Preview Extract
CHAPTER 2
2.1. Three point charges are positioned in the xy plane as follows: 5nC at y = 5 cm, 10 nC at y = 5 cm,
15 nC at x = 5 cm. Find the required xy coordinates of a 20nC fourth charge that will produce a
zero electric field at the origin.
With the charges thus configured, the electric field at the origin will be the superposition of the
individual charge fields:
๏ฃฟ
1
15
E0 =
ax
4โกโ0 (5)2
5
ay
(5)2
10
ay
(5)2
1
=
4โกโ0
โ โ
3
[ax
5
ay ]
nC/m
The field, E20 , associated with the 20nC charge (evaluated at the origin) must exactly cancel
this field, so we write:
1
E20 =
4โกโ0
โ โ
3
[ax
5
20
ay ] =
4โกโ0 โข2
โ
1
p
2
โ
[ax
ay ]
q
p
100/(3 2) = 4.85. The x and
p
y coordinates of the 20nC charge will both be equal in magnitude to 4.85/ 2 = 3.43. The
coodinates of the 20nC charge are then (3.43, 3.43).
From this, we identify the distance from the origin: โข =
2.2. Point charges of 1nC and 2nC are located at (0,0,0) and (1,1,1), respectively, in free space. Determine
the vector force acting on each charge.
First, the electric field intensity associated with the 1nC charge, evalutated at the 2nC charge
location is:
โ
โ
1
1
p
E12 =
(ax + ay + az ) nC/m
4โกโ0 (3)
3
p
in which the distance between charges is 3 m. The force on the 2nC charge is then
F12 = q2 E12 =
2
1
p
(ax + ay + az ) =
(ax + ay + az )
10.4
โกโ0
12 3 โกโ0
The force on the 1nC charge at the origin is just the opposite of this result, or
F21 =
+1
(ax + ay + az )
10.4 โกโ0
12
nN
nN
2.3. Point charges of 50nC each are located at A(1, 0, 0), B( 1, 0, 0), C(0, 1, 0), and D(0, 1, 0) in free
space. Find the total force on the charge at A.
The force will be:
๏ฃฟ
(50 โฅ 10 9 )2 RCA
RDA
RBA
F=
+
+
3
3
4โกโ0
RCA 
RDA 
RBA 3
where RCA = ax ay , RDA = ax + ay , and RBA = 2ax . The magnitudes are RCA  = RDA  =
and RBA  = 2. Substituting these leads to
๏ฃฟ
(50 โฅ 10 9 )2
1
1
2
p + p +
F=
ax = 21.5ax ยตN
4โกโ0
2 2 2 2 8
p
2,
where distances are in meters.
2.4. Eight identical point charges of Q C each are located at the corners of a cube of side length a, with
one charge at the origin, and with the three nearest charges at (a, 0, 0), (0, a, 0), and (0, 0, a). Find
an expression for the total vector force on the charge at P (a, a, a), assuming free space:
The total electric field at P (a, a, a) that produces a force on the charge there will be the sum
of the fields from the other seven charges. This is written below, where the charge locations
associated with each term are indicated:
2
3
Enet (a, a, a) =
6a + a + a
7
q
ay + az ax + az ax + ay
6 x
7
y
z
p
p
p
p
+
+
+
+
a
+
a
+
a
6
x
y
z 7
{z}
{z}
4โกโ0 a2 4
5
{z}
3 3
2 2
2 2
2 2

{z
}  {z }  {z }  {z } (0,a,a) (a,0,a) (a,a,0)
(0,0,0)
(a,0,0)
(0,a,0)
(0,0,a)
The force is now the product of this field and the charge at (a, a, a). Simplifying, we obtain
๏ฃฟ
q2
1
1
1.90 q 2
p
p
F(a, a, a) = qEnet (a, a, a) =
+
+
1
(a
+
a
+
a
)
=
(ax + ay + az )
x
y
z
4โกโ0 a2 3 3
4โกโ0 a2
2
in which the magnitude is F = 3.29 q 2 /(4โกโ0 a2 ).
