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Physics For Scientists And Engineers 6E - part 205

SECTION 26.7 • An Atomic Description of Dielectrics

817

Example 26.8 The H

O Molecule

The water (H

O) molecule has an electric dipole moment

of 6.3 * 10

C + m. A sample contains 10

water mole-

cules, with the dipole moments all oriented in the direction
of an electric field of magnitude 2.5 * 10

N/C. How much

work is required to rotate the dipoles from this orientation
(2 # 0°) to one in which all the moments are perpendicular
to the field (2 # 90°)?

Solution The work required to rotate one molecule 90° is
equal to the difference in potential energy between the 90°
orientation and the 0° orientation. Using Equation 26.19,

we obtain

Because there are 10

molecules in the sample, the total

work required is

1.6 * 10

total

(10

)(1.6 * 10

J) #

# 1.6 * 10

# pE # (6.3 * 10

C+m)(2.5 * 10

N/C)

W # U

905

($pE cos 905) $ ($pE cos 05)

26.7 An Atomic Description of Dielectrics

In Section 26.5 we found that the potential difference !V

between the plates of a

capacitor is reduced to !V

/1 when a dielectric is introduced. The potential difference

is reduced because the magnitude of the electric field decreases between the plates. In
particular, if

is the electric field without the dielectric, the field in the presence of a

dielectric is

(26.21)

Let us first consider a dielectric made up of polar molecules placed in the electric

field between the plates of a capacitor. The dipoles (that is, the polar molecules mak-
ing up the dielectric) are randomly oriented in the absence of an electric field, as
shown in Figure 26.25a. When an external field

due to charges on the capacitor

plates is applied, a torque is exerted on the dipoles, causing them to partially align with
the field, as shown in Figure 26.25b. We can now describe the dielectric as being polar-
ized. The degree of alignment of the molecules with the electric field depends on
temperature and on the magnitude of the field. In general, the alignment increases
with decreasing temperature and with increasing electric field.

If the molecules of the dielectric are nonpolar, then the electric field due to the

plates produces some charge separation and an induced dipole moment. These induced
dipole moments tend to align with the external field, and the dielectric is polarized.
Thus, we can polarize a dielectric with an external field regardless of whether the
molecules are polar or nonpolar.

E #

(b)

(c)

– +

–

– +

–

– +

–

– +

–

(a)

ind

–

ind

–

–
–
–
–

+
+
+
+

–

–
–
–
–

+
+
+
+

Figure 26.25 (a) Polar molecules are randomly oriented in the absence of an external

electric field. (b) When an external electric field is applied, the molecules partially

align with the field. (c) The charged edges of the dielectric can be modeled as an

additional pair of parallel plates establishing an electric field E

ind

in the direction

opposite to that of E

818

C H A P T E R 2 6 • Capacitance and Dielectrics

With these ideas in mind, consider a slab of dielectric material placed between the

plates of a capacitor so that it is in a uniform electric field

, as shown in Figure

26.25b. The electric field due to the plates is directed to the right and polarizes the
dielectric. The net effect on the dielectric is the formation of an induced positive
surface charge density '

ind

on the right face and an equal-magnitude negative surface

charge density $ '

ind

on the left face, as shown in Figure 26.25c. Because we can

model these surface charge distributions as being due to parallel plates, the induced
surface charges on the dielectric give rise to an induced electric field

ind

in the

direction opposite the external field

. Therefore, the net electric field

E in the

dielectric has a magnitude

(26.22)

In the parallel-plate capacitor shown in Figure 26.26, the external field E

is related

to the charge density ' on the plates through the relationship E

. The induced

electric field in the dielectric is related to the induced charge density '

ind

through the

relationship E

ind

. Because E # E

/1 # '/1)

, substitution into Equation

26.22 gives

(26.23)

Because 1 , 1, this expression shows that the charge density '

ind

induced on the

dielectric is less than the charge density ' on the plates. For instance, if 1 # 3 we see
that the induced charge density is two-thirds the charge density on the plates. If no
dielectric is present, then 1 # 1 and '

ind

0 as expected. However, if the dielectric is

replaced by an electrical conductor, for which E # 0, then Equation 26.22 indicates
that E

ind

; this corresponds to '

ind

. That is, the surface charge induced on

the conductor is equal in magnitude but opposite in sign to that on the plates, result-
ing in a net electric field of zero in the conductor (see Fig. 24.16).

We can use the existence of the induced surface charge distributions on the dielec-

tric to explain the result of Example 26.7. As we saw there, the energy of a capacitor not
connected to a battery is lowered when a dielectric is inserted between the plates; this
means that negative work is done on the dielectric by the external agent inserting the
dielectric into the capacitor. This, in turn, implies that a force must be acting on the
dielectric that draws it into the capacitor. This force originates from the nonuniform
nature of the electric field of the capacitor near its edges, as indicated in Figure 26.27.

ind

1 $

)

ind

)

E # E

ind

–

ind

Figure 26.26 Induced charge on a

dielectric placed between the

plates of a charged capacitor. Note

that the induced charge density on

the dielectric is less than the charge

density on the plates.

–Q

–

Figure 26.27 The nonuniform electric field near the edges of a parallel-plate

capacitor causes a dielectric to be pulled into the capacitor. Note that the field acts on

the induced surface charges on the dielectric, which are nonuniformly distributed.

