Engineering Electromagnetics, 8th Edition Solution Manual
Engineering Electromagnetics, 8th Edition Solution Manual provides expert solutions to textbook questions, making complex problems easier to solve.
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CHAPTER 1
1.1. Given the vectors M = 10ax + 4ay 8az and N = 8ax + 7ay 2az , find:
a) a unit vector in the direction of M + 2N.
M + 2N = 10ax 4ay + 8az + 16ax + 14ay 4az = (26, 10, 4)
Thus
a = (26, 10, 4)
|(26, 10, 4)| = (0.92, 0.36, 0.14)
b) the magnitude of 5ax + N 3M:
(5, 0, 0) + (8, 7, 2) ( 30, 12, 24) = (43, 5, 22), and |(43, 5, 22)| = 48.6.
c) |M||2N|(M + N):
|( 10, 4, 8)||(16, 14, 4)|( 2, 11, 10) = (13.4)(21.6)( 2, 11, 10)
= ( 580.5, 3193, 2902)
1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2).
a) Find the unit vector in the direction of (A B): First
A B = (ax + 2ay + 3az ) (2ax + 3ay 2az ) = ( ax ay + 5az )
whose magnitude is |A B| = [( ax ay + 5az ) · ( ax ay + 5az )]1/2 = p1 + 1 + 25 =
3p3 = 5.20. The unit vector is therefore
aAB = ( ax ay + 5az )/5.20
b) find the unit vector in the direction of the line extending from the origin to the midpoint of the
line joining the ends of A and B:
The midpoint is located at
Pmp = [1 + (2 1)/2, 2 + (3 2)/2, 3 + ( 2 3)/2)] = (1.5, 2.5, 0.5)
The unit vector is then
amp = (1.5ax + 2.5ay + 0.5az )
p(1.5)2 + (2.5)2 + (0.5)2 = (1.5ax + 2.5ay + 0.5az )/2.96
1.3. The vector from the origin to the point A is given as (6, 2, 4), and the unit vector directed from
the origin toward point B is (2, 2, 1)/3. If points A and B are ten units apart, find the coordinates
of point B.
With A = (6, 2, 4) and B = 1
3 B(2, 2, 1), we use the fact that |B A| = 10, or
|(6 2
3 B)ax (2 2
3 B)ay (4 + 1
3 B)az | = 10
Expanding, obtain
36 8B + 4
9 B2 + 4 8
3 B + 4
9 B2 + 16 + 8
3 B + 1
9 B2 = 100
or B2 8B 44 = 0. Thus B = 8±p64 176
2 = 11.75 (taking positive option) and so
B = 2
3 (11.75)ax 2
3 (11.75)ay + 1
3 (11.75)az = 7.83ax 7.83ay + 3.92az
1
1.1. Given the vectors M = 10ax + 4ay 8az and N = 8ax + 7ay 2az , find:
a) a unit vector in the direction of M + 2N.
M + 2N = 10ax 4ay + 8az + 16ax + 14ay 4az = (26, 10, 4)
Thus
a = (26, 10, 4)
|(26, 10, 4)| = (0.92, 0.36, 0.14)
b) the magnitude of 5ax + N 3M:
(5, 0, 0) + (8, 7, 2) ( 30, 12, 24) = (43, 5, 22), and |(43, 5, 22)| = 48.6.
c) |M||2N|(M + N):
|( 10, 4, 8)||(16, 14, 4)|( 2, 11, 10) = (13.4)(21.6)( 2, 11, 10)
= ( 580.5, 3193, 2902)
1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2).
a) Find the unit vector in the direction of (A B): First
A B = (ax + 2ay + 3az ) (2ax + 3ay 2az ) = ( ax ay + 5az )
whose magnitude is |A B| = [( ax ay + 5az ) · ( ax ay + 5az )]1/2 = p1 + 1 + 25 =
3p3 = 5.20. The unit vector is therefore
aAB = ( ax ay + 5az )/5.20
b) find the unit vector in the direction of the line extending from the origin to the midpoint of the
line joining the ends of A and B:
The midpoint is located at
Pmp = [1 + (2 1)/2, 2 + (3 2)/2, 3 + ( 2 3)/2)] = (1.5, 2.5, 0.5)
The unit vector is then
amp = (1.5ax + 2.5ay + 0.5az )
p(1.5)2 + (2.5)2 + (0.5)2 = (1.5ax + 2.5ay + 0.5az )/2.96
1.3. The vector from the origin to the point A is given as (6, 2, 4), and the unit vector directed from
the origin toward point B is (2, 2, 1)/3. If points A and B are ten units apart, find the coordinates
of point B.
With A = (6, 2, 4) and B = 1
3 B(2, 2, 1), we use the fact that |B A| = 10, or
|(6 2
3 B)ax (2 2
3 B)ay (4 + 1
3 B)az | = 10
Expanding, obtain
36 8B + 4
9 B2 + 4 8
3 B + 4
9 B2 + 16 + 8
3 B + 1
9 B2 = 100
or B2 8B 44 = 0. Thus B = 8±p64 176
2 = 11.75 (taking positive option) and so
B = 2
3 (11.75)ax 2
3 (11.75)ay + 1
3 (11.75)az = 7.83ax 7.83ay + 3.92az
1
CHAPTER 1
1.1. Given the vectors M = 10ax + 4ay 8az and N = 8ax + 7ay 2az , find:
a) a unit vector in the direction of M + 2N.
M + 2N = 10ax 4ay + 8az + 16ax + 14ay 4az = (26, 10, 4)
Thus
a = (26, 10, 4)
|(26, 10, 4)| = (0.92, 0.36, 0.14)
b) the magnitude of 5ax + N 3M:
(5, 0, 0) + (8, 7, 2) ( 30, 12, 24) = (43, 5, 22), and |(43, 5, 22)| = 48.6.
c) |M||2N|(M + N):
|( 10, 4, 8)||(16, 14, 4)|( 2, 11, 10) = (13.4)(21.6)( 2, 11, 10)
= ( 580.5, 3193, 2902)
1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2).
a) Find the unit vector in the direction of (A B): First
A B = (ax + 2ay + 3az ) (2ax + 3ay 2az ) = ( ax ay + 5az )
whose magnitude is |A B| = [( ax ay + 5az ) · ( ax ay + 5az )]1/2 = p1 + 1 + 25 =
3p3 = 5.20. The unit vector is therefore
aAB = ( ax ay + 5az )/5.20
b) find the unit vector in the direction of the line extending from the origin to the midpoint of the
line joining the ends of A and B:
The midpoint is located at
Pmp = [1 + (2 1)/2, 2 + (3 2)/2, 3 + ( 2 3)/2)] = (1.5, 2.5, 0.5)
The unit vector is then
amp = (1.5ax + 2.5ay + 0.5az )
p(1.5)2 + (2.5)2 + (0.5)2 = (1.5ax + 2.5ay + 0.5az )/2.96
1.3. The vector from the origin to the point A is given as (6, 2, 4), and the unit vector directed from
the origin toward point B is (2, 2, 1)/3. If points A and B are ten units apart, find the coordinates
of point B.
With A = (6, 2, 4) and B = 1
3 B(2, 2, 1), we use the fact that |B A| = 10, or
|(6 2
3 B)ax (2 2
3 B)ay (4 + 1
3 B)az | = 10
Expanding, obtain
36 8B + 4
9 B2 + 4 8
3 B + 4
9 B2 + 16 + 8
3 B + 1
9 B2 = 100
or B2 8B 44 = 0. Thus B = 8±p64 176
2 = 11.75 (taking positive option) and so
B = 2
3 (11.75)ax 2
3 (11.75)ay + 1
3 (11.75)az = 7.83ax 7.83ay + 3.92az
1
1.1. Given the vectors M = 10ax + 4ay 8az and N = 8ax + 7ay 2az , find:
a) a unit vector in the direction of M + 2N.
M + 2N = 10ax 4ay + 8az + 16ax + 14ay 4az = (26, 10, 4)
Thus
a = (26, 10, 4)
|(26, 10, 4)| = (0.92, 0.36, 0.14)
b) the magnitude of 5ax + N 3M:
(5, 0, 0) + (8, 7, 2) ( 30, 12, 24) = (43, 5, 22), and |(43, 5, 22)| = 48.6.
c) |M||2N|(M + N):
|( 10, 4, 8)||(16, 14, 4)|( 2, 11, 10) = (13.4)(21.6)( 2, 11, 10)
= ( 580.5, 3193, 2902)
1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2).
a) Find the unit vector in the direction of (A B): First
A B = (ax + 2ay + 3az ) (2ax + 3ay 2az ) = ( ax ay + 5az )
whose magnitude is |A B| = [( ax ay + 5az ) · ( ax ay + 5az )]1/2 = p1 + 1 + 25 =
3p3 = 5.20. The unit vector is therefore
aAB = ( ax ay + 5az )/5.20
b) find the unit vector in the direction of the line extending from the origin to the midpoint of the
line joining the ends of A and B:
The midpoint is located at
Pmp = [1 + (2 1)/2, 2 + (3 2)/2, 3 + ( 2 3)/2)] = (1.5, 2.5, 0.5)
The unit vector is then
amp = (1.5ax + 2.5ay + 0.5az )
p(1.5)2 + (2.5)2 + (0.5)2 = (1.5ax + 2.5ay + 0.5az )/2.96
1.3. The vector from the origin to the point A is given as (6, 2, 4), and the unit vector directed from
the origin toward point B is (2, 2, 1)/3. If points A and B are ten units apart, find the coordinates
of point B.