2.5. Let a point charge Q1 = 25 nC be located at P1 (4, 2, 7) and a charge Q2 = 60 nC be at P2 ( 3, 4, 2).
a) If โ = โ0 , find E at P3 (1, 2, 3): This field will be
๏ฃฟ
10 9 25R13
60R23
E=
+
3
4โกโ0 R13 
R23 3
where R13 =
So
p
p
4az and R23 = 4ax 2ay + 5az . Also, R13  = 41 and R23  = 45.
๏ฃฟ
10 9 25 โฅ ( 3ax + 4ay 4az ) 60 โฅ (4ax 2ay + 5az )
E=
+
4โกโ0
(41)1.5
(45)1.5
3ax + 4ay
= 4.58ax
0.15ay + 5.51az
b) At what point on the y axis is Ex = 0? P3 is now p
at (0, y, 0), so R13 = 4ax + p
(y + 2)ay
2
and R23 = 3ax + (y 4)ay + 2az . Also, R13  = 65 + (y + 2) and R23  = 13 + (y
Now the x component of E at the new P3 will be:
๏ฃฟ
10 9
25 โฅ ( 4)
60 โฅ 3
Ex =
+
2
1.5
4โกโ0 [65 + (y + 2) ]
[13 + (y 4)2 ]1.5
13
7az
4)2 .
2.5b (continued) To obtain Ex = 0, we require the expression in the large brackets to be zero. This
expression simplifies to the following quadratic:
0.48y 2 + 13.92y + 73.10 = 0
which yields the two values: y =
6.89, 22.11
2.6. Two point charges of equal magnitude q are positioned at z = ยฑd/2.
a) find the electric field everywhere on the z axis: For a point charge at any location, we have
E=
q(r r0 )
4โกโ0 r r0 3
In the case of two charges, we would therefore have
ET =
q1 (r r01 )
q2 (r r02 )
+
0
4โกโ0 r r1 3
4โกโ0 r r02 3
(1)
In the present case, we assign q1 = q2 = q, the observation point position vector as r = zaz , and
the charge position vectors as r01 = (d/2)az , and r02 = (d/2)az Therefore
r01 = [z
(d/2)]az ,
r
r1 3 = [z
(d/2)]3
and r
r
then
r
r02 = [z + (d/2)]az ,
r2 3 = [z + (d/2)]3
Substitute these results into (1) to obtain:
๏ฃฟ
q
1
1
ET (z) =
+
2
4โกโ0 [z (d/2)]
[z + (d/2)]2
az
V/m
(2)
b) find the electric field everywhere on the x axis: We proceed as in part a, except that now r = xax .
Eq. (1) becomes
๏ฃฟ
q
xax (d/2)az
xax + (d/2)az
ET (x) =
+
(3)
4โกโ0 xax (d/2)az 3
xax + (d/2)az 3
where
xax
Therefore (3) becomes
โฅ
โค1/2
(d/2)az  = xax + (d/2)az  = x2 + (d/2)2
ET (x) =
4โกโ0
2qx ax
3/2
2
[x + (d/2)2 ]
c) repeat parts a and b if the charge at z = d/2 is q instead of +q: The field along the z axis is
quickly found by changing the sign of the second term in (2):
๏ฃฟ
q
1
1
ET (z) =
az V/m
4โกโ0 [z (d/2)]2
[z + (d/2)]2
In like manner, the field along the x axis is found from (3) by again changing the sign of the
second term. The result is
2qd az
3/2
4โกโ0 [x2 + (d/2)2 ]
14
2.7. A 2 ยตC point charge is located at A(4, 3, 5) in free space. Find Eโข , E , and Ez at P (8, 12, 2). Have
EP =
๏ฃฟ
2 โฅ 10 6 RAP
2 โฅ 10 6 4ax + 9ay 3az
=
= 65.9ax + 148.3ay
4โกโ0 RAP 3
4โกโ0
(106)1.5
Then, at point P , โข =
p
82 + 122 = 14.4,
49.4az
= tan 1 (12/8) = 56.3 , and z = z. Now,
Eโข = Ep ยท aโข = 65.9(ax ยท aโข ) + 148.3(ay ยท aโข ) = 65.9 cos(56.3 ) + 148.3 sin(56.3 ) = 159.7
and
E = Ep ยท a = 65.9(ax ยท a ) + 148.3(ay ยท a ) =
Finally, Ez =
65.9 sin(56.3 ) + 148.3 cos(56.3 ) = 27.4
49.4 V/m
2.8. A crude device for measuring charge consists of two small insulating spheres of radius a, one of which
is fixed in position. The other is movable along the x axis, and is subject to a restraining force kx,
where k is a spring constant. The uncharged spheres are centered at x = 0 and x = d, the latter
fixed. If the spheres are given equal and opposite charges of Q coulombs:
a) Obtain the expression by which Q may be found as a function of x: The spheres will attract, and
so the movable sphere at x = 0 will move toward the other until the spring and Coulomb forces
balance. This will occur at location x for the movable sphere. With equal and opposite forces,
we have
Q2
= kx
4โกโ0 (d x)2
p
from which Q = 2(d x) โกโ0 kx.
