737
KEY IDEA
Now we apply Gauss’ law in the form of Eq. 25-36 to
Gaussian surface II in Fig. 25-17.
Calculations: Only the free charge Ϫq is in Eq. 25-36, so
(25-37)
The first minus sign in this equation comes from the dot
product
along the top of the Gaussian surface because now the field vector
is directed downward and the
area vector
(which, as always, points outward from the
interior of a closed Gaussian surface) is directed upward.
With 180Њ between the vectors, the dot product is negative.
Now k ϭ 2.61. Thus, Eq. 25-37 gives us
(Answer)
(e) What is the potential difference V between the plates
after the slab has been introduced?
KEY IDEA
We find V by integrating along a straight line directly from
the bottom plate to the top plate.
Calculation: Within the dielectric, the path length is b and
the electric field is E 1 . Within the two gaps above and below
the dielectric, the total path length is d Ϫ b and the electric
field is E 0 . Equation 25-6 then yields
(Answer)
This is less than the original potential difference of 85.5 V.
(f) What is the capacitance with the slab in place?
KEY IDEA
The capacitance C is related to q and V via Eq. 25-1.
Calculation: Taking q from (b) and V from (e), we have
(Answer)
This is greater than the original capacitance of 8.21 pF.
ϭ 1.34 ϫ 10
Ϫ11
F ϭ 13.4 pF.
C ϭ
q
V
ϭ
7.02 ϫ 10
Ϫ10
C
52.3 V
ϭ 52.3 V.
ϩ (2640 V/m)(0.00780 m)
ϭ (6900 V/m)(0.0124 m Ϫ 0.00780 m)
V ϭ ͵
ϩ
Ϫ
E ds ϭ E 0 (d Ϫ b) ϩ E 1 b
ϭ 2.64 kV/m.
E 1 ϭ
q
´ 0 kA
ϭ
E 0
k
ϭ
6.90 kV/m
2.61
dA
:
E
:
1
E
:
1 ؒ dA
:
´ 0 ͶkE
:
1 ؒ dA
: ϭ Ϫ´ 0 kE 1 A ϭ Ϫq.
Calculation: From Eq. 25-9 we have
(Answer)
(b) What free charge appears on the plates?
Calculation: From Eq. 25-1,
(Answer)
Because the battery was disconnected before the slab was
inserted, the free charge is unchanged.
(c) What is the electric field E 0 in the gaps between the
plates and the dielectric slab?
KEY IDEA
We need to apply Gauss’ law, in the form of Eq. 25-36, to
Gaussian surface I in Fig. 25-17.
Calculations: That surface passes through the gap, and so it
encloses only the free charge on the upper capacitor plate.
Electric field pierces only the bottom of the Gaussian surface.
Because there the area vector
and the field vector
are
both directed downward, the dot product in Eq. 25-36 becomes
Equation 25-36 then becomes
The integration now simply gives the surface area A of the
plate. Thus, we obtain
´ 0 kE 0 A ϭ q,
or
We must put k ϭ 1 here because Gaussian surface I does
not pass through the dielectric. Thus, we have
(Answer)
Note that the value of E 0 does not change when the slab is
introduced because the amount of charge enclosed by
Gaussian surface I in Fig. 25-17 does not change.
(d) What is the electric field E 1 in the dielectric slab?
ϭ 6900 V/m ϭ 6.90 kV/m.
E 0 ϭ
q
´ 0 kA
ϭ
7.02 ϫ 10
Ϫ10
C
(8.85 ϫ 10
Ϫ12
F/m)(1)(115 ϫ 10
Ϫ4
m
2
)
E 0 ϭ
q
´ 0 kA
.
´ 0 kE 0 ͶdA ϭ q.
E
:
0 ؒ dA
: ϭ E 0 dA cos 0Њ ϭ E 0 dA.
E
:
0
dA
:
ϭ 7.02 ϫ 10
Ϫ10
C ϭ 702 pC.
q ϭ C 0 V 0 ϭ (8.21 ϫ 10
Ϫ12
F)(85.5 V)
ϭ 8.21 ϫ 10
Ϫ12
F ϭ 8.21 pF.
C 0 ϭ
0 A
d
ϭ
(8.85 ϫ 10
Ϫ12
F/m)(115 ϫ 10
Ϫ4
m
2
)
1.24 ϫ 10
Ϫ2
m
Additional examples, video, and practice available at WileyPLUS
25-6 DI E LECTR ICS AN D GAUSS’ L AW
KEY IDEA
Now we apply Gauss’ law in the form of Eq. 25-36 to
Gaussian surface II in Fig. 25-17.