With A = (6, 2, 4) and B = 1
3 B(2, 2, 1), we use the fact that |B A| = 10, or
|(6 2
3 B)ax (2 2
3 B)ay (4 + 1
3 B)az | = 10
Expanding, obtain
36 8B + 4
9 B2 + 4 8
3 B + 4
9 B2 + 16 + 8
3 B + 1
9 B2 = 100
or B2 8B 44 = 0. Thus B = 8±p64 176
2 = 11.75 (taking positive option) and so
B = 2
3 (11.75)ax 2
3 (11.75)ay + 1
3 (11.75)az = 7.83ax 7.83ay + 3.92az
1
1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine the unit
vector in rectangular components that lies in the xy plane, is tangent to the circle at ( p3, 1, 0),
and is in the general direction of increasing values of y:
A unit vector tangent to this circle in the general increasing y direction is t = a . Its x and
y components are tx = a · ax = sin , and ty = a · ay = cos . At the point ( p3, 1),
= 150 , and so t = sin 150 ax cos 150 ay = 0.5(ax + p3ay ).
1.5. A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z2az . Given two points, P (1, 2, 1)
and Q( 2, 1, 3), find:
a) G at P : G(1, 2, 1) = (48, 36, 18)
b) a unit vector in the direction of G at Q: G( 2, 1, 3) = ( 48, 72, 162), so
aG = ( 48, 72, 162)
|( 48, 72, 162)| = ( 0.26, 0.39, 0.88)
c) a unit vector directed from Q toward P :
aQP = P Q
|P Q| = (3, 1, 4)
p26 = (0.59, 0.20, 0.78)
d) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x2 + 2), 18z2)|, or
10 = |(4xy, 2x2 + 4, 3z2)|, so the equation is
100 = 16x2y2 + 4x4 + 16x2 + 16 + 9z4
1.6. Find the acute angle between the two vectors A = 2ax + ay + 3az and B = ax 3ay + 2az by using
the definition of:
a) the dot product: First, A · B = 2 3 + 6 = 5 = AB cos ✓, where A = p22 + 12 + 32 = p14,
and where B = p12 + 32 + 22 = p14. Therefore cos ✓ = 5/14, so that ✓ = 69.1 .
b) the cross product: Begin with
A ⇥ B =
ax ay az
2 1 3
1 3 2
= 11ax ay 7az
and then |A ⇥ B| = p112 + 12 + 72 = p171. So now, with |A ⇥ B| = AB sin ✓ = p171,
find ✓ = sin 1 p171/14 = 69.1
1.7. Given the vector field E = 4zy2 cos 2xax + 2zy sin 2xay + y2 sin 2xaz for the region |x|, |y|, and |z|
less than 2, find:
a) the surfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0,
with |x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2,
|z| < 2; 4) the plane x = ⇡/2, with |y| < 2, |z| < 2.
b) the region in which Ey = Ez : This occurs when 2zy sin 2x = y2 sin 2x, or on the plane 2z = y,
with |x| < 2, |y| < 2, |z| < 1.
c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy2 cos 2x = zy sin 2x =
y2 sin 2x = 0. This condition is met on the plane y = 0, with |x| < 2, |z| < 2.
2
vector in rectangular components that lies in the xy plane, is tangent to the circle at ( p3, 1, 0),
and is in the general direction of increasing values of y:
A unit vector tangent to this circle in the general increasing y direction is t = a . Its x and
y components are tx = a · ax = sin , and ty = a · ay = cos . At the point ( p3, 1),
= 150 , and so t = sin 150 ax cos 150 ay = 0.5(ax + p3ay ).
1.5. A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z2az . Given two points, P (1, 2, 1)
and Q( 2, 1, 3), find:
a) G at P : G(1, 2, 1) = (48, 36, 18)
b) a unit vector in the direction of G at Q: G( 2, 1, 3) = ( 48, 72, 162), so
aG = ( 48, 72, 162)
|( 48, 72, 162)| = ( 0.26, 0.39, 0.88)
c) a unit vector directed from Q toward P :
aQP = P Q
|P Q| = (3, 1, 4)
p26 = (0.59, 0.20, 0.78)
d) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x2 + 2), 18z2)|, or
10 = |(4xy, 2x2 + 4, 3z2)|, so the equation is
100 = 16x2y2 + 4x4 + 16x2 + 16 + 9z4
1.6. Find the acute angle between the two vectors A = 2ax + ay + 3az and B = ax 3ay + 2az by using
the definition of:
a) the dot product: First, A · B = 2 3 + 6 = 5 = AB cos ✓, where A = p22 + 12 + 32 = p14,
and where B = p12 + 32 + 22 = p14. Therefore cos ✓ = 5/14, so that ✓ = 69.1 .
b) the cross product: Begin with
A ⇥ B =
ax ay az
2 1 3
1 3 2
= 11ax ay 7az
and then |A ⇥ B| = p112 + 12 + 72 = p171. So now, with |A ⇥ B| = AB sin ✓ = p171,
find ✓ = sin 1 p171/14 = 69.1
1.7. Given the vector field E = 4zy2 cos 2xax + 2zy sin 2xay + y2 sin 2xaz for the region |x|, |y|, and |z|
less than 2, find:
a) the surfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0,
with |x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2,
|z| < 2; 4) the plane x = ⇡/2, with |y| < 2, |z| < 2.
b) the region in which Ey = Ez : This occurs when 2zy sin 2x = y2 sin 2x, or on the plane 2z = y,
with |x| < 2, |y| < 2, |z| < 1.
c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy2 cos 2x = zy sin 2x =
y2 sin 2x = 0. This condition is met on the plane y = 0, with |x| < 2, |z| < 2.
2
1.8. Demonstrate the ambiguity that results when the cross product is used to find the angle between
two vectors by finding the angle between A = 3ax 2ay + 4az and B = 2ax + ay 2az . Does this
ambiguity exist when the dot product is used?
We use the relation A ⇥ B = |A||B| sin ✓n. With the given vectors we find
A ⇥ B = 14ay + 7az = 7p5
2ay + az
p5
| {z }
±n
= p9 + 4 + 16p4 + 1 + 4 sin ✓ n
where n is identified as shown; we see that n can be positive or negative, as sin ✓ can be
positive or negative. This apparent sign ambiguity is not the real problem, however, as we
really want the magnitude of the angle anyway. Choosing the positive sign, we are left with
sin ✓ = 7p5/(p29p9) = 0.969. Two values of ✓ (75.7 and 104.3 ) satisfy this equation, and
hence the real ambiguity.
In using the dot product, we find A · B = 6 2 8 = 4 = |A||B| cos ✓ = 3p29 cos ✓, or
cos ✓ = 4/(3p29) = 0.248 ) ✓ = 75.7 . Again, the minus sign is not important, as we
care only about the angle magnitude. The main point is that only one ✓ value results when
using the dot product, so no ambiguity.
1.9. A field is given as
G = 25
(x2 + y2) (xax + yay )
Find:
a) a unit vector in the direction of G at P (3, 4, 2): Have Gp = 25/(9 + 16) ⇥ (3, 4, 0) = 3ax + 4ay ,
and |Gp| = 5. Thus aG = (0.6, 0.8, 0).
b) the angle between G and ax at P : The angle is found through aG · ax = cos ✓. So cos ✓ =
(0.6, 0.8, 0) · (1, 0, 0) = 0.6. Thus ✓ = 53 .
c) the value of the following double integral on the plane y = 7:
Z 4
0
Z 2
0
G · ay dzdx
Z 4
0
Z 2
0
25
x2 + y2 (xax + yay ) · ay dzdx =
Z 4
0
Z 2
0
25
x2 + 49 ⇥ 7 dzdx =
Z 4
0
350
x2 + 49 dx
= 350 ⇥ 1
7
tan 1
✓ 4
7
◆
0
= 26
1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle
between any two diagonals of a cube, where each diagonal connects diametrically opposite corners,
and passes through the center of the cube:
Assuming a side length, b, two diagonal vectors would be A = b(ax + ay + az ) and B =
b(ax ay + az ). Now use A · B = |A||B| cos ✓, or b2(1 1 + 1) = (p3b)(p3b) cos ✓ ) cos ✓ =
1/3 ) ✓ = 70.53 . This result (in magnitude) is the same for any two diagonal vectors.
3
two vectors by finding the angle between A = 3ax 2ay + 4az and B = 2ax + ay 2az . Does this
ambiguity exist when the dot product is used?
We use the relation A ⇥ B = |A||B| sin ✓n. With the given vectors we find
A ⇥ B = 14ay + 7az = 7p5
2ay + az
p5
| {z }
±n
= p9 + 4 + 16p4 + 1 + 4 sin ✓ n
where n is identified as shown; we see that n can be positive or negative, as sin ✓ can be
positive or negative. This apparent sign ambiguity is not the real problem, however, as we
really want the magnitude of the angle anyway. Choosing the positive sign, we are left with
sin ✓ = 7p5/(p29p9) = 0.969. Two values of ✓ (75.7 and 104.3 ) satisfy this equation, and
hence the real ambiguity.
