
where q is the charge on the ring and R is the radius of the ring (see Eq. 22-16). For q positive, the f
73. Studying Sample Problem 22-3, we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc is given by E=
sin 4 0 r
along the symmetry axis, where = q = q r with in radians. Here is the length of the arc,
71. (a) Using the density of water ( = 1000 kg/m3), the weight mg of the spherical drop (of radius r = 6.0 107 m) is W = Vg = 1000 kg m3
c
h FGH 43 c6.0 10 mh IJK c9.8 m s h = 8.87 10
7
3
2
15
N.
(b) Vertical equilibrium of forces leads to mg = qE = neE,