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CHAPTER 1
1.1. Given the vectors M = − 10 a x + 4 a y − 8 a z and N = 8 a x + 7 a y − 2 a z, find: a) a unit vector in the direction of − M + 2 N.
− M + 2 N = 10 a x − 4 a y + 8 a z + 16 a x + 14 a y − 4 a z = ( 26 , 10 , 4 ) Thus a =
b) the magnitude of 5 a x + N − 3 M :
( 5 , 0 , 0 ) + ( 8 , 7 , − 2 ) − (− 30 , 12 , − 24 ) = ( 43 , − 5 , 22 ), and |( 43 , − 5 , 22 )| = 48 .6.
c) | M || 2 N |( M + N ): |(− 10 , 4 , − 8 )||( 16 , 14 , − 4 )|(− 2 , 11 , − 10 ) = ( 13. 4 )( 21. 6 )(− 2 , 11 , − 10 ) = (− 580. 5 , 3193 , − 2902 )
1.2. Given three points, A( 4 , 3 , 2 ), B(− 2 , 0 , 5 ), and C( 7 , − 2 , 1 ):
a) Specify the vector A extending from the origin to the point A.
A = ( 4 , 3 , 2 ) = 4 a x + 3 a y + 2 a z
b) Give a unit vector extending from the origin to the midpoint of line AB. The vector from the origin to the midpoint is given by M = ( 1 / 2 )( A + B ) = ( 1 / 2 )( 4 − 2 , 3 + 0 , 2 + 5 ) = ( 1 , 1. 5 , 3. 5 ) The unit vector will be
m =
c) Calculate the length of the perimeter of triangle ABC: Begin with AB = (− 6 , − 3 , 3 ), BC = ( 9 , − 2 , − 4 ), CA = ( 3 , − 5 , − 1 ). Then
| AB | + | BC | + | CA | = 7. 35 + 10. 05 + 5. 91 = 23. 32
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 = 13 B( 2 , − 2 , 1 ), we use the fact that | B − A | = 10, or |( 6 − 23 B) a x − ( 2 − 23 B) a y − ( 4 + 13 B) a z| = 10 Expanding, obtain 36 − 8 B + 49 B^2 + 4 − 83 B + 49 B^2 + 16 + 83 B + 19 B^2 = 100 or B^2 − 8 B − 44 = 0. Thus B = 8 ±
√ 64 − 176 2 =^11 .75 (taking positive option) and so
B =
( 11. 75 ) a x −
( 11. 75 ) a y +
( 11. 75 ) a z = 7. 83 a x − 7. 83 a y + 3. 92 a z
1.4. given points A( 8 , − 5 , 4 ) and B(− 2 , 3 , 2 ), find:
a) the distance from A to B.
| B − A | = |(− 10 , 8 , − 2 )| = 12. 96
b) a unit vector directed from A towards B. This is found through
a AB =
B − A
| B − A |
c) a unit vector directed from the origin to the midpoint of the line AB.
a 0 M =
( A + B )/ 2
|( A + B )/ 2 |
d) the coordinates of the point on the line connecting A to B at which the line intersects the plane z = 3. Note that the midpoint, ( 3 , − 1 , 3 ), as determined from part c happens to have z coordinate of 3. This is the point we are looking for.
1.5. A vector field is specified as G = 24 xy a x + 12 (x^2 + 2 ) a y + 18 z^2 a z. 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
a G =
c) a unit vector directed from Q toward P :
a QP =
P − Q
| P − Q |
d) the equation of the surface on which | G | = 60: We write 60 = |( 24 xy, 12 (x^2 + 2 ), 18 z^2 )|, or 10 = |( 4 xy, 2 x^2 + 4 , 3 z^2 )|, so the equation is
100 = 16 x^2 y^2 + 4 x^4 + 16 x^2 + 16 + 9 z^4
1.9. A field is given as G =
(x^2 + y^2 )
(x a x + y a y )
Find: a) a unit vector in the direction of G at P ( 3 , 4 , − 2 ): Have G p = 25 /( 9 + 16 ) × ( 3 , 4 , 0 ) = 3 a x + 4 a y , and | G p| = 5. Thus a G = ( 0. 6 , 0. 8 , 0 ). b) the angle between G and a x at P : The angle is found through a G · a x = cos θ. So cos θ = ( 0. 6 , 0. 8 , 0 ) · ( 1 , 0 , 0 ) = 0 .6. Thus θ = 53 ◦. c) the value of the following double integral on the plane y = 7: ∫ (^4)
0
0
G · a y dzdx
0
0
x^2 + y^2
(x a x + y a y ) · a y dzdx =
0
0
x^2 + 49
× 7 dzdx =
0
x^2 + 49
dx
= 350 ×
[
tan−^1
]
1.10. Use the definition of the dot product to find the interior angles at A and B of the triangle defined by the three points A( 1 , 3 , − 2 ), B(− 2 , 4 , 5 ), and C( 0 , − 2 , 1 ): a) Use R AB = (− 3 , 1 , 7 ) and R AC = (− 1 , − 5 , 3 ) to form R AB · R AC = | R AB || R AC | cos θA. Obtain 3 + 5 + 21 =
35 cos θ (^) A. Solve to find θ (^) A = 65. 3 ◦. b) Use R BA = ( 3 , − 1 , − 7 ) and R BC = ( 2 , − 6 , − 4 ) to form R BA · R BC = | R BA|| R BC | cos θB. Obtain 6 + 6 + 28 =
56 cos θ (^) B. Solve to find θ (^) B = 45. 9 ◦.