In using the dot product, we find A · B = 6 2 8 = 4 = |A||B| cos ✓ = 3p29 cos ✓, or
cos ✓ = 4/(3p29) = 0.248 ) ✓ = 75.7 . Again, the minus sign is not important, as we
care only about the angle magnitude. The main point is that only one ✓ value results when
using the dot product, so no ambiguity.
1.9. A field is given as
G = 25
(x2 + y2) (xax + yay )
Find:
a) a unit vector in the direction of G at P (3, 4, 2): Have Gp = 25/(9 + 16) ⇥ (3, 4, 0) = 3ax + 4ay ,
and |Gp| = 5. Thus aG = (0.6, 0.8, 0).
b) the angle between G and ax at P : The angle is found through aG · ax = cos ✓. So cos ✓ =
(0.6, 0.8, 0) · (1, 0, 0) = 0.6. Thus ✓ = 53 .
c) the value of the following double integral on the plane y = 7:
Z 4
0
Z 2
0
G · ay dzdx
Z 4
0
Z 2
0
25
x2 + y2 (xax + yay ) · ay dzdx =
Z 4
0
Z 2
0
25
x2 + 49 ⇥ 7 dzdx =
Z 4
0
350
x2 + 49 dx
= 350 ⇥ 1
7
tan 1
✓ 4
7
◆
0
= 26
1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle
between any two diagonals of a cube, where each diagonal connects diametrically opposite corners,
and passes through the center of the cube:
Assuming a side length, b, two diagonal vectors would be A = b(ax + ay + az ) and B =
b(ax ay + az ). Now use A · B = |A||B| cos ✓, or b2(1 1 + 1) = (p3b)(p3b) cos ✓ ) cos ✓ =
1/3 ) ✓ = 70.53 . This result (in magnitude) is the same for any two diagonal vectors.
3
1.11. Given the points M (0.1, 0.2, 0.1), N ( 0.2, 0.1, 0.3), and P (0.4, 0, 0.1), find:
a) the vector RM N : RM N = ( 0.2, 0.1, 0.3) (0.1, 0.2, 0.1) = ( 0.3, 0.3, 0.4).
b) the dot product RM N · RM P : RM P = (0.4, 0, 0.1) (0.1, 0.2, 0.1) = (0.3, 0.2, 0.2). RM N ·
RM P = ( 0.3, 0.3, 0.4) · (0.3, 0.2, 0.2) = 0.09 + 0.06 + 0.08 = 0.05.
c) the scalar projection of RM N on RM P :
RM N · aRM P = ( 0.3, 0.3, 0.4) · (0.3, 0.2, 0.2)
p0.09 + 0.04 + 0.04 = 0.05
p0.17 = 0.12
d) the angle between RM N and RM P :
✓M = cos 1
✓ RM N · RM P
|RM N ||RM P |
◆
= cos 1
✓ 0.05
p0.34p0.17
◆
= 78
1.12. Write an expression in rectangular components for the vector that extends from (x1, y1, z1) to
(x2, y2, z2) and determine the magnitude of this vector.
The two points can be written as vectors from the origin:
A1 = x1ax + y1ay + z1az and A2 = x2ax + y2ay + z2az
The desired vector will now be the di↵erence:
A12 = A2 A1 = (x2 x1)ax + (y2 y1)ay + (z2 z1)az
whose magnitude is
|A12| = pA12 · A12 = ⇥(x2 x1)2 + (y2 y1)2 + (z2 z1)2⇤1/2
1.13. a) Find the vector component of F = (10, 6, 5) that is parallel to G = (0.1, 0.2, 0.3):
F||G = F · G
|G|2 G = (10, 6, 5) · (0.1, 0.2, 0.3)
0.01 + 0.04 + 0.09 (0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)
b) Find the vector component of F that is perpendicular to G:
FpG = F F||G = (10, 6, 5) (0.93, 1.86, 2.79) = (9.07, 7.86, 2.21)
c) Find the vector component of G that is perpendicular to F:
GpF = G G||F = G G · F
|F|2 F = (0.1, 0.2, 0.3) 1.3
100 + 36 + 25 (10, 6, 5) = (0.02, 0.25, 0.26)
4
a) the vector RM N : RM N = ( 0.2, 0.1, 0.3) (0.1, 0.2, 0.1) = ( 0.3, 0.3, 0.4).
b) the dot product RM N · RM P : RM P = (0.4, 0, 0.1) (0.1, 0.2, 0.1) = (0.3, 0.2, 0.2). RM N ·
RM P = ( 0.3, 0.3, 0.4) · (0.3, 0.2, 0.2) = 0.09 + 0.06 + 0.08 = 0.05.
c) the scalar projection of RM N on RM P :
RM N · aRM P = ( 0.3, 0.3, 0.4) · (0.3, 0.2, 0.2)
p0.09 + 0.04 + 0.04 = 0.05
p0.17 = 0.12
d) the angle between RM N and RM P :
✓M = cos 1
✓ RM N · RM P
|RM N ||RM P |
◆
= cos 1
✓ 0.05
p0.34p0.17
◆
= 78
1.12. Write an expression in rectangular components for the vector that extends from (x1, y1, z1) to
(x2, y2, z2) and determine the magnitude of this vector.
The two points can be written as vectors from the origin:
A1 = x1ax + y1ay + z1az and A2 = x2ax + y2ay + z2az
The desired vector will now be the di↵erence:
A12 = A2 A1 = (x2 x1)ax + (y2 y1)ay + (z2 z1)az
whose magnitude is
|A12| = pA12 · A12 = ⇥(x2 x1)2 + (y2 y1)2 + (z2 z1)2⇤1/2
1.13. a) Find the vector component of F = (10, 6, 5) that is parallel to G = (0.1, 0.2, 0.3):
F||G = F · G
|G|2 G = (10, 6, 5) · (0.1, 0.2, 0.3)
0.01 + 0.04 + 0.09 (0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)
b) Find the vector component of F that is perpendicular to G:
FpG = F F||G = (10, 6, 5) (0.93, 1.86, 2.79) = (9.07, 7.86, 2.21)
c) Find the vector component of G that is perpendicular to F:
GpF = G G||F = G G · F
|F|2 F = (0.1, 0.2, 0.3) 1.3
100 + 36 + 25 (10, 6, 5) = (0.02, 0.25, 0.26)
4
1.14. Given that A + B + C = 0, where the three vectors represent line segments and extend from a
common origin,
a) must the three vectors be coplanar?
In terms of the components, the vector sum will be
A + B + C = (Ax + Bx + Cx)ax + (Ay + By + Cy )ay + (Az + Bz + Cz )az
which we require to be zero. Suppose the coordinate system is configured so that vectors A
and B lie in the x-y plane; in this case Az = Bz = 0. Then Cz has to be zero in order for the
three vectors to sum to zero. Therefore, the three vectors must be coplanar.
b) If A + B + C + D = 0, are the four vectors coplanar?
The vector sum is now
A + B + C + D = (Ax + Bx + Cx + Dx)ax + (Ay + By + Cy + Dy )ay + (Az + Bz + Cz + Dz )az
Now, for example, if A and B lie in the x-y plane, C and D need not, as long as Cz + Dz = 0.
So the four vectors need not be coplanar to have a zero sum.
1.15. Three vectors extending from the origin are given as r1 = (7, 3, 2), r2 = ( 2, 7, 3), and r3 =
(0, 2, 3). Find:
a) a unit vector perpendicular to both r1 and r2:
ap12 = r1 ⇥ r2
|r1 ⇥ r2| = (5, 25, 55)
60.6 = (0.08, 0.41, 0.91)
b) a unit vector perpendicular to the vectors r1 r2 and r2 r3: r1 r2 = (9, 4, 1) and r2 r3 =
( 2, 5, 6). So r1 r2 ⇥ r2 r3 = (19, 52, 37). Then
ap = (19, 52, 37)
|(19, 52, 37)| = (19, 52, 37)
66.6 = (0.29, 0.78, 0.56)
c) the area of the triangle defined by r1 and r2:
Area = 1
2 |r1 ⇥ r2| = 30.3
d) the area of the triangle defined by the heads of r1, r2, and r3:
Area = 1
2 |(r2 r1) ⇥ (r2 r3)| = 1
2 |( 9, 4, 1) ⇥ ( 2, 5, 6)| = 33.3
5
common origin,
a) must the three vectors be coplanar?
In terms of the components, the vector sum will be
A + B + C = (Ax + Bx + Cx)ax + (Ay + By + Cy )ay + (Az + Bz + Cz )az
which we require to be zero. Suppose the coordinate system is configured so that vectors A
and B lie in the x-y plane; in this case Az = Bz = 0. Then Cz has to be zero in order for the
three vectors to sum to zero. Therefore, the three vectors must be coplanar.
b) If A + B + C + D = 0, are the four vectors coplanar?
The vector sum is now
A + B + C + D = (Ax + Bx + Cx + Dx)ax + (Ay + By + Cy + Dy )ay + (Az + Bz + Cz + Dz )az
Now, for example, if A and B lie in the x-y plane, C and D need not, as long as Cz + Dz = 0.
