Prof. Anchordoqui
Problems set # 3
Physics 169
February 24, 2015
1. A point charge q is located at the center of a uniform ring having linear charge density λ
and radius a, as shown in Fig. 1. Determine the total electric flux through a sphere centered at the
point charge and having radius R, where R < a.
Solution Only the charge inside radius R contributes to the total flux, hence ΦE = q/0 .
2. A point charge Q is located just above the center of the flat face of a hemisphere of radius
R as shown in Fig. 2. What is the electric flux (i) through the curved surface and (ii) through the
flat face?
Solution With δ very small, all points on the hemisphere are nearly at a distance R from
1 Q
the charge, so the field everywhere on the curved surface is 4π
2 radially outward (normal to
0 R
the surface). Therefore, the flux is this field strength times the area of half a sphere Φcurved =
R
~ · dA
~ = 1 Q2 2πR2 = Q . (ii) The closed surface encloses zero charge so Gauss’ law gives
E
4π0 R
20
Φcurved + Φflat = 0 or Φflat = −Φcurved = − 2Q0 .
3. The line ag in Fig. 3 is a diagonal of a cube. A point charge q is located on the extension of
line ag, very close to vertex a of the cube. Determine the electric flux through each of the sides of
the cube which meet at the point a.
Solution No charge is inside the cube. The net flux through the cube is zero. Positive flux
comes out through the three faces meeting at g. These three faces together fill a solid angle equal
to one-eighth of a sphere as seen from q. The total flux passing through these faces is then 18 q0 .
Each face containing a intercepts equal flux going into the cube: 0 = ΦE,net = 3ΦE,abcd + 8q0 .
q
Therefore, ΦE,abcd = − 24
.
0
4. A sphere of radius R surrounds a point charge Q , located at its center. (i) Show that the
electric flux through a circular cap of half-angle (see Fig. 4) is ΦE = 2Q0 (1 − cos θ). What is the
flux for (ii) θ = 90◦ and (iii) θ = 180◦ .
~ pointing radially outward, so ΦE = EA. The
Solution The charge creates a uniform E,
R
arc length is ds = Rdθ, and the circumference is 2πr = 2πR sin θ. Hence, A = 2πrds =
Rθ
R
1 Q
2 θ
2
θ
2
0 (2πR sin θ)Rdθ = 2πR 0 sin θ dθ = −2πR cos θ|0 = 2πR (1 − cos θ) and ΦE = 4π0 R2 ·
2πR2 (1 − cos θ) = 2Q0 (1 − cos θ), i.e., independent of R! (ii) For θ = 90◦ (hemisphere), ΦE =
Q
Q
Q
Q
◦
◦
◦
20 (1 − cos 90 ) = 20 (iii) For θ = 180 (entire sphere), ΦE = 20 (1 − cos 180 ) = 0 ; this is a
formal derivation of Gauss’ law.
5. An insulating solid sphere of radius a has a uniform volume charge density and carries a
total positive charge Q. A spherical gaussian surface of radius r, which shares a common center
with the insulating sphere, is inflated starting from r = 0. (i) Find an expression for the electric
flux passing through the surface of the gaussian sphere as a function of r for r < a. (ii) Find an
expression for the electric flux for r > a. (iii) Plot the flux versus r.
4
ρ
3 πa ⇒
4πr3 ρ
qin
0 = 30
Solution The charge density is determined by Q =
ρ =
3Q
.
4πa3
(i) The flux is that
Qr3
= 0 a3 . (ii) ΦE =
created by the enclosed charge within radius r: ΦE =
the answers to parts (i) and (ii) agree at r = a. (iii) This is shown in Fig. 5.
Q
0 .
Note that
6. A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly distributed
throughout its volume. Concentric with this sphere is a conducting spherical shell with inner radius
b and outer radius c, and having a net charge −Q, as shown in Fig. 6. (i) Construct a spherical
gaussian surface of radius r > c and find the net charge enclosed by this surface. (ii) What is the
direction of the electric field at r > c? (iii) Find the electric field at r ≥ c. (iv) Find the electric
field in the region with radius r where b < r < c (v) Construct a spherical gaussian surface of radius
r, where b < r < c, and find the net charge enclosed by this surface. (vi) Construct a spherical
gaussian surface of radius r, where a < r < b, and find the net charge enclosed by this surface.