b) Determine the maximum charge that can be measured in terms of โ0 , k, and d, and state the
separation of the spheres then: With increasing charge, the spheres move toward each other until
they just touch at xmax = d 2a. Using the part a result, we find the maximum measurable
p
charge: Qmax = 4a โกโ0 k(d 2a). Presumably some form of stop mechanism is placed at
x = xmax to prevent the spheres from actually touching.
c) What happens if a larger charge is applied? No further motion is possible, so nothing happens.
2.9. A 100 nC point charge is located at A( 1, 1, 3) in free space.
a) Find the locus of all points P (x, y, z) at which Ex = 500 V/m: The total field at P will be:
EP =
100 โฅ 10 9 RAP
4โกโ0
RAP 3
where RAP = (x+1)ax +(y 1)ay +(z 3)az , and where RAP  = [(x+1)2 +(y 1)2 +(z 3)2 ]1/2 .
The x component of the field will be
๏ฃฟ
100 โฅ 10 9
(x + 1)
Ex =
= 500 V/m
2
4โกโ0
[(x + 1) + (y 1)2 + (z 3)2 ]1.5
And so our condition becomes:
(x + 1) = 0.56 [(x + 1)2 + (y
15
1)2 + (z
3)2 ]1.5
2.9b) Find y1 if P ( 2, y1 , 3) lies on that locus: At point P , the condition of part a becomes
โฅ
โค3
3.19 = 1 + (y1 1)2
from which (y1
1)2 = 0.47, or y1 = 1.69 or 0.31
2.10. A charge of 1 nC is located at the origin in free space. What charge must be located at (2,0,0) to
cause Ex to be zero at (3,1,1)?
The field from two point charges is given generally by
ET =
q1 (r r01 )
q2 (r r02 )
+
4โกโ0 r r01 3
4โกโ0 r r02 3
(1)
where we let q1 = 1nC and q2 is to be found. With q1 at the origin, r01 = 0. The position
vector for q2 is then r02 = 2ax . The observation point at (3,1,1) gives r = 3ax + ay + az . Eq. (1)
becomes
๏ฃฟ
1
1(3ax + ay + az ) q2 [(3 2)ax + ay + az ]
+
4โกโ0
(32 + 1 + 1)3/2
(1 + 1 + 1)3/2
Requiring the x component to be zero leads to
q2 =
35/2
= 0.43 nC
113/2
2.11. A charge Q0 located at the origin in free space produces a field for which Ez = 1 kV/m at point
P ( 2, 1, 1).
a) Find Q0 : The field at P will be
Q0
EP =
4โกโ0
๏ฃฟ
2ax + ay
61.5
Since the z component is of value 1 kV/m, we find Q0 =
az
4โกโ0 61.5 โฅ 103 =
1.63 ยตC.
b) Find E at M (1, 6, 5) in cartesian coordinates: This field will be:
๏ฃฟ
1.63 โฅ 10 6 ax + 6ay + 5az
EM =
4โกโ0
[1 + 36 + 25]1.5
or EM =
30.11ax
180.63ay
150.53az .