So the four vectors need not be coplanar to have a zero sum.
1.15. Three vectors extending from the origin are given as r1 = (7, 3, 2), r2 = ( 2, 7, 3), and r3 =
(0, 2, 3). Find:
a) a unit vector perpendicular to both r1 and r2:
ap12 = r1 ⇥ r2
|r1 ⇥ r2| = (5, 25, 55)
60.6 = (0.08, 0.41, 0.91)
b) a unit vector perpendicular to the vectors r1 r2 and r2 r3: r1 r2 = (9, 4, 1) and r2 r3 =
( 2, 5, 6). So r1 r2 ⇥ r2 r3 = (19, 52, 37). Then
ap = (19, 52, 37)
|(19, 52, 37)| = (19, 52, 37)
66.6 = (0.29, 0.78, 0.56)
c) the area of the triangle defined by r1 and r2:
Area = 1
2 |r1 ⇥ r2| = 30.3
d) the area of the triangle defined by the heads of r1, r2, and r3:
Area = 1
2 |(r2 r1) ⇥ (r2 r3)| = 1
2 |( 9, 4, 1) ⇥ ( 2, 5, 6)| = 33.3
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1.16. If A represents a vector one unit in length directed due east, B represents a vector three units in
length directed due north, and A + B = 2C D and 2A B = C + 2D, determine the length and
direction of C. (di culty 1)
Take north as the positive y direction, and then east as the positive x direction. Then we may write
A + B = ax + 3ay = 2C D
and
2A B = 2ax 3ay = C + 2D
Multiplying the first equation by 2, and then adding the result to the second equation eliminates
D, and we get
4ax + 3ay = 5C ) C = 4
5 ax + 3
5 ay
The length of C is |C| = ⇥(4/5)2 + (3/5)2⇤1/2 = 1
C lies in the x-y plane at angle from due north (the y axis) given by ↵ = tan 1(4/3) = 53.1 (or
36.9 from the x axis). For those having nautical leanings, this is very close to the compass point
NE 3
4 E (not required).
1.17. Point A( 4, 2, 5) and the two vectors, RAM = (20, 18, 10) and RAN = ( 10, 8, 15), define a
triangle.
a) Find a unit vector perpendicular to the triangle: Use
ap = RAM ⇥ RAN
|RAM ⇥ RAN | = (350, 200, 340)
527.35 = (0.664, 0.379, 0.645)
The vector in the opposite direction to this one is also a valid answer.
b) Find a unit vector in the plane of the triangle and perpendicular to RAN :
aAN = ( 10, 8, 15)
p389 = ( 0.507, 0.406, 0.761)
Then
apAN = ap ⇥ aAN = (0.664, 0.379, 0.645) ⇥ ( 0.507, 0.406, 0.761) = ( 0.550, 0.832, 0.077)
The vector in the opposite direction to this one is also a valid answer.
c) Find a unit vector in the plane of the triangle that bisects the interior angle at A: A non-unit
vector in the required direction is (1/2)(aAM + aAN ), where
aAM = (20, 18, 10)
|(20, 18, 10)| = (0.697, 0.627, 0.348)
Now
1
2 (aAM + aAN ) = 1
2 [(0.697, 0.627, 0.348) + ( 0.507, 0.406, 0.761)] = (0.095, 0.516, 0.207)
Finally,
abis = (0.095, 0.516, 0.207)
|(0.095, 0.516, 0.207)| = (0.168, 0.915, 0.367)
6
length directed due north, and A + B = 2C D and 2A B = C + 2D, determine the length and
direction of C. (di culty 1)
Take north as the positive y direction, and then east as the positive x direction. Then we may write
A + B = ax + 3ay = 2C D
and
2A B = 2ax 3ay = C + 2D
Multiplying the first equation by 2, and then adding the result to the second equation eliminates
D, and we get
4ax + 3ay = 5C ) C = 4
5 ax + 3
5 ay
The length of C is |C| = ⇥(4/5)2 + (3/5)2⇤1/2 = 1
C lies in the x-y plane at angle from due north (the y axis) given by ↵ = tan 1(4/3) = 53.1 (or
36.9 from the x axis). For those having nautical leanings, this is very close to the compass point
NE 3
4 E (not required).
1.17. Point A( 4, 2, 5) and the two vectors, RAM = (20, 18, 10) and RAN = ( 10, 8, 15), define a
triangle.
a) Find a unit vector perpendicular to the triangle: Use
ap = RAM ⇥ RAN
|RAM ⇥ RAN | = (350, 200, 340)
527.35 = (0.664, 0.379, 0.645)
The vector in the opposite direction to this one is also a valid answer.
b) Find a unit vector in the plane of the triangle and perpendicular to RAN :
aAN = ( 10, 8, 15)
p389 = ( 0.507, 0.406, 0.761)
Then
apAN = ap ⇥ aAN = (0.664, 0.379, 0.645) ⇥ ( 0.507, 0.406, 0.761) = ( 0.550, 0.832, 0.077)
The vector in the opposite direction to this one is also a valid answer.
c) Find a unit vector in the plane of the triangle that bisects the interior angle at A: A non-unit
vector in the required direction is (1/2)(aAM + aAN ), where
aAM = (20, 18, 10)
|(20, 18, 10)| = (0.697, 0.627, 0.348)
Now
1
2 (aAM + aAN ) = 1
2 [(0.697, 0.627, 0.348) + ( 0.507, 0.406, 0.761)] = (0.095, 0.516, 0.207)
Finally,
abis = (0.095, 0.516, 0.207)
|(0.095, 0.516, 0.207)| = (0.168, 0.915, 0.367)
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1.18. A certain vector field is given as G = (y + 1)ax + xay . a) Determine G at the point (3,-2,4):
G(3, 2, 4) = ax + 3ay .
b) obtain a unit vector defining the direction of G at (3,-2,4).
|G(3, 2, 4)| = [1 + 32]1/2 = p10. So the unit vector is
aG(3, 2, 4) = ax + 3ay
p10
1.19. a) Express the field D = (x2 + y2) 1(xax + yay ) in cylindrical components and cylindrical variables:
Have x = ⇢ cos , y = ⇢ sin , and x2 + y2 = ⇢2. Therefore
D = 1
⇢ (cos ax + sin ay )
Then
D⇢ = D · a⇢ = 1
⇢ [cos (ax · a⇢) + sin (ay · a⇢)] = 1
⇢
⇥cos2 + sin2 ⇤ = 1
⇢
and
D = D · a = 1
⇢ [cos (ax · a ) + sin (ay · a )] = 1
⇢ [cos ( sin ) + sin cos ] = 0
Therefore
D = 1
⇢ a⇢
b) Evaluate D at the point where ⇢ = 2, = 0.2⇡, and z = 5, expressing the result in cylindrical
and cartesian coordinates: At the given point, and in cylindrical coordinates, D = 0.5a⇢. To
express this in cartesian, we use
D = 0.5(a⇢ · ax)ax + 0.5(a⇢ · ay )ay = 0.5 cos 36 ax + 0.5 sin 36 ay = 0.41ax + 0.29ay
1.20. If the three sides of a triangle are represented by the vectors A, B, and C, all directed counter-
clockwise, show that |C|2 = (A + B) · (A + B) and expand the product to obtain the law of cosines.
With the vectors drawn as described above, we find that C = (A + B) and so |C|2 = C2 = C · C =
(A + B) · (A + B) So far so good. Now if we expand the product, obtain
(A + B) · (A + B) = A2 + B2 + 2A · B
where A · B = AB cos(180 ↵) = AB cos ↵ where ↵ is the interior angle at the junction of A
and B. Using this, we have C2 = A2 + B2 2AB cos ↵, which is the law of cosines.
7
G(3, 2, 4) = ax + 3ay .
b) obtain a unit vector defining the direction of G at (3,-2,4).
|G(3, 2, 4)| = [1 + 32]1/2 = p10. So the unit vector is
aG(3, 2, 4) = ax + 3ay
p10
1.19. a) Express the field D = (x2 + y2) 1(xax + yay ) in cylindrical components and cylindrical variables:
Have x = ⇢ cos , y = ⇢ sin , and x2 + y2 = ⇢2. Therefore
D = 1
⇢ (cos ax + sin ay )
Then
D⇢ = D · a⇢ = 1
⇢ [cos (ax · a⇢) + sin (ay · a⇢)] = 1
⇢
⇥cos2 + sin2 ⇤ = 1
⇢
and
D = D · a = 1
⇢ [cos (ax · a ) + sin (ay · a )] = 1
⇢ [cos ( sin ) + sin cos ] = 0
Therefore
D = 1
⇢ a⇢
b) Evaluate D at the point where ⇢ = 2, = 0.2⇡, and z = 5, expressing the result in cylindrical
and cartesian coordinates: At the given point, and in cylindrical coordinates, D = 0.5a⇢. To
express this in cartesian, we use
D = 0.5(a⇢ · ax)ax + 0.5(a⇢ · ay )ay = 0.5 cos 36 ax + 0.5 sin 36 ay = 0.41ax + 0.29ay
1.20. If the three sides of a triangle are represented by the vectors A, B, and C, all directed counter-
clockwise, show that |C|2 = (A + B) · (A + B) and expand the product to obtain the law of cosines.