(vii) Find the electric field in the region a < r < b. (viii) Construct a spherical gaussian surface of
radius r < a, and find an expression for the net charge enclosed by this surface, as a function of r.
Note that the charge inside this surface is less than 3Q. (ix) Find the electric field in the region
r < a. ( x) Determine the charge on the inner surface of the conducting shell. (xi) Determine
the charge on the outer surface of the conducting shell. (xii) Make a plot of the magnitude of the
electric field versus r.
Solution (i) qin = 3Q − Q = 2Q. (ii) The charge distribution is spherically symmetric and
Q
1 qin
qin > 0. Thus, the field is directed radially outward. (iii) For r ≥ c, E = 4π
2 = 2π r 2 .
0 r
0
(iv) Since all points within this region are located inside conducting material, E = 0, for b < r < c.
R
~ · dA
~ = 0 ⇒ qin = 0 ΦE = 0. (vi) qin = 3Q. (vii) For a ≤ r < b, E = 1 qin2 = 3Q 2
(v) ΦE = E
4π0 r
4π0 r
(radially outward). (viii) qin = ρV =
3Q 4
3
3 πr
4
πa3
3
3
= 3Q ar 3 . (ix) For 0 ≤ r ≤ a, E =
qin
4π0 r2
=
3Qr
4π0 a3
(radially outward). (x) From part (iv), for b < r < c, E = 0.Thus, for a spherical gaussian surface with b < r < c, qin = 3Q + qinner = 0 where qinner is the charge on the inner surface of the
conducting shell. This yields qinner = −3Q. (xi) Since the total charge on the conducting shell is
qnet = qouter + qinner = −Q, we have qouter = −Q − qinner = −Q − (−3Q) = 2Q. (xii) This is shown
in Fig. 6.
7.Consider a long cylindrical charge distribution of radius R with a uniform charge density ρ.
Find the electric field at distance r from the axis where r < R.
Solution If is positive, the field must be radially outward. Choose as the gaussian surface a cylinder of length L and radius r, contained inside the charged rod. Its volume is πr2 L and it encloses
charge ρπr2 L, see Fig. 7. Because the charge distribution is long, no electric flux passes through
~ · dA
~ = EdA cos(π/2) = 0. The curved surface has E
~ · dA
~ = EdA cos 0◦ ,
the circular end caps; E
and E must be the same strength everywhere over the curved surface. Gauss’ law,
becomes E
E2πrL =
R
curved
surface
dA =
ρπr2 L/0 .
ρπr2 L
0 .
~ =
Thus, E
H
~ · dA
~=
E
q
0 ,
Now the lateral surface area of the cylinder is 2πrL, yielding
ρr
20
radially away from the cylinder axis.
8. A solid, insulating sphere of radius a has a uniform charge density ρ and a total charge Q.
Concentric with this sphere is an uncharged, conducting hollow sphere whose inner and outer radii
are b and c, as shown in Fig. 8. (i) Find the magnitude of the electric field in the regions r < a,
a < r < b, b < r < c, and r > c. (ii) Determine the induced charge per unit area on the inner and
outer surfaces of the hollow sphere.
~ · dA
~ = E(4πr2 ) = qin . For r < a, qin = ρ 4 πr3 , so E = ρr . For
Solution (i) From Gauss’ law E
0
3
30
a < r < b and c < r, qin = Q, so E = 4πQ0 r2 . For b ≤ r ≤ c, E = 0, since E = 0 inside a conductor.
(ii) Let q1 be the induced charge on the inner surface of the hollow sphere. Since E = 0 inside
the conductor, the total charge enclosed by a spherical surface of radius b ≤ r ≤ c must be zero.
q1
Q
Therefore, q1 +Q = 0 and σ1 = 4πb
2 = − 4πb2 . Let q2 be the induced charge on the outside surface of
Q
q1
the hollow sphere. Since the hollow sphere is uncharged, we require q1 +q2 = 0 and σ2 = 4πc
2 = 4πc2 .
