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Physics For Scientists And Engineers 6E - part 188

SECTION 24.3 • Application of Gauss’s Law to Various Charge Distributions

749

Example 24.8 A Plane of Charge

Find the electric field due to an infinite plane of positive
charge with uniform surface charge density 4.

Solution By symmetry,

E must be perpendicular to the

plane and must have the same magnitude at all points
equidistant from the plane. The fact that the direction of

E is away from positive charges indicates that the direction
of

E on one side of the plane must be opposite its direc-

tion on the other side, as shown in Figure 24.15. A gauss-
ian surface that reflects the symmetry is a small cylinder
whose axis is perpendicular to the plane and whose ends

each have an area A and are equidistant from the plane.
Because

E is parallel to the curved surface—and,

therefore, perpendicular to d

A everywhere on the

surface—condition  (3)  is  satisfied  and  there  is  no  contri-
bution  to  the  surface  integral  from  this  surface.  For  the
flat ends of the cylinder, conditions (1) and (2) are satis-
fied.  The  flux  through  each  end  of  the  cylinder  is  EA;
hence, the total flux through the entire gaussian surface is
just that through the ends, !

2EA.

Noting that the total charge inside the surface is

A, we use Gauss’s law and find that the total flux

through the gaussian surface is

leading to

(24.8)

Because the distance from each flat end of the cylinder

to the plane does not appear in Equation 24.8, we conclude
that E # 4/2/

at any distance from the plane. That is, the

field is uniform everywhere.

What If?

Suppose we place two infinite planes of charge

parallel to each other, one positively charged and the other
negatively charged. Both planes have the same surface
charge density. What does the electric field look like now?

E #

A #

What If?

What if the line segment in this example were not

infinitely long?

Answer If the line charge in this example were of finite
length,  the  result  for E would  not  be  that  given  by  Equa-
tion  24.7.  A  finite  line  charge  does  not  possess  sufficient
symmetry  for  us  to  make  use  of  Gauss’s  law.  This  is
because  the  magnitude  of  the  electric  field  is  no  longer
constant  over  the  surface  of  the  gaussian  cylinder—
the field near the ends of the line would be different from
that far from the ends. Thus, condition (1) would not be
satisfied  in  this  situation.  Furthermore,

E is not perpen-

dicular to the cylindrical surface at all points—the field
vectors near the ends would have a component parallel to
the line. Thus, condition (2) would not be satisfied. For
points close to a finite line charge and far from the ends,
Equation 24.7 gives a good approximation of the value of
the field.

It is left for you to show (see Problem 29) that the

electric field inside a uniformly charged rod of finite radius
and infinite length is proportional to r.

Gaussian

surface

+
+
+

+
+

(a)

(b)

Figure 24.14 (Example 24.7) (a) An infinite line of charge

surrounded by a cylindrical gaussian surface concentric with

the line. (b) An end view shows that the electric field at the

cylindrical surface is constant in magnitude and perpendicular

to the surface.

Figure 24.15 (Example 24.8) A cylindrical gaussian surface

penetrating an infinite plane of charge. The flux is EA through

each end of the gaussian surface and zero through its curved

surface.

+ +

Gaussian

surface

750

CHAPTE R 24 • Gauss’s Law

24.4 Conductors in Electrostatic Equilibrium

As we learned in Section 23.2, a good electrical conductor contains charges (electrons)
that are not bound to any atom and therefore are free to move about within the mater-
ial. When there is no net motion of charge within a conductor, the conductor is in
electrostatic equilibrium. A conductor in electrostatic equilibrium has the following
properties:

1. The electric field is zero everywhere inside the conductor.
2. If an isolated conductor carries a charge, the charge resides on its surface.
3. The electric field just outside a charged conductor is perpendicular to the surface

of the conductor and has a magnitude 4//

, where 4 is the surface charge density

at that point.

4. On an irregularly shaped conductor, the surface charge density is greatest at loca-

tions where the radius of curvature of the surface is smallest.

We verify the first three properties in the discussion that follows. The fourth prop-

erty is presented here so that we have a complete list of properties for conductors in
electrostatic equilibrium, but cannot be verified until Chapter 25.

We can understand the first property by considering a conducting slab placed in

an external field

E (Fig. 24.16). The electric field inside the conductor must be

zero under the assumption that we have electrostatic equilibrium. If the field
were not zero, free electrons in the conductor would experience an electric force
(

F # q E) and would accelerate due to this force. This motion of electrons, however,

would mean that the conductor is not in electrostatic equilibrium. Thus, the
existence of electrostatic equilibrium is consistent only with a zero field in the
conductor.