With the vectors drawn as described above, we find that C = (A + B) and so |C|2 = C2 = C · C =
(A + B) · (A + B) So far so good. Now if we expand the product, obtain
(A + B) · (A + B) = A2 + B2 + 2A · B
where A · B = AB cos(180 ↵) = AB cos ↵ where ↵ is the interior angle at the junction of A
and B. Using this, we have C2 = A2 + B2 2AB cos ↵, which is the law of cosines.
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1.21. Express in cylindrical components:
a) the vector from C(3, 2, 7) to D( 1, 4, 2):
C(3, 2, 7) ! C(⇢ = 3.61, = 33.7 , z = 7) and
D( 1, 4, 2) ! D(⇢ = 4.12, = 104.0 , z = 2).
Now RCD = ( 4, 6, 9) and R⇢ = RCD · a⇢ = 4 cos(33.7) 6 sin(33.7) = 6.66. Then
R = RCD · a = 4 sin(33.7) 6 cos(33.7) = 2.77. So RCD = 6.66a⇢ 2.77a + 9az
b) a unit vector at D directed toward C:
RCD = (4, 6, 9) and R⇢ = RDC · a⇢ = 4 cos( 104.0) + 6 sin( 104.0) = 6.79. Then R =
RDC · a = 4[ sin( 104.0)] + 6 cos( 104.0) = 2.43. So RDC = 6.79a⇢ + 2.43a 9az
Thus aDC = 0.59a⇢ + 0.21a 0.78az
c) a unit vector at D directed toward the origin: Start with rD = ( 1, 4, 2), and so the
vector toward the origin will be rD = (1, 4, 2). Thus in cartesian the unit vector is a =
(0.22, 0.87, 0.44). Convert to cylindrical:
a⇢ = (0.22, 0.87, 0.44) · a⇢ = 0.22 cos( 104.0) + 0.87 sin( 104.0) = 0.90, and
a = (0.22, 0.87, 0.44) · a = 0.22[ sin( 104.0)] + 0.87 cos( 104.0) = 0, so that finally,
a = 0.90a⇢ 0.44az .
1.22. A sphere of radius a, centered at the origin, rotates about the z axis at angular velocity ⌦ rad/s.
The rotation direction is clockwise when one is looking in the positive z direction.
a) Using spherical components, write an expression for the velocity field, v, which gives the tan-
gential velocity at any point within the sphere:
As in problem 1.20, we find the tangential velocity as the product of the angular velocity and
the perperdicular distance from the rotation axis. With clockwise rotation, we obtain
v(r, ✓) = ⌦r sin ✓ a (r < a)
b) Convert to rectangular components:
From here, the problem is the same as part c in Problem 1.20, except the rotation direction is
reversed. The answer is v(x, y) = ⌦ [ y ax + x ay ], where (x2 + y2 + z2)1/2 < a.
1.23. The surfaces ⇢ = 3, ⇢ = 5, = 100 , = 130 , z = 3, and z = 4.5 define a closed surface.
a) Find the enclosed volume:
Vol =
Z 4.5
3
Z 130
100
Z 5
3
⇢ d⇢ d dz = 6.28
NOTE: The limits on the integration must be converted to radians (as was done here, but not
shown).
b) Find the total area of the enclosing surface:
Area = 2
Z 130
100
Z 5
3
⇢ d⇢ d +
Z 4.5
3
Z 130
100
3 d dz
+
Z 4.5
3
Z 130
100
5 d dz + 2
Z 4.5
3
Z 5
3
d⇢ dz = 20.7
8
a) the vector from C(3, 2, 7) to D( 1, 4, 2):
C(3, 2, 7) ! C(⇢ = 3.61, = 33.7 , z = 7) and
D( 1, 4, 2) ! D(⇢ = 4.12, = 104.0 , z = 2).
Now RCD = ( 4, 6, 9) and R⇢ = RCD · a⇢ = 4 cos(33.7) 6 sin(33.7) = 6.66. Then
R = RCD · a = 4 sin(33.7) 6 cos(33.7) = 2.77. So RCD = 6.66a⇢ 2.77a + 9az
b) a unit vector at D directed toward C:
RCD = (4, 6, 9) and R⇢ = RDC · a⇢ = 4 cos( 104.0) + 6 sin( 104.0) = 6.79. Then R =
RDC · a = 4[ sin( 104.0)] + 6 cos( 104.0) = 2.43. So RDC = 6.79a⇢ + 2.43a 9az
Thus aDC = 0.59a⇢ + 0.21a 0.78az
c) a unit vector at D directed toward the origin: Start with rD = ( 1, 4, 2), and so the
vector toward the origin will be rD = (1, 4, 2). Thus in cartesian the unit vector is a =
(0.22, 0.87, 0.44). Convert to cylindrical:
a⇢ = (0.22, 0.87, 0.44) · a⇢ = 0.22 cos( 104.0) + 0.87 sin( 104.0) = 0.90, and
a = (0.22, 0.87, 0.44) · a = 0.22[ sin( 104.0)] + 0.87 cos( 104.0) = 0, so that finally,
a = 0.90a⇢ 0.44az .
1.22. A sphere of radius a, centered at the origin, rotates about the z axis at angular velocity ⌦ rad/s.
The rotation direction is clockwise when one is looking in the positive z direction.
a) Using spherical components, write an expression for the velocity field, v, which gives the tan-
gential velocity at any point within the sphere:
As in problem 1.20, we find the tangential velocity as the product of the angular velocity and
the perperdicular distance from the rotation axis. With clockwise rotation, we obtain
v(r, ✓) = ⌦r sin ✓ a (r < a)
b) Convert to rectangular components:
From here, the problem is the same as part c in Problem 1.20, except the rotation direction is
reversed. The answer is v(x, y) = ⌦ [ y ax + x ay ], where (x2 + y2 + z2)1/2 < a.
1.23. The surfaces ⇢ = 3, ⇢ = 5, = 100 , = 130 , z = 3, and z = 4.5 define a closed surface.
a) Find the enclosed volume:
Vol =
Z 4.5
3
Z 130
100
Z 5
3
⇢ d⇢ d dz = 6.28
NOTE: The limits on the integration must be converted to radians (as was done here, but not
shown).
b) Find the total area of the enclosing surface:
Area = 2
Z 130
100
Z 5
3
⇢ d⇢ d +
Z 4.5
3
Z 130
100
3 d dz
+
Z 4.5
3
Z 130
100
5 d dz + 2
Z 4.5
3
Z 5
3
d⇢ dz = 20.7
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1.23c) Find the total length of the twelve edges of the surfaces:
Length = 4 ⇥ 1.5 + 4 ⇥ 2 + 2 ⇥
30
360 ⇥ 2⇡ ⇥ 3 + 30
360 ⇥ 2⇡ ⇥ 5
= 22.4
d) Find the length of the longest straight line that lies entirely within the volume: This will be
between the points A(⇢ = 3, = 100 , z = 3) and B(⇢ = 5, = 130 , z = 4.5). Performing
point transformations to cartesian coordinates, these become A(x = 0.52, y = 2.95, z = 3)
and B(x = 3.21, y = 3.83, z = 4.5). Taking A and B as vectors directed from the origin, the
requested length is
Length = |B A| = |( 2.69, 0.88, 1.5)| = 3.21
1.24. Two unit vectors, a1 and a2 lie in the xy plane and pass through the origin. They make angles 1
and 2 with the x axis respectively.
a) Express each vector in rectangular components; Have a1 = Ax1ax + Ay1ay , so that Ax1 =
a1 · ax = cos 1. Then, Ay1 = a1 · ay = cos(90 1) = sin 1. Therefore,
a1 = cos 1 ax + sin 1 ay and similarly, a2 = cos 2 ax + sin 2 ay
b) take the dot product and verify the trigonometric identity, cos( 1 2) = cos 1 cos 2 +
sin 1 sin 2: From the definition of the dot product,
a1 · a2 = (1)(1) cos( 1 2)
= (cos 1 ax + sin 1 ay ) · (cos 2 ax + sin 2 ay ) = cos 1 cos 2 + sin 1 sin 2
c) take the cross product and verify the trigonometric identity sin( 2 1) = sin 2 cos 1
cos 2 sin 1: From the definition of the cross product, and since a1 and a2 both lie in the x-y
plane,
a1 ⇥ a2 = (1)(1) sin( 1 2) az =
ax ay az
cos 1 sin 1 0
cos 2 sin 2 0
= [sin 2 cos 1 cos 2 sin 1] az
thus verified.