c) Find E at M (1, 6, 5) in cylindrical coordinates: At M , โข =
80.54 , and z = 5. Now
Eโข = EM ยท aโข =
so that EM =
E = EM ยท a =
183.12aโข
30.11 cos
30.11( sin )
p
1 + 36 = 6.08,
180.63 sin
180.63 cos
=
= tan 1 (6/1) =
183.12
= 0 (as expected)
150.53az .
p
d) Find E at M (1, 6, 5) in spherical coordinates: At M , r = 1 + 36 + 25 = 7.87, = 80.54 (as
before), and โ = cos 1 (5/7.87) = 50.58 . Now, since the charge is at the origin, we expect to
obtain only a radial component of EM . This will be:
Er = EM ยท ar =
30.11 sin โ cos
180.63 sin โ sin
16
150.53 cos โ =
237.1
2.12. Electrons are in random motion in a fixed region in space. During any 1ยตs interval, the probability
of finding an electron in a subregion of volume 10 15 m2 is 0.27. What volume charge density,
appropriate for such time durations, should be assigned to that subregion?
The finite probabilty eโตectively reduces the net charge quantity by the probability fraction. With
e = 1.602 โฅ 10 19 C, the density becomes
0.27 โฅ 1.602 โฅ 10 19
=
10 15
โขv =
43.3 ยตC/m3
2.13. A uniform volume charge density of 0.2 ยตC/m3 is present throughout the spherical shell extending
from r = 3 cm to r = 5 cm. If โขv = 0 elsewhere:
a) find the total charge present throughout the shell: This will be
Q=
๏ฃฟ
.05
r3
0.2 r sin โ dr dโ d = 4โก(0.2)
= 8.21 โฅ 10 5 ยตC = 82.1 pC
3 .03
.03
Z 2โก Z โก Z .05
0
0
2
b) find r1 if half the total charge is located in the region 3 cm < r < r1 : If the integral over r in
part a is taken to r1 , we would obtain
๏ฃฟ
r
r3 1
4โก(0.2)
= 4.105 โฅ 10 5
3 .03
Thus
r1 =
๏ฃฟ
3 โฅ 4.105 โฅ 10 5
+ (.03)3
0.2 โฅ 4โก
1/3
= 4.24 cm
2.14. The electron beam in a certain cathode ray tube possesses cylindrical symmetry, and the charge
density is represented by โขv = 0.1/(โข2 + 10 8 ) pC/m3 for 0 < โข 3 โฅ 10 4 m.
a) Find the total charge per meter along the length of the beam: We integrate the charge density
over the cylindrical volume having radius 3 โฅ 10 4 m, and length 1m.
q=
Z 1 Z 2โก Z 3โฅ10 4
0
0
0
0.1
(โข2 + 10 8 )
โข dโข d dz
From integral tables, this evaluates as
q=
โ โ
1
0.2โก
ln โข2 + 10 8
2
3โฅ10
0
4
= 0.1โก ln(10) =
0.23โก pC/m
b) if the electron velocity is 5 โฅ 107 m/s, and with one ampere defined as 1C/s, find the beam
current:
Current = charge/m โฅ v =
0.23โก [pC/m] โฅ 5 โฅ 107 [m/s] =
17
11.5โก โฅ 106 [pC/s] =
11.5โก ยตA
2.15. A spherical volume having a 2 ยตm radius contains a uniform volume charge density of 105 C/m3 (not
1015 as stated in earlier printings).
a) What total charge is enclosed in the spherical volume?