Let us investigate how this zero field is accomplished. Before the external field is

applied, free electrons are uniformly distributed throughout the conductor. When the
external field is applied, the free electrons accelerate to the left in Figure 24.16, caus-
ing a plane of negative charge to be present on the left surface. The movement of elec-
trons to the left results in a plane of positive charge on the right surface. These planes
of charge create an additional electric field inside the conductor that opposes the
external field. As the electrons move, the surface charge densities on the left and right
surfaces increase until the magnitude of the internal field equals that of the external
field, resulting in a net field of zero inside the conductor. The time it takes a good con-
ductor to reach equilibrium is on the order of 10

s, which for most purposes can be

considered instantaneous.

Properties of a conductor in

electrostatic equilibrium

Figure 24.16 A conducting slab in

an external electric field

E. The

charges induced on the two

surfaces of the slab produce an

electric field that opposes the

external field, giving a resultant

field of zero inside the slab.

+
+
+
+
+
+
+
+

–
–
–
–
–
–
–
–

Answer In this situation, the electric fields due to the two
planes add in the region between the planes, resulting in a
uniform field of magnitude 4//

, and cancel elsewhere to

give a field of zero. This is a practical way to achieve uniform
electric fields, such as those needed in the CRT tube
discussed in Section 23.7.

Conceptual Example 24.9 Don’t Use Gauss’s Law Here!

Explain why Gauss’s law cannot be used to calculate
the electric field near an electric dipole, a charged disk, or a
triangle with a point charge at each corner.

Solution The charge distributions of all these configura-
tions do not have sufficient symmetry to make the use of

Gauss’s law practical. We cannot find a closed surface
surrounding any of these distributions that satisfies one or
more of conditions (1) through (4) listed at the beginning
of this section.

SECTION 24.4 • Conductors in Electrostatic Equilibrium

751

We can use Gauss’s law to verify the second property of a conductor in electrostatic

equilibrium. Figure 24.17 shows an arbitrarily shaped conductor. A gaussian surface is
drawn inside the conductor and can be as close to the conductor’s surface as we wish.
As we have just shown, the electric field everywhere inside the conductor is zero when
it is in electrostatic equilibrium. Therefore, the electric field must be zero at every
point on the gaussian surface, in accordance with condition (4) in Section 24.3. Thus,
the net flux through this gaussian surface is zero. From this result and Gauss’s law, we
conclude that the net charge inside the gaussian surface is zero. Because there can be
no net charge inside the gaussian surface (which is arbitrarily close to the conductor’s
surface),

any net charge on the conductor must reside on its surface. Gauss’s law

does not indicate how this excess charge is distributed on the conductor’s surface, only
that it resides exclusively on the surface.

We can also use Gauss’s law to verify the third property. First, note that if the field

vector

E had a component parallel to the conductor’s surface, free electrons would

experience an electric force and move along the surface; in such a case, the conduc-
tor would not be in equilibrium. Thus, the field vector must be perpendicular to the
surface. To determine the magnitude of the electric field, we draw a gaussian surface
in  the  shape  of  a  small  cylinder  whose  end  faces  are  parallel  to  the  surface  of  the
conductor (Fig. 24.18). Part of the cylinder is just outside the conductor, and part is
inside.  The  field  is  perpendicular  to  the  conductor’s  surface  from  the  condition  of
electrostatic equilibrium. Thus, we satisfy condition (3) in Section 24.3 for the curved
part  of  the  cylindrical  gaussian  surface—there  is  no  flux  through  this  part  of  the
gaussian surface because

E is parallel to the surface. There is no flux through the flat

face of the cylinder inside the conductor because here

E # 0; this satisfies condition

(4). Hence, the net flux through the gaussian surface is that through only the flat face
outside the conductor, where the field is perpendicular to the gaussian surface. Using
conditions (1) and (2) for this face, the flux is EA, where E is the electric field just
outside the conductor and A is the area of the cylinder’s face. Applying Gauss’s law to
this surface, we obtain

where we have used the fact that q

A. Solving for E gives for the electric field just

outside a charged conductor

(24.9)

Figure 24.19 shows electric field lines made visible by pieces of thread floating

in oil. Notice that the field lines are perpendicular to both the cylindrical conducting
surface and the straight conducting surface.

E #

E dA # EA #

Figure 24.17 A conductor of

arbitrary shape. The broken line

represents a gaussian surface that

can be as close to the surface of the

conductor as we wish.

