Ch22 ssm

Chapter 22
The Electric Field 2: Continuous Charge
Distributions
Conceptual Problems
1 • Figure 22-37 shows an L-shaped object that has sides which are equal
in length. Positive charge is distributed uniformly along the length of the object.
What is the direction of the electric field along the dashed 45o
line? Explain your
answer.
Determine the Concept The resultant field is the superposition of the electric
fields due to the charge distributions along the axes and is directed along the
dashed line, pointing away from the intersection of the two sides of the L-shaped
object. This can be seen by dividing each leg of the object into 10 (or more) equal
segments and then drawing the electric field on the dashed line due to the charges
on each pair of segments that are equidistant from the intersection of the legs.
7 •• An electric dipole is completely inside a closed imaginary surface and
there are no other charges. True or False:
(a) The electric field is zero everywhere on the surface.
(b) The electric field is normal to the surface everywhere on the surface.
(c) The electric flux through the surface is zero.
(d) The electric flux through the surface could be positive or negative.
(e) The electric flux through a portion of the surface might not be zero.
(a) False. Near the positive end of the dipole, the electric field, in accordance with
Coulomb’s law, will be directed outward and will be nonzero. Near the negative
end of the dipole, the electric field, in accordance with Coulomb’s law, will be
directed inward and will be nonzero.
(b) False. The electric field is perpendicular to the Gaussian surface only at the
intersections of the surface with a line defined by the axis of the dipole.
(c) True. Because the net charge enclosed by the Gaussian surface is zero, the net
flux, given by inside
S
nnet 4 kQdAE πφ == ∫ , through this surface must be zero.
19

Chapter 2220
(d) False. The flux through the closed surface is zero.
(e) True. All Gauss’s law tells us is that, because the net charge inside the surface
is zero, the net flux through the surface must be zero.
9 •• Suppose that the total charge on the conducting spherical shell in
Figure 22-38 is zero. The negative point charge at the center has a magnitude
given by Q. What is the direction of the electric field in the following regions?
(a) r < R1 , (b) R2 > r > R1 , (c) and r > R2 . Explain your answer.
Determine the Concept We can apply Gauss’s law to determine the electric field
for r < R1, R2 > r > R1, and r > R2. We also know that the direction of an electric
field at any point is determined by the direction of the electric force acting on a
positively charged object located at that point.
(a) From the application of Gauss’s law we know that the electric field in this
region is not zero. A positively charged object placed in the region for which
r < R1 will experience an attractive force from the charge –Q located at the center
of the shell. Hence the direction of the electric field is radially inward.
(b) Because the total charge on the conducting sphere is zero, the charge on its
inner surface is +Q (the positive charges in the conducting sphere are drawn there
by the negative charge at the center of the shell) and the charge on its outer
surface is –Q. Applying Gauss’s law in the region R2 > r > R1 (the net charge
enclosed by a Gaussian surface of radius r is zero) leads to the conclusion that the
electric field in this region is zero. It has no direction.
(c) Because the charge on the outer surface of the conducting shell is negative,
the electric field in the region r > R2 is radially inward.
Calculating E
r
From Coulomb’s Law
13 •• A uniform line charge that has a linear charge density λ equal to
3.5 nC/m is on the x axis between x = 0 and x = 5.0 m. (a) What is its total
charge? Find the electric field on the x axis at (b) x = 6.0 m, (c) x = 9.0 m, and
(d) x = 250 m. (e) Estimate the electric field at x = 250 m, using the
approximation that the charge is a point charge on the x axis at x = 2.5 m, and
compare your result with the result calculated in Part (d). (To do this you will
need to assume that the values given in this problem statement are valid to more

The Electric Field 2: Continuous Charge Distributions 21
than two significant figures.) Is your approximate result greater or smaller than
the exact result? Explain your answer.
Picture the Problem (a) We can use the definition of λ to find the total charge of
the line of charge. (b), (c) and (d) Equation 22-2b gives the electric field on the axis
of a finite line of charge. In Part (e) we can apply Coulomb’s law for the electric
field due to a point charge to approximate the electric field at x = 250 m. In the
following diagram, L = 5.0 m and P is a generic point on the x axis.
+ + + ++ + + + +
5.00 6.0 9.0 250
x, m
r2
1
r
P
(a) Use the definition of linear charge
density to express Q in terms of λ:
( )( )
nC18
nC17.5m5.0nC/m3.5
=
=== LQ λ
The electric field on the axis of a
finite line charge is given by
Equation 22-2b:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=
12
11
rr
kEx λ
(b) Substitute numerical values and evaluate Ex = 6.0 m:
N/C26
m0.6
1
m5.0m0.6
1
m
C
105.3
C
mN
108.988 9
2
2
9
m0.6 =⎟
⎠
⎞
⎜
⎝
⎛
−
−
⎟
⎠
⎞
⎜
⎝
⎛
×⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
×= −
=xE
(c) Substitute numerical values and evaluate Ex = 9.0 m:
N/C4.4
m0.9
1
m5.0m0.9
1
m
C
105.3
C
mN
108.988 9
2
2
9
m0.9 =⎟
⎠
⎞
⎜
⎝
⎛
−
−
⎟
⎠
⎞
⎜
⎝
⎛
×⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
×= −
=xE
(d) Substitute numerical values and evaluate Ex at x = 250 m:
mN/C6.2mN/C56800.2
m502
1
m5.0m502
1
m
C
105.3
C
mN
108.988 9
2
2
9
m502
==
⎟
⎠
⎞
⎜
⎝
⎛
−
−
⎟
⎠
⎞
⎜
⎝
⎛
×⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
×= −
=xE
(e) Using the approximation that the
charge is a point charge on the x axis
at x = 2.5 m, Coulomb’s law gives:
( )2
2
1
1 Lr
kQ
Ex
−
=