SECTION 26.7 • An Atomic Description of Dielectrics

819

(b)

(d – a)/2

(a)

d a

(d – a)/2

–

Figure 26.28 (Example 26.9) (a) A parallel-plate capacitor of

plate separation d partially filled with a metallic slab of thick-

ness a. (b) The equivalent circuit of the device in part (a) con-

sists of two capacitors in series, each having a plate separation

(d $ a)/2.

Example 26.9 Effect of a Metallic Slab

A parallel-plate capacitor has a plate separation d and plate
area A. An uncharged metallic slab of thickness a is inserted
midway between the plates.

(A)

Find the capacitance of the device.

Solution We  can  solve  this  problem  by  noting  that  any
charge  that  appears  on  one  plate  of  the  capacitor  must
induce  a  charge  of  equal  magnitude  and  opposite  sign
on the  near  side  of  the  slab,  as  shown  in  Figure  26.28a.
Consequently,  the  net  charge  on  the  slab  remains  zero,
and  the  electric  field  inside  the  slab  is  zero.  Hence,  the
capacitor  is  equivalent  to  two  capacitors  in  series,  each
having  a  plate  separation  (d $ a)/2,  as  shown  in  Figure
26.28b.

Using Eq. 26.3 and the rule for adding two capacitors in

series (Eq. 26.10), we obtain

Note that C approaches infinity as a approaches d. Why?

)

d $ a

C #

)

(d $ a)/2

(

)

(d $ a)/2

(

(B)

Show that the capacitance of the original capacitor is

unaffected by the insertion of the metallic slab if the slab is
infinitesimally thin.

Solution In the result for part (A), we let a : 0:

which is the original capacitance.

What If?

What if the metallic slab in part (A) is not

midway between the plates? How does this affect the
capacitance?

Answer Let  us  imagine  that  the  slab  in  Figure  26.27a  is
moved upward so that the distance between the upper edge
of  the  slab  and  the  upper  plate  is  b.  Then,  the  distance
between  the  lower  edge  of  the  slab  and  the  lower  plate  is
d $ b $ a.  As  in  part  (A),  we  find  the  total  capacitance  of
the series combination:

This  is  the  same  result  as  in  part  (A).  It  is  independent  of
the value of b, so it does not matter where the slab is located.
In Figure 26.28b, when the central structure is moved up or
down,  the  decrease  in  plate  separation  of  one  capacitor  is
compensated  by  the  increase  in  plate  separation  for  the
other.

C #

)

d $ a

)

d $ b $ a

)

d $ a

)

()

A/b)

)

A/(d $ b $ a)

C # lim

a : 0

)

d $ a

)

Example 26.10 A Partially Filled Capacitor

A parallel-plate capacitor with a plate separation d has a
capacitance C

in the absence of a dielectric. What is the

capacitance when a slab of dielectric material of dielectric
constant 1 and thickness

is inserted between the plates

(Fig. 26.29a)?

Solution In Example 26.9, we found that we could insert a
metallic slab between the plates of a capacitor and consider
the  combination  as  two  capacitors  in  series.  The  resulting
capacitance  was  independent  of  the  location  of  the  slab.
Furthermore,  if  the  thickness  of  the  slab  approaches  zero,

The horizontal component of this fringe field acts on the induced charges on the surface
of the dielectric, producing a net horizontal force directed into the space between the
capacitor plates.

820

C H A P T E R 2 6 • Capacitance and Dielectrics

– d

(a)

(b)

– d

Figure 26.29 (Example 26.10) (a) A parallel-plate capacitor of

plate separation d partially filled with a dielectric of thickness

d/3. (b) The equivalent circuit of the capacitor consists of two

capacitors connected in series.

then  the  capacitance  of  the  system  approaches  the  capaci-
tance when the slab is absent. From this, we conclude that
we can insert an infinitesimally thin metallic slab anywhere
between  the  plates  of  a  capacitor  without  affecting  the
capacitance.  Thus,  let  us  imagine  sliding  an  infinitesimally
thin  metallic  slab  along  the  bottom  face  of  the  dielectric
shown in Figure 26.29a. We can then consider this system to
be  the  series  combination  of  the  two  capacitors  shown  in
Figure 26.29b: one having a plate separation d/3 and filled
with  a  dielectric,  and  the  other  having  a  plate  separation
2d/3 and air between its plates.

From Equations 26.15 and 26.3, the two capacitances

are

Using Equation 26.10 for two capacitors combined in series,
we have

Because the capacitance without the dielectric is C

)

A/d,

we see that

21 & 1

C #

21 & 1

)

1
1

)

1 & 21

d/3

2d/3

)

d/3

and

)

2d/3

capacitor consists of two conductors carrying charges of equal magnitude and

opposite sign. The

capacitance C of any capacitor is the ratio of the charge Q on

either conductor to the potential difference !V between them:

(26.1)

The capacitance depends only on the geometry of the conductors and not on an exter-
nal source of charge or potential difference.

The SI unit of capacitance is coulombs per volt, or the

farad (F), and 1 F #

1 C/V.

Capacitance expressions for various geometries are summarized in Table 26.2.
If two or more capacitors are connected in parallel, then the potential difference is

the same across all of them. The equivalent capacitance of a parallel combination of
capacitors is

(26.8)

If two or more capacitors are connected in series, the charge is the same on all of

them, and the equivalent capacitance of the series combination is given by

& + + +

∆V

S U M M A R Y

Take a practice test for

this chapter by clicking on
the Practice Test link at
http://www.pse6.com.

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