Serway Cap 24

24
Gauss’s Law
CHAPTER OUTLINE
24.1
24.2
24.3

Electric Flux
Gauss’s Law
Application of Gauss’s Law
to Various Charge
Distributions
Conductors in Electrostatic
Equilibrium
Formal Derivation of
Gauss‘s Law

ANSWERS TO QUESTIONS

24.5

The luminous flux on a given area is less when the sun is low in
the sky, because the angle between the rays of the sun and the
local areavector, dA, is greater than zero. The cosine of this
angle is reduced. The decreased flux results, on the average, in
colder weather.

Q24.2

If the region is just a point, line, or plane, no. Consider two
protons in otherwise empty space. The electric field is zero at
the midpoint of the line joining the protons. If the field-free
region is three-dimensional, then it can contain no charges,but
it might be surrounded by electric charge. Consider the interior
of a metal sphere carrying static charge.

Q24.3

24.4

Q24.1

The surface must enclose a positive total charge.

Q24.4

The net flux through any gaussian surface is zero. We can argue it two ways. Any surface contains
zero charge so Gauss’s law says the total flux is zero. The field is uniform, so the field linesentering
one side of the closed surface come out the other side and the net flux is zero.

Q24.5

Gauss’s law cannot tell the different values of the electric field at different points on the surface.
When E is an unknown number, then we can say E cos θdA = E cos θdA . When E x , y , z is an

z

unknown function, then there is no such simplification.

z

b

g

Q24.6

Theelectric flux through a sphere around a point charge is independent of the size of the sphere. A
sphere of larger radius has a larger area, but a smaller field at its surface, so that the product of field
strength and area is independent of radius. If the surface is not spherical, some parts are closer to the
charge than others. In this case as well, smaller projected areas go with stronger fields,so that the
net flux is unaffected.

Q24.7

Faraday’s visualization of electric field lines lends insight to this question. Consider a section of a
1
field lines pointing out from it horizontally to
vertical sheet carrying charge +1 coulomb. It has
∈0
the right and left, all uniformly spaced. The lines have the same uniform spacing close to the sheet
and far away, showing that the fieldhas the same value at all distances.

29

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Gauss’s Law

Q24.8

Consider any point, zone, or object where electric field lines begin. Surround it with a close-fitting
gaussian surface. The lines will go outward through the surface to constitute positive net flux. Then
Gauss’s law asserts that positive net charge must be inside the surface: it is where the lines begin.
Similarly,any place where electric field lines end must be just inside a gaussian surface passing net
negative flux, and must be a negative charge.

Q24.9

Inject some charge at arbitrary places within a conducting object. Every bit of the charge repels
every other bit, so each bit runs away as far as it can, stopping only when it reaches the outer surface
of the conductor.

Q24.10

If the personis uncharged, the electric field inside the sphere is zero. The interior wall of the shell
carries no charge. The person is not harmed by touching this wall. If the person carries a (small)
charge q, the electric field inside the sphere is no longer zero. Charge –q is induced on the inner wall
of the sphere. The person will get a (small) shock when touching the sphere, as all the charge on hisbody jumps to the metal.

Q24.11

The electric fields outside are identical. The electric fields inside are very different. We have E = 0
everywhere inside the conducting sphere while E decreases gradually as you go below the surface of
the sphere with uniform volume charge density.

Q24.12

There is zero force. The huge charged sheet creates a uniform field. The field can polarize...