This will be Q = (4/3)โก(2 โฅ 10 6 )3 โฅ 105 = 3.35 โฅ 10 12 C.
b) Now assume that a large region contains one of these little spheres at every corner of a cubical
grid 3mm on a side, and that there is no charge between spheres. What is the average volume
charge density throughout this large region? Each cube will contain the equivalent of one little
sphere. Neglecting the little sphere volume, the average density becomes
โขv,avg =
3.35 โฅ 10 12
= 1.24 โฅ 10 4 C/m3
(0.003)3
2.16. Within a region of free space, charge density is given as โขv = (โข0 r/a) cos โ C/m3 , where โข0 and a are
constants. Find the total charge lying within:
a) the sphere, r ๏ฃฟ a: This will be
Qa =
Z 2โก Z โก Z a
0
0
0
โข0 r
cos โ r2 sin โ dr dโ d = 2โก
a
Z a
0
โข0 r3
dr = 0
a
b) the cone, r ๏ฃฟ a, 0 ๏ฃฟ โ ๏ฃฟ 0.1โก:
Qb =
Z 2โก Z 0.1โก Z a
0
0
0
โข0 r
โข0 a3 โฅ
cos โ r2 sin โ dr dโ d = โก
1
a
4
c) the region, r ๏ฃฟ a, 0 ๏ฃฟ โ ๏ฃฟ 0.1โก, 0 ๏ฃฟ
Qc =
Z 0.2โก Z 0.1โก Z a
0
0
0
โค
cos2 (0.1โก) = 0.024โกโข0 a3
๏ฃฟ 0.2โก.
โข0 r
cos โ r2 sin โ dr dโ d = 0.024โกโข0 a3
a
โ
0.2โก
2โก
2.17. A uniform line charge of 16 nC/m is located along the line defined by y =
a) Find E at P (1, 2, 3): This will be
EP =
where RP = (1, 2, 3)
โ
= 0.0024โกโข0 a3
2, z = 5. If โ = โ0 :
โขl
RP
2โกโ0 RP 2
(1, 2, 5) = (0, 4, 2), and RP 2 = 20. So
EP =
๏ฃฟ
16 โฅ 10 9 4ay 2az
= 57.5ay
2โกโ0
20
28.8az V/m
b) Find E at that point in the z = 0 plane where the direction of E is given by (1/3)ay
With z = 0, the general field will be
๏ฃฟ
โขl
(y + 2)ay 5az
Ez=0 =
2โกโ0
(y + 2)2 + 25
18
(2/3)az :
We require Ez  =
2Ey , so 2(y + 2) = 5. Thus y = 1/2, and the field becomes:
๏ฃฟ
โขl
2.5ay 5az
Ez=0 =
= 23ay
2โกโ0 (2.5)2 + 25
46az
2.18. a) Find E in the plane z = 0 that is produced by a uniform line charge, โขL , extending along the z
axis over the range L < z < L in a cylindrical coordinate system: We find E through
E=
Z L
โขL dz(r r0 )
r0 3
L 4โกโ0 r
where the observation point position vector is r = โขaโข (anywhere in the xy plane), and where the
position vector that locates any diโตerential charge element on the z axis is r0 = zaz . So r r0 =
โขaโข zaz , and r r0  = (โข2 + z 2 )1/2 . These relations are substituted into the integral to yield:
E=
Z L
โขL dz(โขaโข zaz )
โขL โข aโข
=
2
2
3/2
4โกโ0
L 4โกโ0 (โข + z )
Z L
L
dz
(โข2 + z 2 )3/2
= Eโข aโข
Note that the second term in the lefthand integral (involving zaz ) has eโตectively vanished because it
produces equal and opposite sign contributions when the integral is taken over symmetric limits (odd
parity). Evaluating the integral results in
Eโข =
L
โขL โข
z
โขL
L
โขL
1
p
p
p
=
=
4โกโ0 โข2 โข2 + z 2 L 2โกโ0 โข โข2 + L2
2โกโ0 โข 1 + (โข/L)2
Note that as L ! 1, the expression reduces to the expected field of the infinite line charge in free
space, โขL /(2โกโ0 โข).
b) if the finite line charge is approximated by an infinite line charge (L ! 1), by what percentage
is Eโข in error if โข = 0.5L? The percent error in this situation will be
"
% error = 1
For โข = 0.5L, this becomes % error = 10.6 %
#
1
p
โฅ 100
1 + (โข/L)2
c) repeat b with โข = 0.1L. For this value, obtain % error = 0.496 %.
