2. Flux = EA cos (theta); watch units (ans: -0.015 Nm2/C
6. Flux = enclosed charge/epsilon nought;
19. Use Gaussian surface within conductor
surrounding cavity; (a) since E = 0 everywhere
on surface (it's inside the conductor), net
charge enclosed = 0 (b) net charge on conductor
is charge on cavity wall plus charge on outer
surface
22.(a) Refer to problem 15 to find electric
field on drum. Use Gauss' law to find charge
enclosed by Gaussian surface at drum's surface.
(ans: 0.32 microcoulombs) (b) Use proportion;
both charge to area ratios equal same value
for E. (ans. 0.14 microcoulombs)
24. Use cylindrical Gaussian surface of radius
r and length l. Enclosed charge q = (lambda)l. If r < a, q = 0. For a < r < b, Write Gauss' law, solve for E.
25. For fields to cancel, they must be equal
and opposite at the surface, 1.5 cm from
the center. The expression for E due to the internal charge will be found
by the answer to part (b) in # 24. Using
Gauss' law, write an expression for the field
at the surface of the cylinder due to the
surface charge and set equal to the field
due to the inner charge. Solve for sigma.
32. (a) Use equation for an infinite plate
of charge; ½ of charge will be on
each side of plate. (5.3 x 107 N/C) (b) plate reduces to point charge.
(60 N/C)
37. All excess charge is on inner surface
of plates, so eqn. 25-15 describes the field
from each plate. Total field is sum of the
two plates' fields which returns us to eqn.
25-16, the field from one conducting plate.
42. (a, b, c) shell theorem (0, 2.9 x 104 N/C, 200 N/C)
48. Use spherical Gaussian surface concentric
with sphere of charge. (a) r < a Enclosed charge is total charge on sphere
times ratio of spherical volume of radius
r to spherical volume of radius a. (b) b > r > a Enclosed charge is q. (c) c > r > b Inside conductor so E = 0 (d) r > c Enclosed charge = q + (-q). (e) Consider Gaussian surfface within
shell. Since field on surface = 0, enclosed
charge = 0.