1.25. Given point P (r = 0.8, ✓ = 30 , = 45 ), and
E = 1
r2
✓
cos ar + sin
sin ✓ a
◆
a) Find E at P : E = 1.10a⇢ + 2.21a .
b) Find |E| at P : |E| = p1.102 + 2.212 = 2.47.
c) Find a unit vector in the direction of E at P :
aE = E
|E| = 0.45ar + 0.89a
9
Length = 4 ⇥ 1.5 + 4 ⇥ 2 + 2 ⇥
30
360 ⇥ 2⇡ ⇥ 3 + 30
360 ⇥ 2⇡ ⇥ 5
= 22.4
d) Find the length of the longest straight line that lies entirely within the volume: This will be
between the points A(⇢ = 3, = 100 , z = 3) and B(⇢ = 5, = 130 , z = 4.5). Performing
point transformations to cartesian coordinates, these become A(x = 0.52, y = 2.95, z = 3)
and B(x = 3.21, y = 3.83, z = 4.5). Taking A and B as vectors directed from the origin, the
requested length is
Length = |B A| = |( 2.69, 0.88, 1.5)| = 3.21
1.24. Two unit vectors, a1 and a2 lie in the xy plane and pass through the origin. They make angles 1
and 2 with the x axis respectively.
a) Express each vector in rectangular components; Have a1 = Ax1ax + Ay1ay , so that Ax1 =
a1 · ax = cos 1. Then, Ay1 = a1 · ay = cos(90 1) = sin 1. Therefore,
a1 = cos 1 ax + sin 1 ay and similarly, a2 = cos 2 ax + sin 2 ay
b) take the dot product and verify the trigonometric identity, cos( 1 2) = cos 1 cos 2 +
sin 1 sin 2: From the definition of the dot product,
a1 · a2 = (1)(1) cos( 1 2)
= (cos 1 ax + sin 1 ay ) · (cos 2 ax + sin 2 ay ) = cos 1 cos 2 + sin 1 sin 2
c) take the cross product and verify the trigonometric identity sin( 2 1) = sin 2 cos 1
cos 2 sin 1: From the definition of the cross product, and since a1 and a2 both lie in the x-y
plane,
a1 ⇥ a2 = (1)(1) sin( 1 2) az =
ax ay az
cos 1 sin 1 0
cos 2 sin 2 0
= [sin 2 cos 1 cos 2 sin 1] az
thus verified.
1.25. Given point P (r = 0.8, ✓ = 30 , = 45 ), and
E = 1
r2
✓
cos ar + sin
sin ✓ a
◆
a) Find E at P : E = 1.10a⇢ + 2.21a .
b) Find |E| at P : |E| = p1.102 + 2.212 = 2.47.
c) Find a unit vector in the direction of E at P :
aE = E
|E| = 0.45ar + 0.89a
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1.26. Express the uniform vector field, F = 5 ax in
a) cylindrical components: F⇢ = 5 ax · a⇢ = 5 cos , and F = 5 ax · a = 5 sin . Combining, we
obtain F(⇢, ) = 5(cos a⇢ sin a ).
b) spherical components: Fr = 5 ax ·ar = 5 sin ✓ cos ; F✓ = 5 ax ·a✓ = 5 cos ✓ cos ; F = 5 ax ·a =
5 sin . Combining, we obtain F(r, ✓, ) = 5 [sin ✓ cos ar + cos ✓ cos a✓ sin a ].
1.27. The surfaces r = 2 and 4, ✓ = 30 and 50 , and = 20 and 60 identify a closed surface.
a) Find the enclosed volume: This will be
Vol =
Z 60
20
Z 50
30
Z 4
2
r2 sin ✓drd✓d = 2.91
where degrees have been converted to radians.
b) Find the total area of the enclosing surface:
Area =
Z 60
20
Z 50
30
(42 + 22) sin ✓d✓d +
Z 4
2
Z 60
20
r(sin 30 + sin 50 )drd
+ 2
Z 50
30
Z 4
2
rdrd✓ = 12.61
c) Find the total length of the twelve edges of the surface:
Length = 4
Z 4
2
dr + 2
Z 50
30
(4 + 2)d✓ +
Z 60
20
(4 sin 50 + 4 sin 30 + 2 sin 50 + 2 sin 30 )d
= 17.49
d) Find the length of the longest straight line that lies entirely within the surface: This will be
from A(r = 2, ✓ = 50 , = 20 ) to B(r = 4, ✓ = 30 , = 60 ) or
A(x = 2 sin 50 cos 20 , y = 2 sin 50 sin 20 , z = 2 cos 50 )
to
B(x = 4 sin 30 cos 60 , y = 4 sin 30 sin 60 , z = 4 cos 30 )
or finally A(1.44, 0.52, 1.29) to B(1.00, 1.73, 3.46). Thus B A = ( 0.44, 1.21, 2.18) and
Length = |B A| = 2.53
1.28. State whether or not A = B and, if not, what conditions are imposed on A and B when
a) A · ax = B · ax: For this to be true, both A and B must be oriented at the same angle, ✓, from
the x axis. But this would allow either vector to lie anywhere along a conical surface of angle
✓ about the x axis. Therefore, A can be equal to B, but not necessarily.
b) A ⇥ ax = B ⇥ ax: This is a more restrictive condition because the cross product gives a vector.
For both cross products to lie in the same direction, A, B, and ax must be coplanar. But if A
lies at angle ✓ to the x axis, B could lie at ✓ or at 180 ✓ to give the same cross product. So
again, A can be equal to B, but not necessarily.
10
a) cylindrical components: F⇢ = 5 ax · a⇢ = 5 cos , and F = 5 ax · a = 5 sin . Combining, we
obtain F(⇢, ) = 5(cos a⇢ sin a ).
b) spherical components: Fr = 5 ax ·ar = 5 sin ✓ cos ; F✓ = 5 ax ·a✓ = 5 cos ✓ cos ; F = 5 ax ·a =
5 sin . Combining, we obtain F(r, ✓, ) = 5 [sin ✓ cos ar + cos ✓ cos a✓ sin a ].
1.27. The surfaces r = 2 and 4, ✓ = 30 and 50 , and = 20 and 60 identify a closed surface.
a) Find the enclosed volume: This will be
Vol =
Z 60
20
Z 50
30
Z 4
2
r2 sin ✓drd✓d = 2.91
where degrees have been converted to radians.
b) Find the total area of the enclosing surface:
Area =
Z 60
20
Z 50
30
(42 + 22) sin ✓d✓d +
Z 4
2
Z 60
20
r(sin 30 + sin 50 )drd
+ 2
Z 50
30
Z 4
2
rdrd✓ = 12.61
c) Find the total length of the twelve edges of the surface:
Length = 4
Z 4
2
dr + 2
Z 50
30
(4 + 2)d✓ +
Z 60
20
(4 sin 50 + 4 sin 30 + 2 sin 50 + 2 sin 30 )d
= 17.49
d) Find the length of the longest straight line that lies entirely within the surface: This will be
from A(r = 2, ✓ = 50 , = 20 ) to B(r = 4, ✓ = 30 , = 60 ) or
A(x = 2 sin 50 cos 20 , y = 2 sin 50 sin 20 , z = 2 cos 50 )
to
B(x = 4 sin 30 cos 60 , y = 4 sin 30 sin 60 , z = 4 cos 30 )
or finally A(1.44, 0.52, 1.29) to B(1.00, 1.73, 3.46). Thus B A = ( 0.44, 1.21, 2.18) and
Length = |B A| = 2.53
1.28. State whether or not A = B and, if not, what conditions are imposed on A and B when
a) A · ax = B · ax: For this to be true, both A and B must be oriented at the same angle, ✓, from
the x axis. But this would allow either vector to lie anywhere along a conical surface of angle
✓ about the x axis. Therefore, A can be equal to B, but not necessarily.
b) A ⇥ ax = B ⇥ ax: This is a more restrictive condition because the cross product gives a vector.
For both cross products to lie in the same direction, A, B, and ax must be coplanar. But if A
lies at angle ✓ to the x axis, B could lie at ✓ or at 180 ✓ to give the same cross product. So
again, A can be equal to B, but not necessarily.
10
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1.28c) A · ax = B · ax and A ⇥ ax = B ⇥ ax: In this case, we need to satisfy both requirements in parts
a and b – that is, A, B, and ax must be coplanar, and A and B must lie at the same angle, ✓, to
ax. With coplanar vectors, this latter condition might imply that both +✓ and ✓ would therefore
work. But the negative angle reverses the direction of the cross product direction. Therefore both
vectors must lie in the same plane and lie at the same angle to x; i.e., A must be equal to B.
d) A · C = B · C and A ⇥ C = B ⇥ C where C is any vector except C = 0: This is just the general
case of part c. Since we can orient our coordinate system in any manner we choose, we can
arrange it so that the x axis coincides with the direction of vector C. Thus all the arguments
of part c apply, and again we conclude that A must be equal to B.
