730
CHAPTE R 25 CAPACITANCE
From this we see that the threshold energy corresponds to a potential of
ϭ 3.9 ϫ 10
4
V.
Safety engineers attempt to keep the potential of the personnel below this level
by “bleeding” off the charge through, say, a conducting floor.
Energy Density
In a parallel-plate capacitor, neglecting fringing, the electric field has the same
value at all points between the plates. Thus, the energy density u — that is, the
potential energy per unit volume between the plates — should also be uniform.
We can find u by dividing the total potential energy by the volume Ad of the
space between the plates. Using Eq. 25-22, we obtain
(25-23)
With Eq. 25-9 (C ϭ ´ 0 A/d), this result becomes
(25-24)
However, from Eq. 24-42 (E ϭ Ϫ⌬V/⌬s), V/d equals the electric field magnitude E; so
(energy density).
(25-25)
Although we derived this result for the special case of an electric field of a
parallel-plate capacitor, it holds for any electric field. If an electric field exists at any point in space, that site has an electric potential energy with a density (amount per unit volume) given by Eq. 25-25.
E
:
u ϭ
1
2 ´ 0 E
2
u ϭ
1
2 ´ 0
V
d
2
.
u ϭ
U
Ad
ϭ
CV
2
2Ad
.
V ϭ A
2U t
4p´ 0 R
ϭ A
2(150 ϫ 10
Ϫ3
J)
4p(8.85 ϫ 10
Ϫ12
C
2
/Nиm
2
)(1.8 m)
Additional examples, video, and practice available at WileyPLUS
(b) What is the energy density at the surface of the sphere?
KEY IDEA
The density u of the energy stored in an electric field
depends on the magnitude E of the field, according to
Eq. 25-25
.
Calculations: Here we must first find E at the surface of
the sphere, as given by Eq. 23-15:
The energy density is then
(Answer)
ϭ 2.54 ϫ 10
Ϫ5
J/m
3 ϭ 25.4 mJ/m
3
.
ϭ
(1.25 ϫ 10
Ϫ9
C)
2
(32p
2
)(8.85 ϫ 10
Ϫ12
C
2
/Nиm
2
)(0.0685 m)
4
u ϭ
1
2 ´ 0 E
2 ϭ
q
2
32p
2 ´ 0 R
4
E ϭ
1
4p´ 0
q
R
2
.
(u ϭ
1
2 ´ 0 E
2
)
Sample Problem 25.04 Potential energy and energy density of an electric field
An isolated conducting sphere whose radius R is 6.85 cm
has a charge q ϭ 1.25 nC.
(a) How much potential energy is stored in the electric field
of this charged conductor?
KEY IDEAS
(1) An isolated sphere has capacitance given by Eq. 25-18
(C ϭ 4p´ 0 R). (2) The energy U stored in a capacitor depends on the capacitor’s charge q and capacitance C according to Eq. 25-21 (U ϭ q
2
/2C).
Calculation: Substituting C ϭ 4p´ 0 R into Eq. 25-21 gives us
(Answer)
ϭ 1.03 ϫ 10
Ϫ7
J ϭ 103 nJ.
ϭ
(1.25 ϫ 10
Ϫ9
C)
2
(8p)(8.85 ϫ 10
Ϫ12
F/m)(0.0685 m)
U ϭ
q
2
2C
ϭ
q
2
8p´ 0 R
CHAPTE R 25 CAPACITANCE
From this we see that the threshold energy corresponds to a potential of
ϭ 3.9 ϫ 10
4
V.
Safety engineers attempt to keep the potential of the personnel below this level
by “bleeding” off the charge through, say, a conducting floor.
Energy Density
In a parallel-plate capacitor, neglecting fringing, the electric field has the same
value at all points between the plates. Thus, the energy density u — that is, the
potential energy per unit volume between the plates — should also be uniform.
We can find u by dividing the total potential energy by the volume Ad of the
space between the plates. Using Eq. 25-22, we obtain
(25-23)
With Eq. 25-9 (C ϭ ´ 0 A/d), this result becomes
(25-24)
However, from Eq. 24-42 (E ϭ Ϫ⌬V/⌬s), V/d equals the electric field magnitude E; so
(energy density).
(25-25)
Although we derived this result for the special case of an electric field of a
parallel-plate capacitor, it holds for any electric field. If an electric field exists at any point in space, that site has an electric potential energy with a density (amount per unit volume) given by Eq. 25-25.
E
:
u ϭ
1
2 ´ 0 E
2
u ϭ
1
2 ´ 0
V
d
2
.
u ϭ
U
Ad
ϭ
CV
2
2Ad
.
V ϭ A
2U t
4p´ 0 R
ϭ A
2(150 ϫ 10
Ϫ3
J)
4p(8.85 ϫ 10
Ϫ12
C
2
/Nиm
2
)(1.8 m)
Additional examples, video, and practice available at WileyPLUS
(b) What is the energy density at the surface of the sphere?
KEY IDEA
The density u of the energy stored in an electric field
depends on the magnitude E of the field, according to
Eq. 25-25
.
Calculations: Here we must first find E at the surface of
the sphere, as given by Eq. 23-15:
The energy density is then
(Answer)
ϭ 2.54 ϫ 10
Ϫ5
J/m
3 ϭ 25.4 mJ/m
3
.
ϭ
(1.25 ϫ 10
Ϫ9
C)
2
(32p
2
)(8.85 ϫ 10
Ϫ12
C
2
/Nиm
2
)(0.0685 m)
4
u ϭ
1
2 ´ 0 E
2 ϭ
q
2
32p
2 ´ 0 R
4
E ϭ
1
4p´ 0
q
R
2
.
(u ϭ
1
2 ´ 0 E
2
)
Sample Problem 25.04 Potential energy and energy density of an electric field
An isolated conducting sphere whose radius R is 6.85 cm
has a charge q ϭ 1.25 nC.
(a) How much potential energy is stored in the electric field
of this charged conductor?
KEY IDEAS
(1) An isolated sphere has capacitance given by Eq. 25-18
(C ϭ 4p´ 0 R). (2) The energy U stored in a capacitor depends on the capacitor’s charge q and capacitance C according to Eq. 25-21 (U ϭ q
2
/2C).
Calculation: Substituting C ϭ 4p´ 0 R into Eq. 25-21 gives us
(Answer)
ϭ 1.03 ϫ 10
Ϫ7
J ϭ 103 nJ.
ϭ
(1.25 ϫ 10
Ϫ9
C)
2
(8p)(8.85 ϫ 10
Ϫ12
F/m)(0.0685 m)
U ϭ
q
2
2C
ϭ
q
2
8p´ 0 R