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 R MN : R MN = (− 0. 2 , 0. 1 , 0. 3 ) − ( 0. 1 , − 0. 2 , − 0. 1 ) = (− 0. 3 , 0. 3 , 0. 4 ). b) the dot product R MN · R MP : R MP = ( 0. 4 , 0 , 0. 1 ) − ( 0. 1 , − 0. 2 , − 0. 1 ) = ( 0. 3 , 0. 2 , 0. 2 ). R MN · R MP = (− 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 R MN on R MP :
R MN · a RMP = (− 0. 3 , 0. 3 , 0. 4 ) ·
d) the angle between R MN and R MP :
θ (^) M = cos−^1
R MN · R MP
| R MN || R MP |
= cos−^1
1.12. Given points A( 10 , 12 , − 6 ), B( 16 , 8 , − 2 ), C( 8 , 1 , − 4 ), and D(− 2 , − 5 , 8 ), determine:
a) the vector projection of R AB + R BC on R AD : R AB + R BC = R AC = ( 8 , 1 , 4 ) − ( 10 , 12 , − 6 ) = (− 2 , − 11 , 10 ) Then R AD = (− 2 , − 5 , 8 ) − ( 10 , 12 , − 6 ) = (− 12 , − 17 , 14 ). So the projection will be:
( R AC · a RAD ) a RAD =
[
]
b) the vector projection of R AB + R BC on R DC : R DC = ( 8 , − 1 , 4 ) − (− 2 , − 5 , 8 ) = ( 10 , 6 , − 4 ). The projection is:
( R AC · a RDC ) a RDC =
[
]
c) the angle between R DA and R DC : Use R DA = − R AD = ( 12 , 17 , − 14 ) and R DC = ( 10 , 6 , − 4 ). The angle is found through the dot product of the associated unit vectors, or:
θD = cos−^1 ( a RDA · a RDC ) = cos−^1
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 =
b) Find the vector component of F that is perpendicular to G :
F pG = 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 :
G pF = G − G ||F = G −
G · F
| F |^2
F = ( 0. 1 , 0. 2 , 0. 3 ) −
1.14. The four vertices of a regular tetrahedron are located at O( 0 , 0 , 0 ), A( 0 , 1 , 0 ), B( 0. 5
3 , 0. 5 , 0 ), and C(
a) Find a unit vector perpendicular (outward) to the face ABC: First find
R BA × R BC = [( 0 , 1 , 0 ) − ( 0. 5
3 , 0. 5 , 0 )] × [(
3 , 0. 5 , 0 )]
3 , 0. 5 , 0 ) × (−
The required unit vector will then be:
R BA × R BC | R BA × R BC |
b) Find the area of the face ABC:
Area =
| R BA × R BC | = 0. 43
1.17c. (continued) Now
1 2
( a AM + a AN ) =
[( 0. 697 , 0. 627 , − 0. 348 ) + (− 0. 507 , 0. 406 , 0. 761 )] = ( 0. 095 , 0. 516 , 0. 207 )
Finally, a bis =
1.18. Given points A(ρ = 5 , φ = 70 ◦, z = − 3 ) and B(ρ = 2 , φ = − 30 ◦, z = 1 ), find: a) unit vector in cartesian coordinates at A toward B: A(5 cos 70◦, 5 sin 70◦, − 3 ) = A( 1. 71 , 4. 70 , − 3 ), In the same manner, B( 1. 73 , − 1 , 1 ). So R AB = ( 1. 73 , − 1 , 1 ) − ( 1. 71 , 4. 70 , − 3 ) = ( 0. 02 , − 5. 70 , 4 ) and therefore a AB =
b) a vector in cylindrical coordinates at A directed toward B: a AB · a ρ = 0 .003 cos 70◦^ − 0 .82 sin 70◦^ = − 0 .77. a AB · a φ = − 0 .003 sin 70◦^ − 0 .82 cos 70◦^ = − 0 .28. Thus
a AB = − 0. 77 a ρ − 0. 28 a φ + 0. 57 a z
.