Chapter 2222
Substitute numerical values and evaluate Ex = 250 m:
( )( )
( )( )
mN/C6.2mN/C56774.2
m0.5m250
nC17.5/CmN108.988
2
2
1
229
m250 ==
−
⋅×
==xE
This result is about 0.01% less than the exact value obtained in (d). This suggests
that the line of charge can be modeled to within 0.01% as that due to a point
charge.
17 • A ring that has radius a lies in the z = 0 plane with its center at the
origin. The ring is uniformly charged and has a total charge Q. Find Ez on the z
axis at (a) z = 0.2a, (b) z = 0.5a, (c) z = 0.7a, (d) z = a, and (e) z = 2a. (f) Use your
results to plot Ez versus z for both positive and negative values of z. (Assume that
these distances are exact.)
Picture the Problem The electric field at a distance z from the center of a ring
whose charge is Q and whose radius is a is given by
( ) 2322
az
kQz
Ez
+
= .
(a) Evaluating Ez = 0.2a gives: ( )
( )[ ] 223220.2 189.0
2.0
2.0
a
kQ
aa
akQ
E az =
+
==
(b) Evaluating Ez = 0.5a gives: ( )
( )[ ] 223220.5 358.0
5.0
5.0
a
kQ
aa
akQ
E az =
+
==
(c) Evaluating Ez = 0.7a gives: ( )
( )[ ] 223220.7 385.0
7.0
7.0
a
kQ
aa
akQ
E az =
+
==
(d) Evaluating Ez = a gives:
[ ] 22322
354.0
a
kQ
aa
kQa
E az =
+
==
(e) Evaluating Ez = 2a gives:
( )[ ] 223222 179.0
2
2
a
kQ
aa
kQa
E az =
+
==
(f) The field along the z axis is plotted below. The z coordinates are in units of z/a
and E is in units of kQ/a2
.

-0.4
-0.2
0.0
0.2
0.4
-3 -2 -1 0 1 2 3
z/a
Ez
25 •• Calculate the electric field a distance z from a uniformly charged
infinite flat non-conducting sheet by modeling the sheet as a continuum of infinite
circular rings of charge.
Picture the Problem The field at a
point on the axis of a uniformly
charged ring lies along the axis and is
given by Equation 22-8. The diagram
shows one ring of the continuum of
circular rings of charge. The radius of
the ring is a and the distance from its
center to the field point P is z. The ring
has a uniformly distributed charge Q.
The resultant electric field at P is the
sum of the fields due to the continuum
of circular rings. Note that, by
symmetry, the horizontal components
of the electric field cancel.
P
a
da
Ed
r
Q
z
Express the field of a single
uniformly charged ring with charge
Q and radius a on the axis of the ring
at a distance z away from the plane
of the ring:
iE ˆ
zE=
r
, where
( ) 2322
az
kQz
Ez
+
=
Substitute dq for Q and dEz for Ez to
obtain: ( ) 2322
az
kzdq
dEz
+
=
The resultant electric field at P is the
sum of the fields due to all the circular
rings. Integrate both sides to calculate
( ) ( )∫∫ +
=
+
= 23222322
az
dq
kz
az
kzdq
E

Chapter 2224
the resultant field for the entire plane.
The field point remains fixed, so z is
constant:
To evaluate this integral we change
integration variables from q to a.
The charge dAdq σ= where
daadA π2= is the area of a ring of
radius a and width da:
daadq πσ2=
so
( )
( )∫
∫
∞
∞
+
=
+
=
0
2322
0
2322
2
2
az
daa
kz
ax
daa
kzE
σπ
σπ
To integrate this expression, let
22
azu += . Then:
( ) da
u
a
ada
az
du =
+
= 2
1
222
1
or
uduada =
Noting that when a = 0, u = z,
substitute and simplify to obtain: ∫∫
∞
−
∞
==
xx
duukzdu
u
u
kzE 2
3
22 πσπσ
Evaluating the integral yields:
02
2
1
2
∈
σ
σππσ ==⎟
⎠
⎞
⎜
⎝
⎛
−=
∞
k
u
kzE
z
Gauss’s Law
29 • An electric field is given by ( ) ( )iE ˆN/C300sign ⋅= x
r
, where sign(x)
equals –1 if x < 0, 0 if x = 0, and +1 if x > 0. A cylinder of length 20 cm and
radius 4.0 cm has its center at the origin and its axis along the x axis such that one
end is at x = +10 cm and the other is at x = –10 cm. (a) What is the electric flux
through each end? (b) What is the electric flux through the curved surface of the
cylinder? (c) What is the electric flux through the entire closed surface? (d) What
is the net charge inside the cylinder?
Picture the Problem The field at both circular faces of the cylinder is parallel to
the outward vector normal to the surface, so the flux is just EA. There is no flux
through the curved surface because the normal to that surface is perpendicular
to .E
r
The net flux through the closed surface is related to the net charge inside by
Gauss’s law.

(a) Use Gauss’s law to calculate
the flux through the right circular
surface:
( ) ( )( )
/CmN5.1
m040.0ˆˆN/C300
ˆ
2
2
rightrightright
⋅=
⋅=
⋅=
π
φ
ii
nE A
r
Apply Gauss’s law to the left
circular surface:
( ) ( )( )( )
/CmN5.1
m040.0ˆˆN/C300
ˆ
2
2
leftleftleft
⋅=
−⋅−=
⋅=
π
φ
ii
nE A
r
(b) Because the field lines are
parallel to the curved surface of
the cylinder:
0curved =φ
(c) Express and evaluate the net
flux through the entire cylindrical
surface:
/CmN0.3
0/CmN5.1/CmN5.1
2
22
curvedleftrightnet
⋅=
+⋅+⋅=
++= φφφφ
(d) Apply Gauss’s law to obtain:
insidenet 4 kQπφ = ⇒
k
Q
π
φ
4
net
inside =
Substitute numerical values and
evaluate :insideQ ( )
C107.2
/CmN10988.84
/CmN0.3
11
229
2
inside
−
×=
⋅×
⋅
=
π
Q
33 • A single point charge is placed at the center of an imaginary cube that
has 20-cm-long edges. The electric flux out of one of the cube’s sides is –1.50
kN⋅m2
/C. How much charge is at the center?