Gaussian

surface

Figure 24.18 A gaussian surface

in the shape of a small cylinder is

used to calculate the electric field

just outside a charged conductor.

The flux through the gaussian

surface is EA. Remember that

E is

zero inside the conductor.

+ + +

Figure 24.19 Electric field pattern

surrounding a charged conducting

plate placed near an oppositely

charged conducting cylinder. Small

pieces of thread suspended in oil

align with the electric field lines.

Note that (1) the field lines are

perpendicular to both conductors

and (2) there are no lines inside

the cylinder (E # 0).

Quick Quiz 24.6

Your little brother likes to rub his feet on the carpet

and  then  touch  you  to  give  you  a  shock.  While  you  are  trying  to  escape  the  shock
treatment, you discover a hollow metal cylinder in your basement, large enough to
climb inside. In which of the following cases will you not be shocked? (a) You climb
inside  the  cylinder,  making  contact  with  the  inner  surface,  and  your  charged
brother touches the outer metal surface. (b) Your charged brother is inside touch-
ing the inner metal surface and you are outside, touching the outer metal surface.
(c)  Both  of  you  are  outside  the  cylinder,  touching  its  outer  metal  surface  but  not
touching each other directly.

Courtesy of Harold M. W

aage, Princeton University

Explore the electric field of the system in Figure 24.20 at the Interactive Worked Example link at http://www.pse6.com.

752

CHAPTE R 24 • Gauss’s Law

24.5 Formal Derivation of Gauss’s Law

One way of deriving Gauss’s law involves solid angles. Consider a spherical surface of
radius r containing an area element +A. The solid angle +5 (5: uppercase Greek
omega) subtended at the center of the sphere by this element is defined to be

From this equation, we see that +5 has no dimensions because +A and r

both have

dimensions L

. The dimensionless unit of a solid angle is the

steradian. (You may want

to compare this equation to Equation 10.1b, the definition of the radian.) Because the

Figure 24.20 (Example 24.10) A solid conducting sphere of

radius a and carrying a charge 2Q surrounded by a conducting

spherical shell carrying a charge 'Q.

–Q

Figure 24.21 (Example 24.10) A plot of E versus r for the two-

conductor system shown in Figure 24.20.

E =

Example 24.10 A Sphere Inside a Spherical Shell

A solid conducting sphere of radius a carries a net positive
charge  2Q.  A  conducting  spherical  shell  of  inner  radius  b
and  outer  radius  c is  concentric  with  the  solid  sphere  and
carries a net charge 'Q. Using Gauss’s law, find the electric
field in the regions labeled !, ", #, and $ in Figure 24.20
and  the  charge  distribution  on  the  shell  when  the  entire
system is in electrostatic equilibrium.

Solution First note that the charge distributions on both
the sphere and the shell are characterized by spherical sym-
metry around their common center. To determine the elec-
tric field at various distances r from this center, we construct
a spherical gaussian surface for each of the four regions of
interest. Such a surface for region " is shown in Figure
24.20.

To find E inside the solid sphere (region !), consider a

gaussian surface of radius r , a. Because there can be no
charge inside a conductor in electrostatic equilibrium, we
see that q

0; thus, on the basis of Gauss’s law and sym-

metry, E

0 for r , a.

In region "— between the surface of the solid sphere

and the inner surface of the shell — we construct a spherical
gaussian  surface  of  radius  r where  a , r , b and  note  that
the  charge  inside  this  surface  is  $ 2Q (the  charge  on  the
solid  sphere).  Because  of  the  spherical  symmetry,  the
electric field lines must be directed radially outward and be

constant in magnitude on the gaussian surface. Following
Example 24.4 and using Gauss’s law, we find that

In region $, where r - c, the spherical gaussian surface we
construct surrounds a total charge of q

2Q $ ('Q) #

Q. Therefore, application of Gauss’s law to this surface gives

In  region  #,  the  electric  field  must  be  zero  because  the
spherical  shell  is  also  a  conductor  in  equilibrium.  Figure
24.21  shows  a  graphical  representation  of  the  variation  of
electric field with r.

If we construct a gaussian surface of radius r where

b , r , c, we see that q

must be zero because E

0. From

this  argument,  we  conclude  that  the  charge  on  the  inner
surface  of  the  spherical  shell  must  be  ' 2Q to  cancel  the
charge $ 2Q on the solid sphere. Because the net charge on
the  shell  is  ' Q ,  we  conclude  that  its  outer  surface  must
carry a charge $ Q.

(for r - c)

(for a , r , b)

4(/

A # E

(4(r

) #

Interactive

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