H
9. An early (incorrect) model of the hydrogen atom, suggested by J. J. Thomson, proposed
that a positive cloud of charge e was uniformly distributed throughout the volume of a sphere of
radius R, with the electron an equal-magnitude negative point charge e at the center. (i) Using
Gauss’ law, show that the electron would be in equilibrium at the center and, if displaced from the
center a distance r < R, would experience a restoring force of the form F = −kr, where k is a con2
stant. (ii) Show that k = 4πe0 R3 . (iii) Find an expression for the frequency f of simple harmonic
oscillations that an electron of mass me would undergo if displaced a small distance (< R) from
the center and released. (iv) Calculate a numerical value for R that would result in a frequency of
2.47×1015 Hz, the frequency of the light radiated in the most intense line in the hydrogen spectrum.
Solution First, consider the field at distance r < R from the center of a uniform sphere of positive
charge (Q = +e) with radius R. From Gauss’ law, 4πr2 E =
qin
0
=
ρV
0
=
+e
4
πR3
3
4
πr2
3
0
, so E =
e
r
4π0 R3
directed outward. (i) The force exerted on a point charge q = −e located at distance r from the
2
2
2
center is then F = qE = − 4πe0 R3 r = −kr. (ii) k = 4πe0 R3 (iii) Fr = me ar = − 4πe0 R3 r, so ar =
2
− 4π0eme R3 r = −ω 2 r. Thus, the motion is simple harmonic with frequency f =
q
9
2
2
−19
8.99×10 N·m /C (1.60×10
1
(iv) f = 2.47 × 1015 Hz = 2π
9.11×10−31 kg R3
R = 1.02 × 10−10 m = 102 pm.
C)2
, which yields R3 =
q
e2
.
4π0 me R3
1.05 × 10−30 m3 , or
ω
2π
=
1
2π
10. An infinitely long cylindrical insulating shell of inner radius a and outer radius b has a
uniform volume charge density ρ. A line of uniform linear charge density λ is placed along the axis
of the shell. Determine the electric field everywhere.
Solution The field direction is radially outward perpendicular to the axis. The field strength
depends on r but not on the other cylindrical coordinates θ or z. Choose a Gaussian cylinder of
~ = λ r̂. If a < r < b,
radius r and length L. If r < a, ΦE = qin0 and E2πrL = λL/0 , so E
2πr0
λ+ρπ(r2 −a2 )
2
2
~
E2πrL = [λL + ρπ(r − a )L]/0 and E =
r̂. If r > b, E2πrL = [λL + ρπ(b2 − a2 )L]/0
~ =
and E
λ+ρπ(b2 −a2 )
r̂.
2πr0
2πr0
11. A particle of mass m and charge q moves at high speed along the x axis. It is initially
near x = −∞, and it ends up near x = +∞. A second charge Q is fixed at the point x = 0,
y = −d. As the moving charge passes the stationary charge, its x component of velocity does not
change appreciably, but it acquires a small velocity in the y direction. Determine the angle through
which the moving charge is deflected. [Hint: The integral you encounter in determining vy can be
evaluated by applying Gauss’ law to a long cylinder of radius d, centered on the stationary charge.]
dv
Solution The vertical velocity component of the moving charge increases according to m dty =
dv
q
Fy → m dxy dx
9. Now dx
dt R= qEy , see Fig.
dt = vx , has a nearly constant value v. Hence dvy = mv Ey dx,
R
vy
+∞
q
yielding vy = 0 dvy0 = mv
The radially outward compnent of the electric field varies
−∞ Ey dx.
R +∞
R +∞
along the x axis; it is described by −∞ Ey dA = −∞
Ey 2πd dx = Q/0 . Putting all this together
R +∞
Q
qQ
E
dx
=
and
v
=
.
The
angle
of
deflection
is described by tan θ = vy /v, so
y
y
−∞
2πd0
mv2πd0
qQ
−1
θ = tan 2π0 dmv2 .
12. Two infinite, nonconducting sheets of charge are parallel to each other, as shown in Fig. 10.
The sheet on the left has a uniform surface charge density σ, and the one on the right has a uniform
charge density −σ. Calculate the electric field at points (i) to the left of, (ii) in between, and (iii)
to the right of the two sheets. (iv) Repeat the calculations when both sheets have positive uniform
surface charge densities of value σ.