Chapter 2226
Picture the Problem The net flux through the cube is given by 0insidenet ∈φ Q= ,
where is the charge at the center of the cube.insideQ
The flux through one side of the
cube is one-sixth of the total flux
through the cube:
0
inside
net6
1
faces1
6∈
φφ
Q
==
Solving for yields:insideQ faces20inside 6 φ∈=Q

Substitute numerical values and evaluate :insideQ
nC7.79
C
mkN
50.1
mN
C
10854.86 2
2
2
2
12
inside −=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
×= −
Q
Gauss’s Law Applications in Spherical Symmetry Situations
39 •• A non-conducting sphere of radius 6.00 cm has a uniform volume
charge density of 450 nC/m3
. (a) What is the total charge on the sphere? Find the
electric field at the following distances from the sphere’s center: (b) 2.00 cm,
(c) 5.90 cm, (d) 6.10 cm, and (e) 10.0 cm.
(a) Using the definition of volume
charge density, relate the charge on
the sphere to its volume:
3
3
4
rVQ πρρ ==
evaluate Q:
( )( )
nC407.0nC4072.0
m0600.0nC/m450
33
3
4
==
= πQ
Apply Gauss’s law to a spherical
surface of radius r < R that is
concentric with the spherical shell
to obtain:
inside
0
S
1
QdAEn
∈
=∫ ⇒
0
inside2
4
∈
π
Q
Er n =
Noting that, due to symmetry,
, solve for Ern EE = r to obtain: 2
inside
2
0
inside 1
4 r
kQ
r
Q
Er ==
∈π
Because the charge distribution is
uniform, we can find the charge
inside the Gaussian surface by using
the definition of volume charge
density to establish the proportion:
V'
Q
V
Q inside
=
where V′ is the volume of the Gaussian
surface.
Solve for to obtain:insideQ
3
3
inside
R
r
Q
V
V'
QQ ==
Substitute for to obtain:insideQ
r
R
kQ
r
Q
E Rr 32
0
inside 1
4
==<
∈π

Chapter 2228
(b) Evaluate Er = 2.00 cm:
( )( )
( )
( ) N/C339m0.0200
m0.0600
nC0.4072/CmN10988.8
3
229
cm2.00 =
⋅×
==rE
(c) Evaluate Er = 5.90 cm:
( )( )
( )
( ) kN/C00.1m0.0590
m0.0600
nC0.4072/CmN10988.8
3
229
cm5.90 =
⋅×
==rE
Apply Gauss’s law to the Gaussian
surface with r > R: 0
inside2
4
∈
π
Q
Er r = ⇒ 22
inside
r
kQ
r
kQ
Er ==
(d) Evaluate Er = 6.10 cm:
( )( )
( )
N/C983
m0.0610
nC0.4072/CmN10988.8
2
229
cm6.10 =
⋅×
==rE
(e) Evaluate Er = 10.0 cm:
( )( )
( )
N/C366
m0.100
nC0.4072/CmN10988.8
2
229
cm10.0 =
⋅×
==rE
43 •• A sphere of radius R has volume charge density ρ = B/r for r < R ,
where B is a constant and ρ = 0 for r > R. (a) Find the total charge on the sphere.
(b) Find the expressions for the electric field inside and outside the charge
distribution (c) Sketch the magnitude of the electric field as a function of the
distance r from the sphere’s center.
Picture the Problem We can find the total charge on the sphere by expressing the
charge dq in a spherical shell and integrating this expression between r = 0 and
r = R. By symmetry, the electric fields must be radial. To find Er inside the
charged sphere we choose a spherical Gaussian surface of radius r < R. To find Er
outside the charged sphere we choose a spherical Gaussian surface of radius r > R.
On each of these surfaces, Er is constant. Gauss’s law then relates Er to the total
charge inside the surface.
(a) Express the charge dq in a shell
of thickness dr and volume 4πr2
dr:
Brdr
dr
r
B
rdrrdq
π
πρπ
4
44 22
=
==

Integrate this expression from
r = 0 to R to find the total charge
on the sphere:
[ ]
2
0
2
0
2
24
BR
BrdrrBQ
R
R
π
ππ
=
=== ∫
(b) Apply Gauss’s law to a spherical
surface of radius r > R that is
concentric with the nonconducting
sphere to obtain:
inside
0
S
1
QdAEr
∈
=∫ or
0
inside2
4
∈
π
Q
Er r =
Solving for Er yields:
2
0
2
2
2
2
inside
2
0
inside
2
2
1
4
r
BR
r
BRk
r
kQ
r
Q
E Rr
∈
π
∈π
==
==>
surface of radius r < R that is
concentric with the nonconducting
sphere to obtain:
inside
0
S
1
QdAEr
∈
=∫ ⇒
0
inside2
4
∈
π
Q
Er r =
Solving for Er yields:
00
2
2
0
2
inside
24
2
4 ∈∈π
π
∈π
B
r
Br
r
Q
E Rr ===<
(c) The following graph of Er versus r/R, with Er in units of B/(2∈0), was plotted
using a spreadsheet program.
0.0
0.2
0.4
0.6
0.8
1.0
1.2
0.0 0.5 1.0 1.5 2.0 2.5 3.0
r /R
E r
Remarks: Note that our results for (a) and (b) agree at r = R.

Chapter 2230
Gauss’s Law Applications in Cylindrical Symmetry Situations
51 •• A solid cylinder of length 200 m and radius 6.00 cm has a uniform
volume charge density of 300 nC/m3
. (a) What is the total charge of the cylinder?
Use the formulas given in Problem 50 to calculate the electric field at a point
equidistant from the ends at the following radial distances from the cylindrical
axis: (b) 2.00 cm, (c) 5.90 cm, (d) 6.10 cm, and (e) 10.0 cm.
Picture the Problem We can use the definition of volume charge density to find
the total charge on the cylinder. From symmetry, the electric field tangent to the
surface of the cylinder must vanish. We can construct a Gaussian surface in the
shape of a cylinder of radius r and length L and apply Gauss’s law to find the
electric field as a function of the distance from the centerline of the uniformly
charged cylinder.
(a) Use the definition of volume
charge density to express the
total charge of the cylinder:
( )LRVQ 2
total πρρ ==
Substitute numerical values to
obtain:
( )( ) ( )
nC679
m200m0.0600nC/m300
23
total
=
= πQ
(b) From Problem 50, for r < R, we
have:
rE Rr
02∈
ρ
=<
For r = 2.00 cm:
( )( )
( ) N/C339
m/NC108.8542
m0.0200nC/m300
2212
3
cm2.00 =
⋅×
= −=rE
(c) For r = 5.90 cm:
( )( )
( ) kN/C00.1
m/NC108.8542
m0.0590nC/m300
2212
3
cm5.90 =
⋅×
= −=rE
From Problem 50, for r > R, we have:
r
R
E Rr
0
2
2∈
ρ
=>