2.19. A uniform line charge of 2 ยตC/m is located on the z axis. Find E in rectangular coordinates at
P (1, 2, 3) if the charge extends from
a) 1 < z < 1: With the infinite line, we know that the field will have only a radial component
in cylindrical coordinates (or x and y components in cartesian). The field from an infinite line
on the z axis is generally E = [โขl /(2โกโ0 โข)]aโข . Therefore, at point P :
EP =
โขl RzP
(2 โฅ 10 6 ) ax + 2ay
=
= 7.2ax + 14.4ay kV/m
2
2โกโ0 RzP 
2โกโ0
5
where RzP is the vector that extends from the line charge to point P , and is perpendicular to
the z axis; i.e., RzP = (1, 2, 3) (0, 0, 3) = (1, 2, 0).
19
b)
4 ๏ฃฟ z ๏ฃฟ 4: Here we use the general relation
EP =
Z
โขl dz r r0
4โกโ0 r r0 3
2.19b (continued) where r = ax + 2ay + 3az and r0 = zaz . So the integral becomes
Z 4
(2 โฅ 10 6 )
EP =
4โกโ0
ax + 2ay + (3 z)az
dz
[5 + (3 z)2 ]1.5
4
Using integral tables, we obtain:
EP = 3597
๏ฃฟ
4
(ax + 2ay )(z 3) + 5az
(z 2 6z + 14)
V/m = 4.9ax + 9.8ay + 4.9az kV/m
4
The student is invited to verify that when evaluating the above expression over the limits 1 < z <
1, the z component vanishes and the x and y components become those found in part a.
2.20. A line charge of uniform charge density โข0 C/m and of length `, is oriented along the z axis at
`/2 < z < `/2.
a) Find the electric field strength, E, in magnitude and direction at any position along the x axis:
This follows the method in Problem 2.18. We find E through
E=
Z `/2
โข0 dz(r r0 )
r0 3
`/2 4โกโ0 r
where the observation point position vector is r = xax (anywhere on the x axis), and where
the position vector that locates any diโตerential charge element on the z axis is r0 = zaz . So
r r0 = xax zaz , and r r0  = (x2 + z 2 )1/2 . These relations are substituted into the integral
to yield:
Z `/2
Z
โข0 dz(xax zaz )
โข0 x ax `/2
dz
E=
=
= Ex ax
2 + z 2 )3/2
2 + z 2 )3/2
4โกโ
4โกโ
(x
(x
0
0
`/2
`/2
Note that the second term in the lefthand integral (involving zaz ) has eโตectively vanished because
it produces equal and opposite sign contributions when the integral is taken over symmetric limits
(odd parity). Evaluating the integral results in
Ex =
`/2
โข0 x
z
โข0
`/2
โข0
1
p
p
p
=
=
2
2
2
2
2
4โกโ0 x x + z
2โกโ0 x x + (`/2)
2โกโ0 x 1 + (2x/`)2
`/2
b) with the given line charge in position, find the force acting on an identical line charge that is
oriented along the x axis at `/2 < x < 3`/2: The diโตerential force on an element of the xdirected
line charge will be dF = dqE = (โข0 dx)E, where E is the field as determined in part a. The net
force is then the integral of the diโตerential force over the length of the horizontal line charge, or
F=
Z 3`/2
`/2
โข20
1
p
dx ax
2โกโ0 x 1 + (2x/`)2
20
This can be rewritten and then evaluated using integral tables as
0
"
#3`/2 1
p
Z 3`/2
2
2
2
2
`/2 + x + (`/2)
โข ` ax
dx
โข0 ` ax @ 1
A
p
F= 0
=
ln
2
2
4โกโ0 `/2 x x + (`/2)
4โกโ0
(`/2)
x
`/2
"
#
"
#
p
p
(`/2) 1 + 10
โข20 ax
โข2 ax
3(1 + 2)
0.55โข20
p
p
=
ln
= 0
ln
=
ax N
2โกโ0
2โกโ0
2โกโ0
3(`/2) 1 + 2
1 + 10
2.21. Two identical uniform line charges with โขl = 75 nC/m are located in free space at x = 0, y = ยฑ0.4
m. What force per unit length does each line charge exert on the other? The charges are parallel to
the z axis and are separated by 0.8 m. Thus the field from the charge at y = 0.4 evaluated at the
location of the charge at y = +0.4 will be E = [โขl /(2โกโ0 (0.8))]ay . The force on a diโตerential length
of the line at the positive y location is dF = dqE = โขl dzE. Thus the force per unit length acting on
the line at postive y arising from the charge at negative y is
F=
Z 1
0
โข2l dz
ay = 1.26 โฅ 10 4 ay N/m = 126 ay ยตN/m
2โกโ0 (0.8)
The force on the line at negative y is of course the same, but with
ay .