Solution Consider the field due to a single sheet and let E+ and E− represent the fields due
to the positive and negative sheets, see Fig. 10. The field at any distance from each sheet has a
magnitude given by |E+ | = |E− | = 2σ0 . (i) To the left of the positive sheet, E+ is directed toward
~ = 0. (ii) In the region
the left and E− toward the right and the net field over this region is E
~ = σ/0
between the sheets, E+ and E− are both directed toward the right and the net field is E
to the right. (iii) To the right of the negative sheet, E− and E+ are again oppositely directed and
~ = 0. (iv) If both charges are positive (see Fig. 10), in the region to the left of the pair of sheets,
E
~ = σ/0 to the left; in the region
both fields are directed toward the left and the net field is E
between the sheets, the fields due to the individual sheets are oppositely directed and the net field
~ = 0; in the region to the right of the pair of sheets, both are fields are directed toward the
is E
~ = σ/0 to the right.
right and the net field is E
13. A sphere of radius 2a is made of a nonconducting material that has a uniform volume charge
density ρ. (Assume that the material does not affect the electric field.) A spherical cavity of radius
a is now removed from the sphere, as shown in Fig. 11. Show that the electric field within the
ρa
cavity is uniform and is given by Ex = 0 and Ey = 3
. [Hint: The field within the cavity is the
0
superposition of the field due to the original uncut sphere, plus the field due to a sphere the size of
the cavity with a uniform negative charge density −ρ].
~ + due to a
Solution The resultant field within the cavity is the superposition of two fields, one E
~
uniform sphere of positive charge of radius 2a, and the other E− due to a sphere of negative charge
3
~ + = ρr r̂ =
of radius a centered within the cavity. From Gauss’ law we have 43 πr0 ρ = 4πr2 E+ , so E
30
3
πr ρ
ρ
~ − = ρr1 (−r̂1 ) = − ρ ~r1 . It is easily seen in
~r. Using again Gauss’ law, − 4 1 = 4πr2 E− , so E
1
3 0
30
ρ(~
r−~a)
~
~
~
Fig. 11 that ~r = a + ~r1 , so E= − 30 , yielding E = E+ + E−
ρa
Therefore Ex = 0 and Ey = 3
at all points within the cavity.
0
30
=
ρ~
r
30
−
30
ρ~
r
ρ~a
30 + 30
=
ρ~a
30
= 0ı̂ +
ρa
30 ̂.
14. A solid insulating sphere of radius R has a nonuniform charge density that varies with r
according to the expression ρ = Ar2 , where A is a constant and r < R is measured from the
center of the sphere. (i) Show that the magnitude of the electric field outside (r > R) the sphere
AR5
is E = 5
(ii) Show that the magnitude of the electric field inside (r < R) the sphere is
2.
0r
3
E = Ar
50 . [Hint: The total charge Q on the sphere is equal to the integral of ρdV , where r extends
from 0 to R; also, the charge q within a radius r < R is less than Q. To evaluate the integrals,
note that the volume element dV for a spherical shell of radius r and thickness dr is equal to 4r2 dr.]
~ · dA
~ = E4πr2 = qin /0 . (i) For r > R, qin = R Ar2 4πr2 dr =
Solution From Gauss’ law E
0
Rr
5
AR
2 4πr 2 dr = 4πAr 5 /5, and E = Ar3 .
4πAR5 /5, and E = 5
.
(ii)
For
r
<
R,
q
=
Ar
in
2
0
50
0r
H
R
15. A slab of insulating material (infinite in two of its three dimensions) has a uniform positive
charge density ρ. An edge view of the slab is shown in Fig.12. (i) Show that the magnitude of
the electric field a distance x from its center and inside the slab is E = ρx/0 . (ii) Suppose an
electron of charge −e and mass me can move freely within the slab. It is released from rest at a
distance x from
q the center. Show that the electron exhibits simple harmonic motion with a fre1
quency s = 2π mρe
. (iii) A slab of insulating material has a nonuniform positive charge density
e 0
ρ = Cx2 , where x is measured from the center of the slab as shown in Fig. 12, and C is a constant.
The slab is infinite in the y and z directions. Derive expressions for the electric field in the exterior
regions and the interior region of the slab (−d/2 < x < d/2).