(d) For r = 6.10 cm:
( )( )
( )( )
kN/C00.1
m0610.0m/NC108.8542
m0600.0nC/m300
2212
23
cm6.10 =
⋅×
= −=rE
(e) For r = 10.0 cm:
( )( )
( )( )
N/C610
m100.0m/NC108.8542
m0600.0nC/m300
2212
23
cm10.0 =
⋅×
= −=rE
55 •• An infinitely long non-conducting solid cylinder of radius a has a non-
uniform volume charge density. This density varies with R, the perpendicular
distance from its axis, according to ρ(R) = bR2
, where b is a constant. (a) Show
that the linear charge density of the cylinder is given by λ = πba4
/2. (b) Find
expressions for the electric field for R < a and R > a.
Picture the Problem From symmetry; the field tangent to the surface of the
cylinder must vanish. We can construct a Gaussian surface in the shape of a
cylinder of radius R and length L and apply Gauss’s law to find the electric field
as a function of the distance from the centerline of the infinitely long
nonconducting cylinder.
(a) Apply Gauss’s law to a cylindrical
surface of radius R and length L that is
concentric with the infinitely long
nonconducting cylinder:
inside
0
S
1
QdAEn
∈
=∫ ⇒
0
inside
2
∈
π
Q
RLEn =
where we’ve neglected the end areas
because there is no flux through them.
, solve for ERn EE = R to obtain: 0
inside
2 ∈πRL
Q
ER = (1)
Express for ρ(R) = bRinsidedQ 2
: ( ) ( )
LdRbR
dRRLbRdVRdQ
3
2
inside
2
2
π
πρ
=
==
Integrate from 0 to a to
obtain:
insidedQ
4
0
4
0
3
inside
2
4
22
a
bL
R
bLdrRbLQ
aR
π
ππ
=
⎥
⎦
⎤
⎢
⎣
⎡
== ∫

Chapter 2232
Divide both sides of this equation by
L to obtain an expression for the
charge per unit length λ of the
cylinder:
2
4
inside ba
L
Q π
λ ==
(b) Substitute for in equation
(1) and simplify to obtain:
insideQ
3
00
4
42
2 R
b
LR
R
bL
E aR
∈∈π
π
==<
For R > a: 4
inside
2
a
bL
Q
π
=
Substitute for in equation (1)
and simplify to obtain:
insideQ
R
ba
RL
a
bL
E aR
0
4
0
4
42
2
∈∈π
π
==>
57 ••• The inner cylinder of Figure 22-42 is made of non-conducting material
and has a volume charge distribution given by ρ(R) = C/R, where C = 200 nC/m2
.
The outer cylinder is metallic, and both cylinders are infinitely long. (a) Find the
charge per unit length (that is, the linear charge density) on the inner cylinder.
(b) Calculate the electric field for all values of R.
Picture the Problem We can integrate the density function over the radius of the
inner cylinder to find the charge on it and then calculate the linear charge density
from its definition. To find the electric field for all values of r we can construct a
Gaussian surface in the shape of a cylinder of radius R and length L and apply
Gauss’s law to each region of the cable to find the electric field as a function of
the distance from its centerline.
(a) Letting the radius of the
inner cylinder be a, find the
charge on the inner
cylinder:
innerQ
( )
CLadRCL
RLdR
R
C
VdRQ
a
aa
ππ
πρ
22
2
0
00
inner
==
==
∫
∫∫
Relate this charge to the linear
charge density:
Ca
L
CLa
L
Q
π
π
λ 2
2inner
===
evaluate λ:
( )( )
nC/m8.18
m0.0150nC/m2002
=
= πλ

(b) Apply Gauss’s law to a
cylindrical surface of radius r and
length L that is concentric with the
infinitely long nonconducting
cylinder:
inside
0
S
1
QdAEn
∈
=∫ ⇒
0
inside
2
∈
π
Q
rLEn =
where we’ve neglected the end areas
because there is no flux through them.
, solve forRn EE = RE to obtain: 0
inside
2 ∈πRL
Q
ER =
Substitute to obtain, for
R < 1.50 cm: 00
cm50.1
2
2
∈∈π
π C
LR
CLR
ER ==<
evaluate En(R < 1.50 cm):
kN/C22.6
m/NC108.854
nC/m200
2212
2
cm50.1
=
⋅×
= −<RE
Express forinsideQ
1.50 cm < R < 4.50 cm:
CLaQ π2inside =
Substitute to obtain, for
1.50 cm < R < 4.50 cm: R
Ca
RL
aLC
E R
00
cm50.4cm50.1
2
2
∈
=
∈
=<<
π
π
where R = 1.50 cm.
Substitute numerical values and evaluate :cm50.4cm50.1 <<RE
( )( )
( ) Rr
E R
m/CN339
m/NC108.854
m0.0150nC/m200
2212
2
cm50.4cm50.1
⋅
=
⋅×
= −<<
Because the outer cylindrical shell
is a conductor:
0cm50.6cm50.4 =<<RE
For R > 6.50 cm, CLRQ π2inside =
and: R
ER
m/CN339
cm50.6
⋅
=>
Electric Charge and Field at Conductor Surfaces
63 •• A positive point charge of 2.5 μC is at the center of a conducting
spherical shell that has a net charge of zero, an inner radius equal to 60 cm, and an
outer radius equal to 90 cm. (a) Find the charge densities on the inner and outer
surfaces of the shell and the total charge on each surface. (b) Find the electric