2.22. Two identical uniform sheet charges with โขs = 100 nC/m2 are located in free space at z = ยฑ2.0 cm.
What force per unit area does each sheet exert on the other?
The field from the top sheet is E = โขs /(2โ0 ) az V/m. The diโตerential force produced by this
field on the bottom sheet is the charge density on the bottom sheet times the diโตerential area
there, multiplied by the electric field from the top sheet: dF = โขs daE. The force per unit area is
then just F = โขs E = (100 โฅ 10 9 )( 100 โฅ 10 9 )/(2โ0 ) az = 5.6 โฅ 10 4 az N/m2 .
2.23. Given the surface charge density, โขs = 2 ยตC/m2 , in the region โข > .04. Continuing:
. โขs โฅ
Ez =
1
2โ0
[1
โค 0.04โขs
โก(0.2)2 โขs
(1/2)(0.04)/z 2 ] =
=
4โ0 z 2
4โกโ0 z 2
This the point charge field, where we identify q = โก(0.2)2 โขs as the total charge on the disk (which
now looks like a point).
22
2.24. a) Find the electric field on the z axis produced by an annular ring of uniform surface charge density
โขs in free space. The ring occupies the region z = 0, a ๏ฃฟ โข ๏ฃฟ b, 0 ๏ฃฟ ๏ฃฟ 2โก in cylindrical coordinates:
We find the field through
Z Z
โขs da(r r0 )
E=
4โกโ0 r r0 3
where the integral is taken over the surface of the annular ring, and where r = zaz and r0 = โขaโข . The
integral then becomes
Z 2โก Z b
โขs โข dโข d (zaz โขaโข )
E=
4โกโ0 (z 2 + โข2 )3/2
0
a
In evaluating this integral, we first note that the term involving โขaโข integrates to zero over the
integration range of 0 to 2โก. This is because we need to introduce the dependence in aโข by writing
it as aโข = cos ax + sin ay , where ax and ay are invariant in their orientation as varies. So the
integral now simplifies to
”
#b
Z
2โกโขs z az b
โข dโข
โขs z az
1
p
E=
=
2
2 3/2
4โกโ0
2โ0
z 2 + โข2 a
a (z + โข )
”
#
โขs
1
1
p
p
=
az
2โ0
1 + (a/z)2
1 + (b/z)2
b) from your part a result, obtain the field of an infinite uniform sheet charge by taking appropriate
limits. The infinite sheet is obtained by letting a ! 0 and b ! 1, in which case E ! โขs /(2โ0 ) az
as expected.
2.25. Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC
at P (2, 0, 6); uniform line charge density, 3nC/m at x = 2, y = 3; uniform surface charge density,
0.2 nC/m2 at x = 2. The sum of the fields at the origin from each charge in order is:
๏ฃฟ
(12 โฅ 10 9 ) ( 2ax 6az )
(3 โฅ 10 9 ) (2ax 3ay )
E=
+
4โกโ0
(4 + 36)1.5
2โกโ0
(4 + 9)
=
๏ฃฟ
3.9ax
12.4ay
2.5az V/m
23
๏ฃฟ
(0.2 โฅ 10 9 )ax
2โ0
2.26. Radiallydependent surface charge is distributed on an infinite flat sheet in the xy plane, and is characterized in cylindrical coordinates by surface density โขs = โข0 /โข, where โข0 is a constant. Determine
the electric field strength, E, everywhere on the z axis.
We find the field through
E=
Z Z
โขs da(r r0 )
4โกโ0 r r0 3
where the integral is taken over the surface of the annular ring, and where r = zaz and r0 = โขaโข .