1.29. Express the unit vector ax in spherical components at the point:
a) r = 2, ✓ = 1 rad, = 0.8 rad: Use
ax = (ax · ar)ar + (ax · a✓)a✓ + (ax · a )a =
sin(1) cos(0.8)ar + cos(1) cos(0.8)a✓ + ( sin(0.8))a = 0.59ar + 0.38a✓ 0.72a
b) x = 3, y = 2, z = 1: First, transform the point to spherical coordinates. Have r = p14,
✓ = cos 1( 1/p14) = 105.5 , and = tan 1(2/3) = 33.7 . Then
ax = sin(105.5 ) cos(33.7 )ar + cos(105.5 ) cos(33.7 )a✓ + ( sin(33.7 ))a
= 0.80ar 0.22a✓ 0.55a
c) ⇢ = 2.5, = 0.7 rad, z = 1.5: Again, convert the point to spherical coordinates. r = p⇢2 + z2 =p8.5, ✓ = cos 1(z/r) = cos 1(1.5/p8.5) = 59.0 , and = 0.7 rad = 40.1 . Now
ax = sin(59 ) cos(40.1 )ar + cos(59 ) cos(40.1 )a✓ + ( sin(40.1 ))a
= 0.66ar + 0.39a✓ 0.64a
1.30. Consider a problem analogous to the varying wind velocities encountered by transcontinental aircraft.
We assume a constant altitude, a plane earth, a flight along the x axis from 0 to 10 units, no vertical
velocity component, and no change in wind velocity with time. Assume ax to be directed to the
east and ay to the north. The wind velocity at the operating altitude is assumed to be:
v(x, y) = (0.01x2 0.08x + 0.66)ax (0.05x 0.4)ay
1 + 0.5y2
a) Determine the location and magnitude of the maximum tailwind encountered: Tailwind would
be x-directed, and so we look at the x component only. Over the flight range, this function
maximizes at a value of 0.86/(1 + 0.5y2) at x = 10 (at the end of the trip). It reaches a local
minimum of 0.50/(1 + 0.5y2) at x = 4, and has another local maximum of 0.66/(1 + 0.5y2) at
the trip start, x = 0.
b) Repeat for headwind: The x component is always positive, and so therefore no headwind exists
over the travel range.
c) Repeat for crosswind: Crosswind will be found from the y component, which is seen to maximize
over the flight range at a value of 0.4/(1 + 0.5y2) at the trip start (x = 0).
d) Would more favorable tailwinds be available at some other latitude? If so, where? Minimizing
the denominator accomplishes this; in particular, the lattitude associated with y = 0 gives the
strongest tailwind.
11
a and b – that is, A, B, and ax must be coplanar, and A and B must lie at the same angle, ✓, to
ax. With coplanar vectors, this latter condition might imply that both +✓ and ✓ would therefore
work. But the negative angle reverses the direction of the cross product direction. Therefore both
vectors must lie in the same plane and lie at the same angle to x; i.e., A must be equal to B.
d) A · C = B · C and A ⇥ C = B ⇥ C where C is any vector except C = 0: This is just the general
case of part c. Since we can orient our coordinate system in any manner we choose, we can
arrange it so that the x axis coincides with the direction of vector C. Thus all the arguments
of part c apply, and again we conclude that A must be equal to B.
1.29. Express the unit vector ax in spherical components at the point:
a) r = 2, ✓ = 1 rad, = 0.8 rad: Use
ax = (ax · ar)ar + (ax · a✓)a✓ + (ax · a )a =
sin(1) cos(0.8)ar + cos(1) cos(0.8)a✓ + ( sin(0.8))a = 0.59ar + 0.38a✓ 0.72a
b) x = 3, y = 2, z = 1: First, transform the point to spherical coordinates. Have r = p14,
✓ = cos 1( 1/p14) = 105.5 , and = tan 1(2/3) = 33.7 . Then
ax = sin(105.5 ) cos(33.7 )ar + cos(105.5 ) cos(33.7 )a✓ + ( sin(33.7 ))a
= 0.80ar 0.22a✓ 0.55a
c) ⇢ = 2.5, = 0.7 rad, z = 1.5: Again, convert the point to spherical coordinates. r = p⇢2 + z2 =p8.5, ✓ = cos 1(z/r) = cos 1(1.5/p8.5) = 59.0 , and = 0.7 rad = 40.1 . Now
ax = sin(59 ) cos(40.1 )ar + cos(59 ) cos(40.1 )a✓ + ( sin(40.1 ))a
= 0.66ar + 0.39a✓ 0.64a
1.30. Consider a problem analogous to the varying wind velocities encountered by transcontinental aircraft.
We assume a constant altitude, a plane earth, a flight along the x axis from 0 to 10 units, no vertical
velocity component, and no change in wind velocity with time. Assume ax to be directed to the
east and ay to the north. The wind velocity at the operating altitude is assumed to be:
v(x, y) = (0.01x2 0.08x + 0.66)ax (0.05x 0.4)ay
1 + 0.5y2
a) Determine the location and magnitude of the maximum tailwind encountered: Tailwind would
be x-directed, and so we look at the x component only. Over the flight range, this function
maximizes at a value of 0.86/(1 + 0.5y2) at x = 10 (at the end of the trip). It reaches a local
minimum of 0.50/(1 + 0.5y2) at x = 4, and has another local maximum of 0.66/(1 + 0.5y2) at
the trip start, x = 0.
b) Repeat for headwind: The x component is always positive, and so therefore no headwind exists
over the travel range.
c) Repeat for crosswind: Crosswind will be found from the y component, which is seen to maximize
over the flight range at a value of 0.4/(1 + 0.5y2) at the trip start (x = 0).
d) Would more favorable tailwinds be available at some other latitude? If so, where? Minimizing
the denominator accomplishes this; in particular, the lattitude associated with y = 0 gives the
strongest tailwind.
11
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CHAPTER 2
2.1. Three point charges are positioned in the x-y plane as follows: 5nC at y = 5 cm, -10 nC at y = 5 cm,
15 nC at x = 5 cm. Find the required x-y coordinates of a 20-nC 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:
E0 = 1
4⇡✏0
15
(5)2 ax 5
(5)2 ay 10
(5)2 ay
= 1
4⇡✏0
✓ 3
5
◆
[ax ay ] nC/m
The field, E20, associated with the 20-nC charge (evaluated at the origin) must exactly cancel
this field, so we write:
E20 = 1
4⇡✏0
✓ 3
5
◆
[ax ay ] = 20
4⇡✏0⇢2
✓ 1
p2
◆
[ax ay ]
From this, we identify the distance from the origin: ⇢ =
q
100/(3p2) = 4.85. The x and
y coordinates of the 20-nC charge will both be equal in magnitude to 4.85/p2 = 3.43. The
coodinates of the 20-nC charge are then (3.43, 3.43).
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:
E12 = 1
4⇡✏0(3)
✓ 1
p3
◆
(ax + ay + az ) nC/m
in which the distance between charges is p3 m. The force on the -2nC charge is then
F12 = q2E12 = 2
12p3 ⇡✏0
(ax + ay + az ) = 1
10.4 ⇡✏0
(ax + ay + az ) nN
The force on the 1nC charge at the origin is just the opposite of this result, or
F21 = +1
10.4 ⇡✏0
(ax + ay + az ) nN
12
2.1. Three point charges are positioned in the x-y plane as follows: 5nC at y = 5 cm, -10 nC at y = 5 cm,
15 nC at x = 5 cm. Find the required x-y coordinates of a 20-nC 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:
E0 = 1
4⇡✏0
15
(5)2 ax 5
(5)2 ay 10
(5)2 ay
= 1
4⇡✏0
✓ 3
5
◆
[ax ay ] nC/m
The field, E20, associated with the 20-nC charge (evaluated at the origin) must exactly cancel
this field, so we write:
E20 = 1
4⇡✏0
✓ 3
5
◆
[ax ay ] = 20
4⇡✏0⇢2
✓ 1
p2
◆
[ax ay ]
From this, we identify the distance from the origin: ⇢ =
q
100/(3p2) = 4.85. The x and
y coordinates of the 20-nC charge will both be equal in magnitude to 4.85/p2 = 3.43. The
coodinates of the 20-nC charge are then (3.43, 3.43).
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:
E12 = 1
4⇡✏0(3)
✓ 1
p3
◆
(ax + ay + az ) nC/m
in which the distance between charges is p3 m. The force on the -2nC charge is then
F12 = q2E12 = 2
12p3 ⇡✏0
(ax + ay + az ) = 1
10.4 ⇡✏0
(ax + ay + az ) nN
The force on the 1nC charge at the origin is just the opposite of this result, or
F21 = +1
10.4 ⇡✏0
(ax + ay + az ) nN
12
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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:
F = (50 ⇥ 10 9)2
4⇡✏0
RCA
|RCA|3 + RDA
|RDA|3 + RBA
|RBA|3
where RCA = ax ay , RDA = ax + ay , and RBA = 2ax. The magnitudes are |RCA| = |RDA| = p2,
and |RBA| = 2. Substituting these leads to
F = (50 ⇥ 10 9)2
4⇡✏0
1
2p2 + 1
2p2 + 2
8
ax = 21.5ax μN
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:
Enet(a, a, a) = q
4⇡✏0a2
2
6
6
6
4
ax + ay + az
3p3
| {z }
(0,0,0)
+ ay + az
2p2
| {z }
(a,0,0)
+ ax + az
2p2
| {z }
(0,a,0)
+ ax + ay
2p2
| {z }
(0,0,a)
+ ax
|{z}
(0,a,a)
+ ay
|{z}
(a,0,a)
+ az
|{z}
(a,a,0)
3
7
7
7
5
The force is now the product of this field and the charge at (a, a, a). Simplifying, we obtain
F(a, a, a) = qEnet(a, a, a) = q2
4⇡✏0a2
1
3p3 + 1
p2 + 1
(ax + ay + az ) = 1.90 q2
4⇡✏0a2 (ax + ay + az )
in which the magnitude is |F| = 3.29 q2/(4⇡✏0a2).