Solution (i) Consider a cylindrical shaped gaussian surface perpendicular to the yz plane with
H
~ · dA
~ = qin /0
one end in the yz plane and the other end containing the point x: Use Gauss law: E
~
By symmetry, the electric field is zero in the yz plane and is perpendicular to dA over the wall
of the gaussian cylinder. Therefore, the only contribution to the integral is over the end cap containing the point x. Hence EA = ρAx/0 , so that at distance x from the mid-line of the slab,
E = ρx/0 . (ii) Use Newton’s law to obtain a = mFe = − mρe
x. The acceleration of the electron
e 0
is of the form a = −ω 2 x with ω =
f=
ω
2π
=
1
2π
q
ρe
me 0 .
q
ρe
me 0 .
Thus, the motion is simple harmonic with frequency
C is applied
ough a recg assuming
) the plane
s the y axis,
x axis.
6. A point charge q is located at the center of a uniform ring
having linear charge density % and radius a, as shown in
Figure P24.6. Determine the total electric flux through a
sphere centered at the point charge and having radius R,
where R & a.
λ
R
N/C exists
derstorm is
m by 3.00 m
rd at 10.0°.
m of the car.
electric field
found. The
05 N " m2/C.
n a horizon104 N/C as
ux through
he slanted
a
q
Figure P24.6
Figure 1: Problem 1.
7. A pyramid with horizontal square base, 6.00 m on each
side, and a height of 4.00 m is placed in a vertical electric
field of 52.0 N/C. Calculate the total electric flux through
the pyramid’s four slanted surfaces.
8. A cone with base radius R and height h is located on a
horizontal table. A horizontal uniform field E penetrates
the cone, as shown in Figure P24.8. Determine the electric
flux that enters the left-hand side of the cone.
δ
Q
0
h
E
R
R
Figure P24.8
face of area
surface lies
he xy plane?
Section 24.2 Gauss’s Law
are
located inside a submarine:
9. The following charges
Figure
P24.15
Figure 2: Problem
2.
5.00 'C, ( 9.00 'C, 27.0 'C, and ( 84.0 'C. (a) Calculate
16. In the air over a particular region at an altitude of 500 m
above the ground the electric field is 120 N/C directed
re
of
surface. Find
the electric q
field
(a) just outside
the
point
charge
isat points
located
on the
extension of line ag, very
shell and (b) inside the shell.
tosquare
vertex
aplate
of 50.0
the
cube.
49.close
A thin
conducting
cm on
a side liesDetermine the electric flux
in the xy plane. A total charge of 4.00 % 10 C is placed
through
each
thedensity
sides
the cube which meet at the
on the plate. Find
(a) theof
charge
on theof
plate,
(b) the electric field just above the plate, and (c) the
point
a.just below the plate. You may assume that the
electric field
!8
m
a
ric
he
cu-
ut
ce
charge density is uniform.
50. A conducting spherical shell of inner radius a and outer
radius b carries a net charge Q. A point charge q is placed
at the center of this shell. Determine the surface charge
density on (a) the inner surface of the shell and (b) the
outer surface of the shell.
d
q
51. A hollow conducting sphere is surrounded by a larger
lectric
field lines
concentric spherical conducting shell. The inner sphere
a has net charge & 3Q .
has charge !Q , and the outer shell
sa
utside.
The charges are in electrostatic equilibrium. Using
es.
he
re
er
of
m,
hat
re
net
eld
nd
he
52. A positive point charge is at a distance R/2 from the center
of an uncharged thin conducting spherical shell of radius
R. Sketch the electric field lines set up by this arrangement
both inside and outside the shell.
h
g
Section 24.5 Formal Derivation of Gauss’s Law
FIG. P24.52
53. A sphere of radius R surrounds a point charge Q , located
at its center. (a) Show that the electric flux through a circular cap of half-angle ' (Fig. P24.53) is
e
f
Figure P24.22
Q
Figure 3: Problem 3.
(1 ! cos ')
2* 0
What is the flux for (b) ' # 90° and (c) ' # 180°?
Section 24.3 Application of Gauss’s Law to Various
Charge Distributions
The
arc length
ds Rd ,of
23. Determine
theis magnitude
of
ctonge
he
ed
ate
ce
est
2
b
Gauss’s law, find the charges and the electric fields
everywhere.