c) a unit vector in cylindrical coordinates at B directed toward A: Use a BA = (− 0 , 003 , 0. 82 , − 0. 57 ). Then a BA · a ρ = − 0 .003 cos(− 30 ◦) + 0 .82 sin(− 30 ◦) = − 0 .43, and a BA · a φ = 0 .003 sin(− 30 ◦) + 0 .82 cos(− 30 ◦) = 0 .71. Finally,
a BA = − 0. 43 a ρ + 0. 71 a φ − 0. 57 a z
1.19 a) Express the field D = (x^2 + y^2 )−^1 (x a x + y a y ) in cylindrical components and cylindrical variables: Have x = ρ cos φ, y = ρ sin φ, and x^2 + y^2 = ρ^2. Therefore
D =
ρ
(cos φ a x + sin φ a y )
Then Dρ = D · a ρ =
ρ
[
cos φ( a x · a ρ ) + sin φ( a y · a ρ )
]
ρ
[
cos^2 φ + sin 2 φ
]
ρ and
Dφ = D · a φ =
ρ
[
cos φ( a x · a φ ) + sin φ( a y · a φ )
]
ρ
[cos φ(− sin φ) + sin φ cos φ] = 0
Therefore D =
ρ
a ρ
1.19b. Evaluate D at the point where ρ = 2, φ = 0. 2 π, and z = 5, expressing the result in cylindrical and cartesian coordinates: At the given point, and in cylindrical coordinates, D = 0. 5 a ρ. To express this in cartesian, we use
D = 0. 5 ( a ρ · a x ) a x + 0. 5 ( a ρ · a y ) a y = 0 .5 cos 36◦ a x + 0 .5 sin 36◦ a y = 0. 41 a x + 0. 29 a y
1.20. Express in cartesian components: a) the vector at A(ρ = 4 , φ = 40 ◦, z = − 2 ) that extends to B(ρ = 5 , φ = − 110 ◦, z = 2 ): We have A(4 cos 40◦, 4 sin 40◦, − 2 ) = A( 3. 06 , 2. 57 , − 2 ), and B(5 cos(− 110 ◦), 5 sin(− 110 ◦), 2 ) = B(− 1. 71 , − 4. 70 , 2 ) in cartesian. Thus R AB = (− 4. 77 , − 7. 30 , 4 ). b) a unit vector at B directed toward A: Have R BA = ( 4. 77 , 7. 30 , − 4 ), and so
a BA =
c) a unit vector at B directed toward the origin: Have r B = (− 1. 71 , − 4. 70 , 2 ), and so − r B = ( 1. 71 , 4. 70 , − 2 ). Thus
a =
1.21. Express in cylindrical components: a) the vector from C( 3 , 2 , − 7 ) to D(− 1 , − 4 , 2 ): C( 3 , 2 , − 7 ) → C(ρ = 3. 61 , φ = 33. 7 ◦, z = − 7 ) and D(− 1 , − 4 , 2 ) → D(ρ = 4. 12 , φ = − 104. 0 ◦, z = 2 ). Now R CD = (− 4 , − 6 , 9 ) and Rρ = R CD · a ρ = −4 cos( 33. 7 ) − 6 sin( 33. 7 ) = − 6 .66. Then Rφ = R CD · a φ = 4 sin( 33. 7 ) − 6 cos( 33. 7 ) = − 2 .77. So R CD = − 6. 66 a ρ − 2. 77 a φ + 9 a z b) a unit vector at D directed toward C: R CD = ( 4 , 6 , − 9 ) and Rρ = R DC · a ρ = 4 cos(− 104. 0 ) + 6 sin(− 104. 0 ) = − 6 .79. Then Rφ = R DC · a φ = 4[− sin(− 104. 0 )] + 6 cos(− 104. 0 ) = 2 .43. So R DC = − 6. 79 a ρ + 2. 43 a φ − 9 a z Thus a DC = − 0. 59 a ρ + 0. 21 a φ − 0. 78 a z c) a unit vector at D directed toward the origin: Start with r D = (− 1 , − 4 , 2 ), and so the vector toward the origin will be − r D = ( 1 , 4 , − 2 ). Thus in cartesian the unit vector is a = ( 0. 22 , 0. 87 , − 0. 44 ). Convert to cylindrical: aρ = ( 0. 22 , 0. 87 , − 0. 44 ) · a ρ = 0 .22 cos(− 104. 0 ) + 0 .87 sin(− 104. 0 ) = − 0 .90, and aφ = ( 0. 22 , 0. 87 , − 0. 44 ) · a φ = 0 .22[− sin(− 104. 0 )] + 0 .87 cos(− 104. 0 ) = 0, so that finally, a = − 0. 90 a ρ − 0. 44 a z.