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 ✏ = ✏0, find E at P3(1, 2, 3): This field will be
E = 10 9
4⇡✏0
25R13
|R13|3 + 60R23
|R23|3
where R13 = 3ax + 4ay 4az and R23 = 4ax 2ay + 5az . Also, |R13| = p41 and |R23| = p45.
So
E = 10 9
4⇡✏0
25 ⇥ ( 3ax + 4ay 4az )
(41)1.5 + 60 ⇥ (4ax 2ay + 5az )
(45)1.5
= 4.58ax 0.15ay + 5.51az
b) At what point on the y axis is Ex = 0? P3 is now at (0, y, 0), so R13 = 4ax + (y + 2)ay 7az
and R23 = 3ax + (y 4)ay + 2az . Also, |R13| = p65 + (y + 2)2 and |R23| = p13 + (y 4)2.
Now the x component of E at the new P3 will be:
Ex = 10 9
4⇡✏0
25 ⇥ ( 4)
[65 + (y + 2)2]1.5 + 60 ⇥ 3
[13 + (y 4)2]1.5
13
space. Find the total force on the charge at A.
The force will be:
F = (50 ⇥ 10 9)2
4⇡✏0
RCA
|RCA|3 + RDA
|RDA|3 + RBA
|RBA|3
where RCA = ax ay , RDA = ax + ay , and RBA = 2ax. The magnitudes are |RCA| = |RDA| = p2,
and |RBA| = 2. Substituting these leads to
F = (50 ⇥ 10 9)2
4⇡✏0
1
2p2 + 1
2p2 + 2
8
ax = 21.5ax μN
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:
Enet(a, a, a) = q
4⇡✏0a2
2
6
6
6
4
ax + ay + az
3p3
| {z }
(0,0,0)
+ ay + az
2p2
| {z }
(a,0,0)
+ ax + az
2p2
| {z }
(0,a,0)
+ ax + ay
2p2
| {z }
(0,0,a)
+ ax
|{z}
(0,a,a)
+ ay
|{z}
(a,0,a)
+ az
|{z}
(a,a,0)
3
7
7
7
5
The force is now the product of this field and the charge at (a, a, a). Simplifying, we obtain
F(a, a, a) = qEnet(a, a, a) = q2
4⇡✏0a2
1
3p3 + 1
p2 + 1
(ax + ay + az ) = 1.90 q2
4⇡✏0a2 (ax + ay + az )
in which the magnitude is |F| = 3.29 q2/(4⇡✏0a2).
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 ✏ = ✏0, find E at P3(1, 2, 3): This field will be
E = 10 9
4⇡✏0
25R13
|R13|3 + 60R23
|R23|3
where R13 = 3ax + 4ay 4az and R23 = 4ax 2ay + 5az . Also, |R13| = p41 and |R23| = p45.
So
E = 10 9
4⇡✏0
25 ⇥ ( 3ax + 4ay 4az )
(41)1.5 + 60 ⇥ (4ax 2ay + 5az )
(45)1.5
= 4.58ax 0.15ay + 5.51az
b) At what point on the y axis is Ex = 0? P3 is now at (0, y, 0), so R13 = 4ax + (y + 2)ay 7az
and R23 = 3ax + (y 4)ay + 2az . Also, |R13| = p65 + (y + 2)2 and |R23| = p13 + (y 4)2.
Now the x component of E at the new P3 will be:
Ex = 10 9
4⇡✏0
25 ⇥ ( 4)
[65 + (y + 2)2]1.5 + 60 ⇥ 3
[13 + (y 4)2]1.5
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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:
0.48y2 + 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 = ±d/2.
a) find the electric field everywhere on the z axis: For a point charge at any location, we have
E = q(r r0)
4⇡✏0|r r0|3
In the case of two charges, we would therefore have
ET = q1(r r0
1)
4⇡✏0|r r0
1|3 + q2(r r0
2)
4⇡✏0|r r0
2|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 r0
1 = (d/2)az , and r0
2 = (d/2)az Therefore
r r0
1 = [z (d/2)]az , r r0
2 = [z + (d/2)]az ,
then
|r r1|3 = [z (d/2)]3 and |r r2|3 = [z + (d/2)]3
Substitute these results into (1) to obtain:
ET (z) = q
4⇡✏0
1
[z (d/2)]2 + 1
[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
ET (x) = q
4⇡✏0
xax (d/2)az
|xax (d/2)az |3 + xax + (d/2)az
|xax + (d/2)az |3
(3)
where
|xax (d/2)az | = |xax + (d/2)az | = ⇥x2 + (d/2)2⇤1/2
Therefore (3) becomes
ET (x) = 2qx ax
4⇡✏0 [x2 + (d/2)2]3/2
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):
ET (z) = q
4⇡✏0
1
[z (d/2)]2 1
[z + (d/2)]2
az V/m
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
4⇡✏0 [x2 + (d/2)2]3/2
14
expression simplifies to the following quadratic:
0.48y2 + 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 = ±d/2.
a) find the electric field everywhere on the z axis: For a point charge at any location, we have
E = q(r r0)
4⇡✏0|r r0|3
In the case of two charges, we would therefore have
ET = q1(r r0
1)
4⇡✏0|r r0
1|3 + q2(r r0
2)
4⇡✏0|r r0
2|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 r0
1 = (d/2)az , and r0
2 = (d/2)az Therefore
r r0
1 = [z (d/2)]az , r r0
2 = [z + (d/2)]az ,
then
|r r1|3 = [z (d/2)]3 and |r r2|3 = [z + (d/2)]3
Substitute these results into (1) to obtain:
ET (z) = q
4⇡✏0
1
[z (d/2)]2 + 1
[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
ET (x) = q
4⇡✏0
xax (d/2)az
|xax (d/2)az |3 + xax + (d/2)az
|xax + (d/2)az |3
(3)
where
|xax (d/2)az | = |xax + (d/2)az | = ⇥x2 + (d/2)2⇤1/2
Therefore (3) becomes
ET (x) = 2qx ax
4⇡✏0 [x2 + (d/2)2]3/2
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):
ET (z) = q
4⇡✏0
1
[z (d/2)]2 1
[z + (d/2)]2
az V/m
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
4⇡✏0 [x2 + (d/2)2]3/2
14
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2.7. A 2 μC point charge is located at A(4, 3, 5) in free space. Find E⇢, E , and Ez at P (8, 12, 2). Have
EP = 2 ⇥ 10 6
4⇡✏0
RAP
|RAP |3 = 2 ⇥ 10 6
4⇡✏0
4ax + 9ay 3az
(106)1.5
= 65.9ax + 148.3ay 49.4az
Then, at point P , ⇢ = p82 + 122 = 14.4, = tan 1(12/8) = 56.3 , and z = z. Now,
E⇢ = Ep · a⇢ = 65.9(ax · a⇢) + 148.3(ay · a⇢) = 65.9 cos(56.3 ) + 148.3 sin(56.3 ) = 159.7
and
E = Ep · a = 65.9(ax · a ) + 148.3(ay · a ) = 65.9 sin(56.3 ) + 148.3 cos(56.3 ) = 27.4
Finally, Ez = 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
4⇡✏0(d x)2 = kx
from which Q = 2(d x)p⇡✏0kx.
b) Determine the maximum charge that can be measured in terms of ✏0, 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
charge: Qmax = 4ap⇡✏0k(d 2a). Presumably some form of stop mechanism is placed at
x = x
max 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 ⇥ 10 9
4⇡✏0
RAP
|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
Ex = 100 ⇥ 10 9
4⇡✏0
(x + 1)
[(x + 1)2 + (y 1)2 + (z 3)2]1.5
= 500 V/m
And so our condition becomes:
(x + 1) = 0.56 [(x + 1)2 + (y 1)2 + (z 3)2]1.5
15
EP = 2 ⇥ 10 6
4⇡✏0
RAP
|RAP |3 = 2 ⇥ 10 6
4⇡✏0
4ax + 9ay 3az
(106)1.5
= 65.9ax + 148.3ay 49.4az
Then, at point P , ⇢ = p82 + 122 = 14.4, = tan 1(12/8) = 56.3 , and z = z. Now,
E⇢ = Ep · a⇢ = 65.9(ax · a⇢) + 148.3(ay · a⇢) = 65.9 cos(56.3 ) + 148.3 sin(56.3 ) = 159.7
and
E = Ep · a = 65.9(ax · a ) + 148.3(ay · a ) = 65.9 sin(56.3 ) + 148.3 cos(56.3 ) = 27.4
Finally, Ez = 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
4⇡✏0(d x)2 = kx
from which Q = 2(d x)p⇡✏0kx.
b) Determine the maximum charge that can be measured in terms of ✏0, 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
charge: Qmax = 4ap⇡✏0k(d 2a). Presumably some form of stop mechanism is placed at
x = x
max 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 ⇥ 10 9
4⇡✏0
RAP
|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
Ex = 100 ⇥ 10 9
4⇡✏0
(x + 1)
[(x + 1)2 + (y 1)2 + (z 3)2]1.5
= 500 V/m
And so our condition becomes:
(x + 1) = 0.56 [(x + 1)2 + (y 1)2 + (z 3)2]1.5
15
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Subject
Electrical Engineering & Electronics