() #
ge
us
he
he
#
his
c
a
θ
the electric field at the
surface of a lead-208
nucleus, which contains 82 protons
Q
and 126 neutrons. Assume the lead nucleus has a volume
208 times that of one proton, and consider a proton to be
2 1.20 % 10&15 m.
a sphere of2Figure
radius
cos
RP24.53
1 cos
R
f
0
b
g
Figure 4: Problem 4.
24. A solid sphere of radius 40.0 cm FIG.
has a total
positive charge
P24.53
Additional
Problems
of 26.0
$C uniformly distributed throughout its volume.
54. A nonuniform electric field is given by the expression
Calculate
magnitude
of the electric field (a) 0 cm,
E # ayî & bzĵ & cxthe
k̂, where
a,
b, and c are constants.
[independent
of
R!]
Determine the electric flux through a rectangular surface
(b)
(c)
40.0
in the 10.0
xy plane,cm,
extending
from
x # 0 tocm,
x # w and
and from(d) 60.0 cm from the center
inre
sa
on,
on
he
he
Q
" Gauss’s
r " Law
b. (e) Construct
a spherical gaussian surface
h
(b)
.
Note
that
the
answers
to
parts
(a)
and
dditional Problems E
s r, where c " r 0" b, and find the net charge
59. A
z
4.54
In general,
E ay i bz j cxk
by this surface. (f) Construct a spherical gaussian
y=0
y =h
al
In the xy plane, z 0 and
E ay i cxk
=
0
x
of radius
r, aywhere
b " r " a, and find the net y
u
E
d
A
i
cx
k
k
dA
(c)
e
j
z
z
E
nclosed by this
surface.
(g) Find the
x = w electric field
dA = hdx
x
chw
x
x
ch z xdx ch
2
gion b " r " a.
(h)2 Construct a spherical gaussian
ch
FIG. P24.54
Q
of radius r # a, and find an expression for the net
ci
3Q Q
2Q
q
4.55
(a)
0
nclosed by this surface, as a function of r. Note
D
(b)
The charge distribution is spherically symmetric and q
0 . Thus, the field is directed
charge inside
this surface is less than 3Q. (i) Find
radially outward .
d
ric field in
the
region r # a. ( j) Determine the
m
k q
2k Q
E
for r c .
(c)
r
rsurface of the conducting shell.
on the inner
lo
(d)
Since all charge
points within this
regionthe
are located
inside conducting
material, of
E 0 the
for
rmine
the
on
outer
surface
r
0
ch
b r c.
ng shell. (l) Make 0a plot of the
a magnitude of the
0
60. R
field (e)versus zr.E dA 0 q
Figure 5: Problem 5.
ge
3Q
(f)
q
FIG. P24.34(c)
it
–Q
k q
3k Q
(g)
(radially outward) for a r b .
E
r
r
th
F 3Q I FG 4 r IJ 3Q r
el
(h)
q
V G
J
H
K
3
a
H a K 3Q
th
k q
k F
r I
r
(i)
E
3Q J
3k Q
(radially outward) for 0 r a .
w
G
r
r H
a K
a
fr
(j)
From part (d), E 0 for b r c . Thus, for a
asurface with b rc c ,
E
spherical gaussian
re
3Q q
0 where q
is the
q
charge on the inner surface of the
(b
conducting shell. This yields
3Q .
bq
fr
(k)
Since the total charge on the conducting
a
b c
r
shell is q
q
q
Q , we have
el
2Q . 6: Problem 6.
q
Q q
QP24.55
b 3Qg Figure
FIG. P24.55(l)
Figure
d
(l)
This is shown in the figure to the right.
a
r two identical conducting spheres whose surfaces
2.
E
2 w
w
2
E
x 0
x 0
in
in
e in
2
e
2
E
in
0
E
in
e in
2
e
2
in
4
3
e in
2
3
3
3
e
2
in
3
3
3
inner
e
3
inner
inner
net
outer
outer
inner
inner
b
the center of the wall to the field point is
Figure P24.55
56. Consider two identical conducting spheres whose surfaces
are separated by a small distance. One sphere is given a
s the
large net positive charge while the other is given a small
ned insidenet positive charge. It is found that the force between
them is attractive even though both spheres have net
2
r L . charges of the same sign. Explain how this is possible.