1.22. A field is given in cylindrical coordinates as
F =
[
ρ^2 + 1
]
a ρ + 3 (cos φ − sin φ) a φ − 2 a z
where the magnitude of F is found to be:
| F | =
F · F =
[
(ρ^2 + 1 )^2
ρ^2 + 1
(cos φ + sin φ) + 22
] 1 / 2
1.23. The surfaces ρ = 3, ρ = 5, φ = 100 ◦, φ = 130 ◦, z = 3, and z = 4 .5 define a closed surface. a) Find the enclosed volume:
Vol =
3
100 ◦
3
ρ dρ dφ dz = 6. 28
NOTE: The limits on the φ integration must be converted to radians (as was done here, but not shown).
b) Find the total area of the enclosing surface:
Area = 2
100 ◦
3
ρ dρ dφ +
3
100 ◦
3 dφdz
3
100 ◦
5 dφdz + 2
3
3
dρ dz = 20. 7
c) Find the total length of the twelve edges of the surfaces:
Length = 4 × 1. 5 + 4 × 2 + 2 ×
[
360 ◦^
× 2 π × 3 +
360 ◦^
× 2 π × 5
]
d) Find the length of the longest straight line that lies entirely within the volume: This will be between the points A(ρ = 3, φ = 100 ◦, z = 3) and B(ρ = 5, φ = 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. At point P (− 3 , 4 , 5 ), express the vector that extends from P to Q( 2 , 0 , − 1 ) in:
a) rectangular coordinates. R P Q = Q − P = 5 a x − 4 a y − 6 a z
Then | R P Q| =
b) cylindrical coordinates. At P , ρ = 5, φ = tan−^1 ( 4 / − 3 ) = − 53. 1 ◦, and z = 5. Now,
R P Q · a ρ = ( 5 a x − 4 a y − 6 a z ) · a ρ = 5 cos φ − 4 sin φ = 6. 20
R P Q · a φ = ( 5 a x − 4 a y − 6 a z ) · a φ = −5 sin φ − 4 cos φ = 1. 60 Thus R P Q = 6. 20 a ρ + 1. 60 a φ − 6 a z
and | R P Q| =
c) spherical coordinates. At P , r =
50 = 7 .07, θ = cos−^1 ( 5 / 7. 07 ) = 45 ◦, and φ = tan−^1 ( 4 / − 3 ) = − 53. 1 ◦.
R P Q · a r = ( 5 a x − 4 a y − 6 a z ) · a r = 5 sin θ cos φ − 4 sin θ sin φ − 6 cos θ = 0. 14
R P Q · a θ = ( 5 a x − 4 a y − 6 a z ) · a θ = 5 cos θ cos φ − 4 cos θ sin φ − (− 6 ) sin θ = 8. 62 R P Q · a φ = ( 5 a x − 4 a y − 6 a z ) · a φ = −5 sin φ − 4 cos φ = 1. 60
1.24. (continued)
Thus R P Q = 0. 14 a r + 8. 62 a θ + 1. 60 a φ
and | R P Q| =
d) Show that each of these vectors has the same magnitude. Each does, as shown above.
