675
23-6 APPLYI NG GAUSS’ L AW: SPH E R ICAL SYM M ETRY
Applying Gauss’ Law: Spherical Symmetry
Here we use Gauss’ law to prove the two shell theorems presented without proof
in Module 21-1:
Additional examples, video, and practice available at WileyPLUS
via the superposition principle. (We can add the fields algebraically because they are parallel to each other.)
Calculations: At any point, the electric field
due to
the positive sheet is directed away from the sheet and, from
Eq. 23-13, has the magnitude
Similarly, at any point, the electric field
due to the negative
sheet is directed toward that sheet and has the magnitude
ϭ 2.43 ϫ 10
5
N/C.
E (Ϫ) ϭ
s (Ϫ)
2´ 0
ϭ
4.3 ϫ 10
Ϫ6
C/m
2
(2)(8.85 ϫ 10
Ϫ12
C
2
/Nиm
2
)
E
:
(Ϫ)
ϭ 3.84 ϫ 10
5
N/C.
E (ϩ) ϭ
s (ϩ)
2 0
ϭ
6.8 ϫ 10
Ϫ6
C/m
2
(2)(8.85 ϫ 10
Ϫ12
C
2
/Nиm
2
)
E
:
(ϩ)
Figure 23-19b shows the fields set up by the sheets to the left of
the sheets (L), between them (B), and to their right (R).
The resultant fields in these three regions follow from the
superposition principle.To the left, the field magnitude is
(Answer)
Because E (ϩ) is larger than E (Ϫ) , the net electric field
in this
region is directed to the left, as Fig. 23-19c shows.To the right of
the sheets, the net electric field has the same magnitude but is
directed to the right, as Fig. 23-19c shows.
Between the sheets, the two fields add and we have
(Answer)
The electric field
is directed to the right.
E
:
B
ϭ 6.3 ϫ 10
5
N/C.
ϭ 3.84 ϫ 10
5
N/C ϩ 2.43 ϫ 10
5
N/C
E B ϭ E (ϩ) ϩ E (Ϫ)
E
:
L
ϭ 1.4 ϫ 10
5
N/C.
ϭ 3.84 ϫ 10
5
N/C Ϫ 2.43 ϫ 10
5
N/C
E L ϭ E (ϩ) Ϫ E (Ϫ)
23-6 APPLYING GAUSS’ LAW: SPHERICAL SYMMETRY
Learning Objectives
charge, apply the relationship between the electric field
magnitude E, the charge q on the shell, and the distance r
from the shell’s center.
23.29 Identify the magnitude of the electric field for points enclosed by a spherical shell with uniform charge.
23.30 For a uniform spherical charge distribution (a uniform
ball of charge), determine the magnitude and direction of
the electric field at interior and exterior points.
● Outside a spherical shell of uniform charge q, the electric
field due to the shell is radial (inward or outward, depending
on the sign of the charge) and has the magnitude
(outside spherical shell),
where r is the distance to the point of measurement from the
center of the shell. The field is the same as though all of the
charge is concentrated as a particle at the center of the shell.
E ϭ
1
4p´ 0
q
r
2
● Inside the shell, the field due to the shell is zero.
● Inside a sphere with a uniform volume charge density, the
field is radial and has the magnitude
(inside sphere of charge),
where q is the total charge, R is the sphere’s radius, and r is
the radial distance from the center of the sphere to the point
of measurement.
E ϭ
1
4p´ 0
q
R
3
r
After reading this module, you should be able to . . .
23.26 Identify that a shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the
shell’s charge is concentrated at the center of the shell.
23.27 Identify that if a charged particle is enclosed by a shell
of uniform charge, there is no electrostatic force on the
particle from the shell.
23.28 For a point outside a spherical shell with uniform
Key Ideas
A shell of uniform charge attracts or repels a charged particle that is outside the
shell as if all the shell’s charge were concentrated at the center of the shell.