Chapter 2234
field everywhere. (c) Repeat Part (a) and Part (b) with a net charge of +3.5 μC
placed on the shell.
Picture the Problem Let the inner and outer radii of the uncharged spherical
conducting shell be R1 and R2 and q represent the positive point charge at the
center of the shell. The positive point charge at the center will induce a negative
charge on the inner surface of the shell and, because the shell is uncharged, an
equal positive charge will be induced on its outer surface. To solve Part (b), we
can construct a Gaussian surface in the shape of a sphere of radius r with the
same center as the shell and apply Gauss’s law to find the electric field as a
function of the distance from this point. In Part (c) we can use a similar strategy
with the additional charge placed on the shell.
(a) Express the charge density on the
inner surface: A
qinner
inner =σ
Express the relationship between the
positive point charge q and the
charge induced on the inner surface
:innerq
0inner =+ qq ⇒ qq −=inner
Substitute for and A to obtain:innerq
2
1
inner
4 R
q
π
σ
−
=
evaluate σinner: ( )
2
2inner C/m55.0
m60.04
C5.2
μ−=
−
=
π
μ
σ
Express the charge density on the
outer surface: A
qouter
outer =σ
Because the spherical shell is
uncharged:
0innerouter =+ qq
Substitute for qouter to obtain:
2
2
inner
outer
4 R
q
π
σ
−
=
evaluate σouter: ( )
2
2outer C/m25.0
m90.04
C5.2
μ==
π
μ
σ

(b) Apply Gauss’s law to a
spherical surface of radius r that
is concentric with the point
charge:
inside
0
S
1
QdAEn
∈
=∫ ⇒
0
inside2
4
∈
π
Q
Er n =
, solve for to obtain:rn EE = rE 0
2
inside
4 ∈π r
Q
Er = (1)
For r < R1 = 60 cm, Qinside = q.
Substitute in equation (1) to
obtain:
2
0
2cm60
4 r
kq
r
q
Er ==<
∈π
Substitute numerical values and evaluate :cm60<rE
( )( ) ( ) 2
24
2
229
cm60
1
/CmN102.2
C5.2/CmN10988.8
rr
Er ⋅×=
⋅×
=<
μ
Because the spherical shell is a
conductor, a charge –q will be
induced on its inner surface. Hence,
for 60 cm < r < 90 cm:
0inside =Q
and
0cm90cm60 =<<rE
For r > 90 cm, the net charge inside
the Gaussian surface is q and:
( ) 2
24
2cm90
1
/CmN103.2
rr
kq
Er ⋅×==>
(c) Because E = 0 in the conductor: C5.2inner μ−=q
and
2
inner C/m55.0 μ−=σ as before.
Express the relationship between
the charges on the inner and
outer surfaces of the spherical
shell:
C5.3innerouter μ=+ qq
and
C0.6-C5.3 innerouter μμ == qq
σouter is now given by:
( )
2
2outer C/m59.0
m90.04
C0.6
μ==
π
μ
σ
For r < R1 = 60 cm, Qinside = q and
is as it was in (a):cm60<rE
( ) 2
24
cm60
1
/CmN103.2
r
Er ⋅×=<

Chapter 2236
Because the spherical shell is a
conductor, a charge –q will be
induced on its inner surface. Hence,
for 60 cm < r < 90 cm:
0inside =Q
and
0cm90cm60 =<<rE
For r > 0.90 m, the net charge inside the Gaussian surface is 6.0 μC and:
( )( ) ( ) 2
24
2
229
2cm90
1
/CmN104.5
1
C0.6/CmN10988.8
rrr
kq
Er ⋅×=⋅×==> μ
65 ••• [SSM] A thin square conducting sheet that has 5.00-m-long edges
has a net charge of 80.0 μC. The square is in the x = 0 plane and is centered at the
origin. (Assume the charge on each surface is uniformly distributed.) (a) Find the
charge density on each side of the sheet and find the electric field on the x axis in
the region |x| << 5.00 m. (b) A thin but infinite nonconducting sheet that has a
uniform charge density of 2.00 μC/m2
is now placed in the x = –2.50 m plane.
Find the electric field on the x axis on each side of the square sheet in the region
|x| << 2.50 m. Find the charge density on each surface of the square sheet.
Picture the Problem (a) One half of the total charge is on each side of the square
thin conducting sheet and the electric field inside the sheet is zero. The electric
field intensity just outside the surface of a conductor is given by 0∈σ=E .
Typical field points to the left and right of the square thin conducting sheet are
shown in the following diagram.
Q
left
E
r
right
E
r
x, m
0
y, m
2.50 m
−2.50 m
net
μ= 80.0 C
thin square conducting sheet
right
E
r
left
E
r
left rightσσ
σ σσσ
(b) We can use the fact that the net charge on the conducting sheet is the sum of
the charges Qleft and Qright on its left and right surfaces to obtain an equation
relating these charges. Because the resultant electric field is zero inside the sheet,
we can obtain a second equation in Qleft and Qright that we can solve
simultaneously with the first equation to find Qleft and Qright. The resultant electric
field is the superposition of three fields–the field due to the charges on the infinite
nonconducting sheet and the fields due to the charges on the surfaces of the thin
square conducting sheet. The electric field intensity due to a uniformly charged

nonconducting infinite sheet is given by 02∈σ=E . Typical field points for
each of the four regions of interest are shown in the following diagram.
σ μ 2
00.2=
leftQ rightQ
leftQE
r
rightQE
r
infinite charged nonconducting sheet
x, m
0
sheet
infiniteE
r
thin square conducting sheet
rightQE
r
sheet
infiniteE
r
leftQE
r
P PI II
rightQE
r
leftQE
r
sheet
infiniteE
r
Q = 80.0 Cnet μ
PIV
I II III IV
C/m
−2.50 m
sheet
infiniteE
r
leftQE
r
rightQE
r
PIII
Note: The vectors in this figure are drawn consistent with the charges Qleft and
Qright both being positive. If either Qleft or Qright are negative then the solution will
produce a negative value for either Qleft or Qright.
(a) Because the square sheet is a
conductor, half the charge on each
surface is half the net charge on the
sheet:
A
Qnet2
1
rightleft == σσ
evaluate leftσ and :rightσ
( )
( ) 22
2
1
rightleft
m
C
60.1
m00.5
C0.80 μμ
σσ ===
For |x| << 5.00 m, the electric field is
given by the expression for the field
just outside a conductor:
0
m00.5
∈
σ
=<<x
E
evaluate m00.5<<x
E :
kN/C181kN/C7.180
m/NC108.854
C/m60.1
2212
2
m00.5
==
⋅×
= −<<
μ
x
E
For x > 0, m00.5<<x
E is in the +x
direction and for x < 0, m00.5<<x
E is in
the −x direction.