57.
A solid, insulating sphere of radius a has a uniform
es
charge density $ and a total charge Q. Concentric with this
e curved sphere is an uncharged, conducting hollow sphere whose
inner and outer radii are b and c, as shown in Figure
Figure 7: Problem 7.
ngth
P24.57. (a) Find the magnitude of the electric field in the
regions r # a, a #FIG.
r # b, P24.29
b # r # c, and r " c. (b) Deter-
L
.
mine the induced charge per unit area on the inner and
outer surfaces of the hollow sphere.
Insulator
Conductor
a
ly away from the cylinder axis .
phere at r1
b
c
5 cm
we
have
Figure
P24.57
Problems 57 and 58.
Figure 8: Problem 8.
61.
62.
e
e
charge.
eb
j
60. Review problem. An early (incorrect) model of the hydro2
2
gen atom, suggested by J. J. Thomson, proposed that a positive cloud of charge ( e was uniformly distributed
throughout the volume of a sphere of radius R, with the
0electron an equal-magnitude negative point charge ! e at
the center. (a) Using Gauss’s law, show that the electron
2 would be in equilibrium at the center and, if displaced
from the center a distance r # R, would experience a
restoring force of the form F % !Kr, where K is a constant.
(b) Show that K % k e e 2/R 3. (c) Find an expression for the
frequency f of simple harmonic oscillations that an
electron of mass me would undergo if displaced a small
distance (#R) from the center and released. (d) Calculate
a numerical value for R that would result in a frequency of
2.47
& 1015RHz,
frequency of
the light radiated in the
C HAPTE
24 the
• Gauss’s
Law
most intense line in the hydrogen spectrum.
a L
j
a due to the each sheet given by
field
r b .
r
ng charge
0
a f
y
vy
vx
x
v
directed perpendicular to the sheet.
0
0et and let E and E
s
760
q
ae and negative sheets. The
d
l the
to
left
of
the
pair
of
sheets,
both
fields
are
has
a
magnitude
given
by
61. An infinitely long cylindrical insulating shell of inner
n
radius a and outer radius b has a uniform volume charge
Q
trd the left and the net field is
density $. A line of uniform linear charge density ) is
the right has a uniform charge density !". Calculate
placed along the axis of the shell. Determine the electric
.m Soon
the electric field at points (a) to the left of, (b) in between,
67. P24.63
FIG.
A solid insulating sphere of ra
charge density that varies with r a
to the left .
# $ Ar 2, where A is a constant a
the center of the sphere. (a) Show
FIG. P24.59
electric field outside (r ( R) the
. E is directed toward the
et,
Show that the magnitude o
between
sheets,
the fields due to the individual sheets are (b)
oppositely
sities ofthe
value
".
(r ' R) the sphere is E $ Ar 3/
andnet
thefield
net field
over this
the
is
64.
A
sphere
of
radius
2a
is
made
of
a
nonconducting
material
nitude of the field due to the each sheet given by
charge Q on the sphere is equal to
that
has
a
uniform
volume
charge
density
#
.
(Assume
that
.
c
field
varies
along
the
x
axis,
but
is
described
by
n 24.8 is
r extends from 0 to R; also, the
the material does not affect the electric field.) A spherical
r ' R is less than Q. To evaluate
E of radius
directed
the sheet.
cavity
a is perpendicular
now removed to
from
the sphere, as
volume element dV for a spher
2 0 E are both
and
eets,
Eright
shown
in Figure
that
the are
electric
field
to the
of the P24.64.
pair of Show
sheets,
both
fields
arewithin
directed toward
the right
thickness dr is equal to 4) r 2dr.)
deld
the
field isis uniform and is given by Ex $ 0 and Ey $
the
isnetcavity
σ
68. A point charge Q is located on
In the
region
to the left
of the
pair
of sheets,
bothisfields
are
#a/3
% 0. (Suggestion:
The
field
within
the cavity
the super–σ
R at a distance b from the plane
directed
left due
and
the
net
field isuncut sphere, plus
of the
the
original
Figure
P24.62to
toposition
the toward
right
.thefield
FIG. that
P24.63
Show
if one fourth of th
FIG.