The integral then becomes
Z 2โก Z 1
(โข0 /โข) โข dโข d (zaz โขaโข )
E=
4โกโ0 (z 2 + โข2 )3/2
0
0
In evaluating this integral, we first note that the term involving โขaโข integrates to zero over the
integration range of 0 to 2โก. This is because we need to introduce the dependence in aโข
by writing it as aโข = cos ax + sin ay , where ax and ay are invariant in their orientation as
varies. So the integral now simplifies to
”
#1
Z
2โกโขs z az 1
dโข
โขs z az
โข
โขs
p
E=
=
=
az
2
2
3/2
4โกโ0
2โ0
2โ0 z
(z + โข )
z 2 z 2 + โข2
0
โข=0
2.27. Given the electric field E = (4x 2y)ax (2x + 4y)ay , find:
a) the equation of the streamline that passes through the point P (2, 3, 4): We write
dy
Ey
(2x + 4y)
=
=
dx
Ex
(4x 2y)
Thus
2(x dy + y dx) = y dy
or
2 d(xy) =
1
d(y 2 )
2
So
C1 + 2xy =
or
y2
1 2
y
2
x dx
1
d(x2 )
2
1 2
x
2
x2 = 4xy + C2
Evaluating at P (2, 3, 4), obtain:
9
4 = 24 + C2 , or C2 =
19
Finally, at P , the requested equation is
y2
x2 = 4xy
19
b) a unit vector specifying the direction
p of E at Q(3, 2, 5): Have EQ = [4(3) + 2(2)]ax
4(2)]ay = 16ax + 2ay . Then E = 162 + 4 = 16.12 So
aQ =
16ax + 2ay
= 0.99ax + 0.12ay
16.12
24
[2(3)
2.28 An electric dipole (discussed in detail in Sec. 4.7) consists of two point charges of equal and opposite
magnitude ยฑQ spaced by distance d. With the charges along the z axis at positions z = ยฑd/2 (with
the positive
โฅ charge at the
โค positive z location), the electric field in spherical coordinates is given by
E(r, โ) = Qd/(4โกโ0 r3 ) [2 cos โar + sin โaโ ], where r >> d. Using rectangular coordinates, determine
expressions for the vector force on a point charge of magnitude q:
a) at (0, 0, z): Here, โ = 0, ar = az , and r = z. Therefore
F(0, 0, z) =
b) at (0, y, 0): Here, โ = 90 , aโ =
qQd az
N
4โกโ0 z 3
az , and r = y. The force is
qQd az
N
4โกโ0 y 3
F(0, y, 0) =
2.29. If E = 20e 5y (cos 5xax sin 5xay ), find:
a) E at P (โก/6, 0.1, 2): Substituting this point, we obtain EP =
12.2.
10.6ax
6.1ay , and so EP  =
b) a unit vector in the direction of EP : The unit vector associated with E is (cos 5xax
which evaluated at P becomes aE = 0.87ax 0.50ay .
sin 5xay ),
c) the equation of the direction line passing through P : Use
dy
sin 5x
=
=
dx
cos 5x
tan 5x ) dy =
tan 5x dx
Thus y = 15 ln cos 5x + C. Evaluating at P , we find C = 0.13, and so
y=
1
ln cos 5x + 0.13
5
2.30. For fields that do not vary with z in cylindrical coordinates, the equations of the streamlines are
obtained by solving the diโตerential equation Eโข /E = dโข(โขd ). Find the equation of the line passing
through the point (2, 30 , 0) for the field E = โข cos 2 aโข โข sin 2 a :
Eโข
dโข
โข cos 2
=
=
=
E
โขd
โข sin 2
cot 2
)
dโข
=
โข
cot 2 d
Integrate to obtain
2 ln โข = ln sin 2 + ln C = ln
๏ฃฟ
C
sin 2
) โข2 =
C
sin 2
p
At the given point, we have
4 = C/ sin(60 ) ) C = 4 sin 60 = 2 3. Finally, the equation for
p
the streamline is โข2 = 2 3/ sin 2 .
25
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