Chapter 2238
(b) The resultant electric field in
Region II is the sum of the fields due
to the infinite nonconducting sheet
and the charge on the surfaces of the
thin square conducting sheet:
i
iii
EEEE
ˆ
2
ˆ
2
ˆ
2
ˆ
2
0
rightleft
sheet
infinite
0
right
0
left
0
sheet
infinite
sheet
infiniteII rightleft
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛ −−
=
−−=
++=
∈
σσσ
∈
σ
∈
σ
∈
σ
QQ
rrrr
Due to the presence of the infinite
nonconducting sheet, the charges on
the thin square conducting sheet are
redistributed on the left and right
surfaces. The net charge on the thin
square conducting sheet is the sum
of the charges on its left- and right-
hand surfaces:
C0.80rightleft μ=+ QQ
where we’ve assumed that and
are both positive.
leftQ
rightQ
Writing this equation in terms of the
surface charge densities yields:
( )
2
2
rightleft
rightleft
rightleft
C/m20.3
m5.00
C0.80
μ
μ
σσ
=
=
+
=
+=+
A
QQ
A
Q
A
Q
(1)
where A is the area of one side of the
thin square conducting sheet.
Because the electric field is zero
inside the thin square conducting
sheet:
0
222 0
right
0
left
0
sheet
infinite
=−+
∈
σ
∈
σ
∈
σ
or
0C/m00.2 rightleft
2
=−+ σσμ (2)
Solving equations (1) and (2)
simultaneously yields:
2
left C/m60.0 μσ =
and
2
right C/m60.2 μσ =

Substitute numerical values and evaluate IIE
r
:
iiE ˆ
C
kN
8.67ˆ
mN
C
108.8542
m
C
60.2
m
C
60.0
m
C
00.2
2
2
12
222
II ⎟
⎠
⎞
⎜
⎝
⎛
−=
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
×
−−
=
−
μμμ
r
The resultant electric field in Region
IV is the sum of the fields due to the
charge on the infinite nonconducting
sheet and the charges on the two
surfaces of the thin square
conducting sheet:
i
iii
EEEE
ˆ
2
ˆ
2
ˆ
2
ˆ
2
0
rightleft
sheet
infinite
0
right
0
left
0
sheet
infinite
sheet
infiniteIV rightleft
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛ ++
=
++=
++=
∈
σσσ
∈
σ
∈
σ
∈
σ
QQ
rrrr
Substitute numerical values and evaluate IVE
r
:
iiE ˆ
C
kN
294ˆ
mN
C
108.8542
m
C
60.2
m
C
60.0
m
C
00.2
2
2
12
222
IV ⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
×
++
=
−
μμμ
r
Substitute numerical values and evaluate IVE
r
:
iiE ˆ
C
kN
294ˆ
mN
C
108.8542
m
C
60.2
m
C
60.0
m
C
00.2
2
2
12
222
IV ⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
×
++
=
−
μμμ
r
General Problems
67 •• A large, flat, nonconducting, non-uniformly charged surface lies in the
x = 0 plane. At the origin, the surface charge density is +3.10 μC/m2
. A small
distance away from the surface on the positive x axis, the x component of the
electric field is 4.65 × 105
N/C. What is Ex a small distance away from the surface
on the negative x axis?

Chapter 2240
Picture the Problem The electric field just to the right of the large, flat,
nonconducting, nonuniformly charged surface is 02∈σ and the electric field just
to the left of the surface is 02∈σ− . We can express the electric field on both
sides of the surface in terms of E0, the electric field in the region in the absence of
the charged surface, and then eliminate E0 from these equations to obtain an
expression for Ex a small distance away from the surface on the negative x axis.
The electric field on the positive x
axis is given by: 0
00
2∈
σ
+=> EEx ⇒
0
00
2∈
σ
−= >xEE
The electric field on the negative x
axis is given by: 0
00
2∈
σ
−=< EEx
Substituting for E0 in the expression
for and simplifying gives:0<xE
0
0
00
00
22
∈
σ
∈
σ
∈
σ
−=
−−=
>
><
x
xx
E
EE
Substitute numerical values and evaluate :neg,xE
kN/C115
m/NC108.854
C/m3.10
N/C1065.4 2212
2
5
neg, =
⋅×
−×= −
μ
Ex
69 •• A thin, non-conducting, uniformly charged spherical shell of radius R
(Figure 22-44a) has a total positive charge of Q. A small circular plug is removed
from the surface. (a) What is the magnitude and direction of the electric field at
the center of the hole? (b) The plug is now put back in the hole (Figure 22-44b).
Using the result of Part (a), find the electric force acting on the plug. (c) Using the
magnitude of the force, calculate the ″electrostatic pressure″ (force/unit area) that
tends to expand the sphere.
Picture the Problem If the patch is small enough, the field at the center of the
patch comes from two contributions. We can view the field in the hole as the sum
of the field from a uniform spherical shell of charge Q plus the field due to a
small patch with surface charge density equal but opposite to that of the patch cut
out.
(a) Express the magnitude of the
electric field at the center of the hole:
hole
shell
spherical EEE +=
gaussian surface just outside the
given sphere:
( )
00
enclosed2
shell
spherical 4
∈∈
π
QQ
rE ==