23-6 APPLYI NG GAUSS’ L AW: SPH E R ICAL SYM M ETRY
Applying Gauss’ Law: Spherical Symmetry
Here we use Gauss’ law to prove the two shell theorems presented without proof
in Module 21-1:
Additional examples, video, and practice available at WileyPLUS
via the superposition principle. (We can add the fields algebraically because they are parallel to each other.)
Calculations: At any point, the electric field
due to
the positive sheet is directed away from the sheet and, from
Eq. 23-13, has the magnitude
Similarly, at any point, the electric field
due to the negative
sheet is directed toward that sheet and has the magnitude
ϭ 2.43 ϫ 10
5
N/C.
E (Ϫ) ϭ
s (Ϫ)
2´ 0
ϭ
4.3 ϫ 10
Ϫ6
C/m
2
(2)(8.85 ϫ 10
Ϫ12
C
2
/Nиm
2
)
E
:
(Ϫ)
ϭ 3.84 ϫ 10
5
N/C.
E (ϩ) ϭ
s (ϩ)
2 0
ϭ
6.8 ϫ 10
Ϫ6
C/m
2
(2)(8.85 ϫ 10
Ϫ12
C
2
/Nиm
2
)
E
:
(ϩ)
Figure 23-19b shows the fields set up by the sheets to the left of
the sheets (L), between them (B), and to their right (R).
The resultant fields in these three regions follow from the
superposition principle.To the left, the field magnitude is
(Answer)
Because E (ϩ) is larger than E (Ϫ) , the net electric field
in this
region is directed to the left, as Fig. 23-19c shows.To the right of
the sheets, the net electric field has the same magnitude but is
directed to the right, as Fig. 23-19c shows.
Between the sheets, the two fields add and we have
(Answer)
The electric field
is directed to the right.
E
:
B
ϭ 6.3 ϫ 10
5
N/C.
ϭ 3.84 ϫ 10
5
N/C ϩ 2.43 ϫ 10
5
N/C
E B ϭ E (ϩ) ϩ E (Ϫ)
E
:
L
ϭ 1.4 ϫ 10
5
N/C.
ϭ 3.84 ϫ 10
5
N/C Ϫ 2.43 ϫ 10
5
N/C
E L ϭ E (ϩ) Ϫ E (Ϫ)
23-6 APPLYING GAUSS’ LAW: SPHERICAL SYMMETRY
Learning Objectives
charge, apply the relationship between the electric field
magnitude E, the charge q on the shell, and the distance r
from the shell’s center.
23.29 Identify the magnitude of the electric field for points enclosed by a spherical shell with uniform charge.
23.30 For a uniform spherical charge distribution (a uniform
ball of charge), determine the magnitude and direction of
the electric field at interior and exterior points.
● Outside a spherical shell of uniform charge q, the electric
field due to the shell is radial (inward or outward, depending
on the sign of the charge) and has the magnitude
(outside spherical shell),
where r is the distance to the point of measurement from the
center of the shell. The field is the same as though all of the
charge is concentrated as a particle at the center of the shell.
E ϭ
1
4p´ 0
q
r
2
● Inside the shell, the field due to the shell is zero.
● Inside a sphere with a uniform volume charge density, the
field is radial and has the magnitude
(inside sphere of charge),
where q is the total charge, R is the sphere’s radius, and r is
the radial distance from the center of the sphere to the point
of measurement.
E ϭ
1
4p´ 0
q
R
3
r
After reading this module, you should be able to . . .
23.26 Identify that a shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the
shell’s charge is concentrated at the center of the shell.
23.27 Identify that if a charged particle is enclosed by a shell
of uniform charge, there is no electrostatic force on the
particle from the shell.
23.28 For a point outside a spherical shell with uniform
Key Ideas
A shell of uniform charge attracts or repels a charged particle that is outside the
shell as if all the shell’s charge were concentrated at the center of the shell.