Calculations: Only the free charge Ϫq is in Eq. 25-36, so
(25-37)
The first minus sign in this equation comes from the dot
product
along the top of the Gaussian surface because now the field vector
is directed downward and the
area vector
(which, as always, points outward from the
interior of a closed Gaussian surface) is directed upward.
With 180Њ between the vectors, the dot product is negative.
Now k ϭ 2.61. Thus, Eq. 25-37 gives us
(Answer)
(e) What is the potential difference V between the plates
after the slab has been introduced?
KEY IDEA
We find V by integrating along a straight line directly from
the bottom plate to the top plate.
Calculation: Within the dielectric, the path length is b and
the electric field is E 1 . Within the two gaps above and below
the dielectric, the total path length is d Ϫ b and the electric
field is E 0 . Equation 25-6 then yields
(Answer)
This is less than the original potential difference of 85.5 V.
(f) What is the capacitance with the slab in place?
KEY IDEA
The capacitance C is related to q and V via Eq. 25-1.
Calculation: Taking q from (b) and V from (e), we have
(Answer)
This is greater than the original capacitance of 8.21 pF.
ϭ 1.34 ϫ 10
Ϫ11
F ϭ 13.4 pF.
C ϭ
q
V
ϭ
7.02 ϫ 10
Ϫ10
C
52.3 V
ϭ 52.3 V.
ϩ (2640 V/m)(0.00780 m)
ϭ (6900 V/m)(0.0124 m Ϫ 0.00780 m)
V ϭ ͵
ϩ
Ϫ
E ds ϭ E 0 (d Ϫ b) ϩ E 1 b
ϭ 2.64 kV/m.
E 1 ϭ
q
´ 0 kA
ϭ
E 0
k
ϭ
6.90 kV/m
2.61
dA
:
E
:
1
E
:
1 ؒ dA
:
´ 0 ͶkE
:
1 ؒ dA
: ϭ Ϫ´ 0 kE 1 A ϭ Ϫq.
Calculation: From Eq. 25-9 we have
(Answer)
(b) What free charge appears on the plates?
Calculation: From Eq. 25-1,
(Answer)
Because the battery was disconnected before the slab was
inserted, the free charge is unchanged.
(c) What is the electric field E 0 in the gaps between the
plates and the dielectric slab?
KEY IDEA
We need to apply Gauss’ law, in the form of Eq. 25-36, to
Gaussian surface I in Fig. 25-17.
Calculations: That surface passes through the gap, and so it
encloses only the free charge on the upper capacitor plate.
Electric field pierces only the bottom of the Gaussian surface.
Because there the area vector
and the field vector
are
both directed downward, the dot product in Eq. 25-36 becomes
Equation 25-36 then becomes
The integration now simply gives the surface area A of the
plate. Thus, we obtain
´ 0 kE 0 A ϭ q,
or
We must put k ϭ 1 here because Gaussian surface I does
not pass through the dielectric. Thus, we have
(Answer)
Note that the value of E 0 does not change when the slab is
introduced because the amount of charge enclosed by
Gaussian surface I in Fig. 25-17 does not change.
(d) What is the electric field E 1 in the dielectric slab?
ϭ 6900 V/m ϭ 6.90 kV/m.
E 0 ϭ
q
´ 0 kA
ϭ
7.02 ϫ 10
Ϫ10
C
(8.85 ϫ 10
Ϫ12
F/m)(1)(115 ϫ 10
Ϫ4
m
2
)
E 0 ϭ
q
´ 0 kA
.
´ 0 kE 0 ͶdA ϭ q.
E
:
0 ؒ dA
: ϭ E 0 dA cos 0Њ ϭ E 0 dA.
E
:
0
dA
:
ϭ 7.02 ϫ 10
Ϫ10
C ϭ 702 pC.
q ϭ C 0 V 0 ϭ (8.21 ϫ 10
Ϫ12
F)(85.5 V)
ϭ 8.21 ϫ 10
Ϫ12
F ϭ 8.21 pF.
C 0 ϭ
0 A
d
ϭ
(8.85 ϫ 10
Ϫ12
F/m)(115 ϫ 10
Ϫ4
m
2
)
1.24 ϫ 10
Ϫ2
m
Additional examples, video, and practice available at WileyPLUS
25-6 DI E LECTR ICS AN D GAUSS’ L AW