Solve for to obtain:
shell
sphericalE
2
0shell
spherical
4 r
Q
E
∈π
=
The electric field due to the small
hole (small enough so that we can
treat it as a plane surface) is:
0
hole
2∈
σ−
=E
Substitute for and and
simplify to obtain:
shell
sphericalE holeE
( )
outwardradially
8
424
24
2
0
2
0
2
0
0
2
0
r
Q
r
Q
r
Q
r
Q
E
∈π
π∈∈π
∈
σ
∈π
=
−=
−
+=
(b) Express the force on the patch: qEF =
where q is the charge on the patch.
Assuming that the patch has radius
a, express the proportion between
its charge and that of the spherical
shell:
22
4 r
Q
a
q
ππ
= or Q
r
a
q 2
2
4
=
Substitute for q and E in the
expression for F to obtain:
outwardradially
32
84
4
0
22
2
0
2
2
r
aQ
r
Q
Q
r
a
F
∈π
∈π
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
(c) The pressure is the force exerted
on the patch divided by the area of
the patch: 4
0
2
2
2
4
0
22
32
32
r
Q
a
r
aQ
P
∈ππ
∈π
==
71 •• Two identical square parallel metal plates each have an area of 500 cm2
.
They are separated by 1.50 cm. They are both initially uncharged. Now a charge
of +1.50 nC is transferred from the plate on the left to the plate on the right and
the charges then establish electrostatic equilibrium. (Neglect edge effects.)
(a) What is the electric field between the plates at a distance of 0.25 cm from the
plate on the right? (b) What is the electric field between the plates a distance of
1.00 cm from the plate on the left? (c) What is the electric field just to the left of
the plate on the left? (d) What is the electric field just to the right of the plate to
the right?

Chapter 2242
Picture the Problem The transfer of charge from the plate on the left to the plate
on the right leaves the plates with equal but opposite charges. The symbols for the
four surface charge densities are shown in the figure. The x component of the
electric field due to the charge on surface 1L is ( )01L 2∈σ− at points to the left of
surface 1L and is ( )01L 2∈σ+ at points to the right of surface 1L, where the +x
direction is to the right. Similar expressions describe the electric fields due to the
other three surface charges. We can use superposition of electric fields to find the
electric field in each of the three regions.
σ σ σ σ
1L 1R 2L 2R
I II III
x
1 2
Define σ1 and σ2 so that: 1R1L1 σσσ +=
and
2R2L2 σσσ +=
(a) and (b) In the region between the
plates (region II):
0
21
0
2
0
1
0
2R
0
2L
0
1R
0
1L
II,
0
22
0
2222
∈
σσ
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
−
=−−+=
−−+=xE
Let σσσ =−= 12 . Then: σσσσσ 221 −=−−=−
Substituting for 21 σσ − and
using the definition of σ yields: 00
II,
2
∈Α∈
σ Q
Ex −=
−
=
Substitute numerical values and evaluate :II,xE
( )
leftthetowardkN/C339
m10500
mN
C
10854.8
nC50.1
26-
2
2
12
II, =
×⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
×
−=
−
xE

(c) The electric field strength just to
the left of the plate on the left (region
I) is given by:
0
22
0
22
0
2222
00
0
2
0
1
0
2R
0
2L
0
1R
0
1L
I,
=−
−
−=
−−−=
−−−−=
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
xE
(d) The electric field strength just to
the right of the plate on the right
(region III) is given by:
0
22
0
22
0
2222
00
0
2
0
1
0
2R
0
2L
0
1R
0
1L
III,
=+
−
=
−++=
+++=
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
∈
σ
xE
Remarks: If we start with the fact that free charges are only found on the
surfaces of the plates facing each other, then a much simpler solution is
possible. Any plane of charge produces a field 02∈σ perpendicular to the
plane. The field in region III directed everywhere away from the plane and
the field of the left plane is everywhere directed toward it.
73 ••• A quantum-mechanical treatment of the hydrogen atom shows that the
electron in the atom can be treated as a smeared-out distribution of negative
charge of the form ρ(r) = −ρ0e–2r/a
. Here r represents the distance from the center
of the nucleus and a represents the first Bohr radius which has a numerical value
of 0.0529 nm. Recall that the nucleus of a hydrogen atom consists of just one
proton and treat this proton as a positive point charge. (a) Calculate ρ0, using the
fact that the atom is neutral. (b) Calculate the electric field at any distance r from
the nucleus.
Picture the Problem Because the atom is uncharged, we know that the integral of
the electron’s charge distribution over all of space must equal its charge qe.
Evaluation of this integral will lead to an expression for ρ0. In (b) we can express
the resultant electric field at any point as the sum of the electric fields due to the
proton and the electron cloud.
(a) Because the atom is uncharged,
the integral of the electron’s charge
distribution over all of space must
equal its charge e:
( ) ( )∫∫
∞∞
==
0
2
0
4 drrrdVre πρρ

Chapter 2244
Substitute for ρ(r) and simplify to
obtain:
∫
∫
∞
−
∞
−
−=
−=
0
22
0
0
22
0
4
4
drer
drree
ar
ar
πρ
πρ
Use integral tables or integration by
parts to obtain:
4
3
0
22 a
drer ar
=∫
∞
−
Substitute for ∫
∞
−
0
22
drer ar
to obtain: 0
3
3
0
4
4 ρππρ a
a
e −=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=
Solving for 0ρ yields:
30
a
e
π
ρ −=
(b) The field will be the sum of the
field due to the proton and that of
the electron charge cloud:
cloudp EEE +=
Express the field due to the electron
cloud: ( ) ( )
2cloud
r
rkQ
rE =
where Q(r) is the net negative charge
enclosed a distance r from the proton.
Substitute for and to obtain:pE cloudE
( ) ( )
22
r
rkQ
r
ke
rE += (1)
Q(r) is given by:
''4
')'('4)(
'2
0
0
2
0
2
drer
drrrrQ
ar
r
r
−
∫
∫
=
=
ρπ
ρπ
From Part (a), 30
a
e
π
ρ
−
= :
''
4
''4)(
'2
0
2
3
'2
0
2
3
drer
a
e
drer
a
e
rQ
ar
r
ar
r
−
−
∫
∫
−
=
⎟
⎠
⎞
⎜
⎝
⎛ −
=
π
π