P24.62
Figure
10: Problem
the field due to a sphere the size of the
cavity
with a 12.
uniangle
of deflection
is described
by
charge passes through the disk, t
form
charge
!#.)
E negative
to the
left density
.
field everywhere.
s
62. Two
infinite,
nonconducting
of charge
are parallel
and
(c)
to the
right of sheets
the two
sheets.
e
to each other, as shown in Figure P24.62. The sheet onFigure
the
9: Problem 11.
e
63.
What
If?
Repeat
the
calculations
left has
a uniform
surface
charge
density
*, and the one for Problem 62
e
when both sheets have positive uniform surface charge dend
0
thin
theagain
superposition
two
heet,the
E cavity
and E is are
oppositelyofdirected
and E
aqQ
uniform sphere of positive charge of radius
0 .
In the region between
the sheets,
the fields due to the individual sheets are oppositely
y
.
due
to aand
sphere
of negative
2 the net
directed
field is charge of radius a
dmv
0
avity.
E 0 .
r
r
a
R
m
center
ofright
a uniform
of positive
charge
x both
Eto the
r of the pair sphere
In the
theso
region
of sheets,
are fields are
directed toward the right
and the net field is3
F
GH 4
E
so
e
E
0
0
3
0
2a
to therright .
b g
I r directed
outward
JK
1
3
r1
3
r1 .
0 Figure P24.64
0
Figure 11: Problem 13.
3
ltant field R
within
the cavity is the superposition
of two
0a
r
ne E due to a uniform sphere of positive charge of radius
a f
b
FIG. P24.64
Q
Figure P24.
3 0E due to a sphere of negative charge of radius a
he other
65. A uniformly charged spherical shell with surface charge
r
a
a center is then
cavity.
69. A spherically symmetric charge
qwithine the
located
atr distance
ra from the
b
this becomes E 2 r
r, length
r2
0
0
I FG a
JK H 2
g
0
Fz G a r IJ dV . The element of volume is a cylindrical
H bK
r
0 0
, and thickness dr so that dV
r
3b
2 r dr .
IJ so inside the cylinder, E
K
0r
2
0
FG a
H
Gauss’s law becomes
Fz G a r IJ b2
H bK
R
0R
g
r dr or outside the cylinder, E
2
0
0
2r
3b
FG a
rH
2
lindrical shaped gaussian surface perpendicular
e with one end in the yz plane and the other end
e point x:
aw:
z
qin
E dA
ance x from the mid-line ofdthe slab, E
x
x
0
.
.
Answers tox Quick Quizzes
z
O
IJ
K
the gaussian surface. (b) De
at a distance
gaussian r from the cent
surfacethat the Earth’s ma
assuming
0
0
2R
3b
A
y
a Axf
.
y
the electric field is zero in the yz plane and is
r to dA over the wall of the gaussian cylinder.
only contribution to the integral is over the end
g the point x :
or EA
IJ
K
24.1
(e). The same number o
x
sphere of any size. Because
sphere are closer to the cha
24.2 (d). All field lines that ente
container so that the total
nature of the field or the co
24.3 (b) and (d). Statement (a)
an equal number of positiv
E
e
Figure
P24.71
Problems
71
and
72.
x
be present inside the surfa
FIG. P24.71
Figure 12: Problem 15.
me 0
sarily true, as can be seen
72. A slab of insulating material has a nonuniform positive
electric field exists everyw
2
e charge is not enclosed wit
charge density ! " Cx , where x is measured
2 from the
with
.
on of the
electron
is
of
the
form
a
x
center of the slab as shown in Figure P24.71, and C is a
m e 0 flux is zero.
constant. The slab is infinite in the y and z directions.
24.4 (c). The charges q1 and q
Derive expressions for the electric field in (a) the exterior
contribute zero net flux thr
e
1
regionsharmonic
and (b)with
thefrequency
interior region
of the slab
f
.
ion is simple
2
2
m e 0 24.5 (d). We don’t need the sur
(#d/2 $ x $ d/2).
point in space will experien
73. (a) Using the mathematical similarity between Coulomb’s
local source charges.
law and Newton’s law of universal gravitation, show that
24.6 (a). Charges added to the m
Gauss’s law for gravitation can be written as
will reside on the outer su
FG
H
IJ
K