From a table of integrals:
( )[ ]
( )
( ) ⎥
⎦
⎤
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−−=
⎥
⎦
⎤
⎢
⎣
⎡
−−−=
−−−=
−−
−−
−−−
∫
2
2
22
3
2
2
232
4
1
2222
4
12
0
2
21
4
221
221
a
r
a
r
ee
a
a
r
a
r
eae
raraeaedxex
arar
arar
ararax
r
Substituting for '' '2
0
2
drer ar
r
−
∫ in the expression for and simplifying yields:)(rQ
( ) ⎥
⎦
⎤
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−−
−
= −−
2
2
22
21
4
)(
a
r
a
r
ee
e
rQ arar
Substitute for in equation (1) and simplify to obtain:)(rQ
( ) ( )
( ) ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎦
⎤
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−−−=
⎥
⎦
⎤
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−−−=
−−
−−
2
2
22
2
2
2
22
22
21
4
1
1
21
4
a
r
a
r
ee
r
ke
a
r
a
r
ee
r
ke
r
ke
rE
arar
arar
79 •• A uniformly charged, infinitely long line of negative charge has a
linear charge density of –λ and is located on the z axis. A small positively charged
particle that has a mass m and a charge q is in a circular orbit of radius R in the xy
plane centered on the line of charge. (a) Derive an expression for the speed of the
particle. (b) Obtain an expression for the period of the particle’s orbit.
Picture the Problem (a) We can apply
Newton’s second law to the particle to
express its speed as a function of its
mass m, charge q, and the radius of its
path R, and the strength of the electric
field due to the infinite line charge E.
(b) The period of the particle’s motion
is the ratio of the circumference of the
circle in which it travels divided by its
orbital speed. x
y
z
λ
R
−
qm,

Chapter 2246
(a) Apply Newton’s second law
to the particle to obtain: ∑ ==
R
v
mqEF
2
radial
where the inward direction is positive.
Solving for v yields:
m
qRE
v =
The strength of the electric field at a
distance R from the infinite line
charge is given by:
R
k
E
λ2
=
Substitute for E and simplify to
obtain: m
kq
v
λ2
=
(b) The speed of the particle is equal
to the circumference of its orbit
divided by its period:
T
R
v
π2
= ⇒
v
R
T
π2
=
Substitute for v and simplify to
obtain: λ
π
kq
m
RT
2
=
81 •• The charges Q and q of Problem 80 are +5.00 μC and –5.00 μC,
respectively, and the radius of the ring is 8.00 cm. When the particle is given a
small displacement in the x direction, it oscillates about its equilibrium position at
a frequency of 3.34 Hz. (a) What is the particle’s mass? (b) What is the
frequency if the radius of the ring is doubled to 16.0 cm and all other parameters
remain unchanged?
Picture the Problem Starting with the equation for the electric field on the axis
of a ring charge, we can factor the denominator of the expression to show that, for
x << a, Ex is proportional to x. We can use Fx = qEx to express the force acting on
the particle and apply Newton’s second law to show that, for small displacements
from equilibrium, the particle will execute simple harmonic motion. Finally, we
can find the angular frequency of the motion from the differential equation and
use this expression to find the frequency of the motion when the radius of the ring
is doubled and all other parameters remain unchanged.
(a) Express the electric field on the
axis of the ring of charge: ( ) 2322
ax
kQx
Ex
+
=

Factor a2
from the denominator of Ex
to obtain:
x
a
kQ
a
x
a
kQx
a
x
a
kQx
Ex
323
2
2
3
23
2
2
2
1
1
≈
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
=
⎥
⎦
⎤
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
=
provided x << a.
Express the force acting on the
particle as a function of its charge
and the electric field:
x
a
kqQ
qEF xx 3
−=−=
Because the negatively charged
particle experiences a linear
restoring force, we know that its
motion will be simple harmonic.
Apply Newton’s second law to the
negatively charged particle to
obtain:
x
a
kqQ
dt
xd
m 32
2
−=
or
032
2
=+ x
ma
kqQ
dt
xd
the differential equation of simple
harmonic motion.
The angular frequency of the simple
harmonic motion of the particle is
given by:
3
ma
kqQ
=ω (1)
Solving for m yields:
32232
4 af
kqQ
a
kqQ
m
πω
==
Substitute numerical values and evaluate m:
( )( )
( ) ( )
kg997.0
cm00.8s34.34
C00.5C00.5
C
mN
10988.8
3212
2
2
9
=
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
×
=
−
π
μμ
m
(b) Express the angular frequency of
the motion if the radius of the ring is
doubled:
( )3
2
'
am
kqQ
=ω (2)

Chapter 2248
Divide equation (2) by equation (1)
to obtain: ( )
8
12
2
2
3
3
===
ma
kqQ
am
kqQ
f
f''
π
π
ω
ω
Solve for f ′ to obtain:
Hz18.1
8
Hz3.34
8
===
f
'f
87 ••• Consider a simple but surprisingly accurate model for the hydrogen
molecule: two positive point charges, each having charge +e, are placed inside a
uniformly charged sphere of radius R, which has a charge equal to –2e. The two
point charges are placed symmetrically, equidistant from the center of the sphere
(Figure 22-48). Find the distance from the center, a, where the net force on either
point charge is zero.
Picture the Problem We can find the distance from the center where the net
force on either charge is zero by setting the sum of the forces acting on either
point charge equal to zero. Each point charge experiences two forces; one a
Coulomb force of repulsion due to the other point charge, and the second due to
that fraction of the sphere’s charge that is between the point charge and the center
of the sphere that creates an electric field at the location of the point charge.
Apply to either of the
point charges:
0=∑F 0fieldCoulomb =− FF (1)
Express the Coulomb force on the
proton:
( ) 2
2
2
2
Coulomb
42 a
ke
a
ke
F ==
The force exerted by the field E is: eEF =field
surface of radius a centered at the
origin:
( )
0
enclosed2
4
∈
π
Q
aE =
Relate the charge density of the
electron sphere to :enclosedQ 3
3
4
enclosed
3
3
4
2
a
Q
R
e
ππ
= ⇒ 3
3
enclosed
2
R
ea
Q =
Substitute for :enclosedQ
( ) 3
0
3
2 2
4
R
ea
aE
∈
π =
Solve for E to obtain:
3
02 R
ea
E
∈π
= ⇒ 3
0
2
field
2 R
ae
F
∈π
=

Substitute for and in
equation (1):
CoulombF fieldF
0
24 3
0
2
2
2
=−
R
ae
a
ke
∈π
or
0
2
4 3
2
2
2
=−
R
ake
a
ke
⇒ RRa 2
13
8
1
==

Ch22 ssm

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