1.25. Given point P (r = 0. 8 , θ = 30 ◦, φ = 45 ◦), and
E =
r^2
cos φ a r +
sin φ sin θ
a φ
a) Find E at P : E = 1. 10 a ρ + 2. 21 a φ. b) Find | E | at P : | E | =
c) Find a unit vector in the direction of E at P :
a E =
E
| E |
= 0. 45 a r + 0. 89 a φ
1.26. a) Determine an expression for a y in spherical coordinates at P (r = 4 , θ = 0. 2 π, φ = 0. 8 π): Use a y · a r = sin θ sin φ = 0 .35, a y · a θ = cos θ sin φ = 0 .48, and a y · a φ = cos φ = − 0 .81 to obtain
a y = 0. 35 a r + 0. 48 a θ − 0. 81 a φ
b) Express a r in cartesian components at P : Find x = r sin θ cos φ = − 1 .90, y = r sin θ sin φ = 1 .38, and z = r cos θ = − 3 .24. Then use a r · a x = sin θ cos φ = − 0 .48, a r · a y = sin θ sin φ = 0 .35, and a r · a z = cos θ = 0 .81 to obtain
a r = − 0. 48 a x + 0. 35 a y + 0. 81 a z
1.27. The surfaces r = 2 and 4, θ = 30 ◦^ and 50◦, and φ = 20 ◦^ and 60◦^ identify a closed surface. a) Find the enclosed volume: This will be
Vol =
20 ◦
30 ◦
2
r^2 sin θdrdθdφ = 2. 91
where degrees have been converted to radians. b) Find the total area of the enclosing surface:
Area =
20 ◦
30 ◦
( 4 2 + 2 2 ) sin θdθdφ +
2
20 ◦
r(sin 30◦^ + sin 50◦)drdφ
30 ◦
2
rdrdθ = 12. 61
c) Find the total length of the twelve edges of the surface:
Length = 4
2
dr + 2
30 ◦
( 4 + 2 )dθ +
20 ◦
(4 sin 50◦^ + 4 sin 30◦^ + 2 sin 50◦^ + 2 sin 30◦)dφ
= 17. 49
1.29 (continued) Express the unit vector a x in spherical components at the point:
b) x = 3, y = 2, z = −1: First, transform the point to spherical coordinates. Have r =
θ = cos−^1 (− 1 /
14 ) = 105. 5 ◦, and φ = tan−^1 ( 2 / 3 ) = 33. 7 ◦. Then
a x = sin( 105. 5 ◦) cos( 33. 7 ◦) a r + cos( 105. 5 ◦) cos( 33. 7 ◦) a θ + (− sin( 33. 7 ◦)) a φ = 0. 80 a r − 0. 22 a θ − 0. 55 a φ
c) ρ = 2 .5, φ = 0 .7 rad, z = 1 .5: Again, convert the point to spherical coordinates. r =
√ ρ^2 +^ z^2 = 8 .5, θ = cos−^1 (z/r) = cos−^1 ( 1. 5 /
- 5 ) = 59. 0 ◦, and φ = 0 .7 rad = 40. 1 ◦. Now
a x = sin( 59 ◦) cos( 40. 1 ◦) a r + cos( 59 ◦) cos( 40. 1 ◦) a θ + (− sin( 40. 1 ◦)) a φ = 0. 66 a r + 0. 39 a θ − 0. 64 a φ
1.30. Given A(r = 20 , θ = 30 ◦, φ = 45 ◦) and B(r = 30 , θ = 115 ◦, φ = 160 ◦), find:
a) | R AB |: First convert A and B to cartesian: Have xA = 20 sin( 30 ◦) cos( 45 ◦) = 7 .07, yA = 20 sin( 30 ◦) sin( 45 ◦) = 7 .07, and z (^) A = 20 cos( 30 ◦) = 17 .3. x (^) B = 30 sin( 115 ◦) cos( 160 ◦) = − 25 .6, y (^) B = 30 sin( 115 ◦) sin( 160 ◦) = 9 .3, and zB = 30 cos( 115 ◦) = − 12 .7. Now R AB = R B − R A = (− 32. 6 , 2. 2 , − 30. 0 ), and so | R AB | = 44 .4.
b) | R AC |, given C(r = 20 , θ = 90 ◦, φ = 45 ◦). Again, converting C to cartesian, obtain x (^) C = 20 sin( 90 ◦) cos( 45 ◦) = 14 .14, y (^) C = 20 sin( 90 ◦) sin( 45 ◦) = 14 .14, and z (^) C = 20 cos( 90 ◦) = 0. So R AC = R C − R A = ( 7. 07 , 7. 07 , − 17. 3 ), and | R AC | = 20 .0.
c) the distance from A to C on a great circle path: Note that A and C share the same r and φ coordinates; thus moving from A to C involves only a change in θ of 60◦. The requested arc length is then
distance = 20 ×
[
2 π 360
)]
CHAPTER 2
2.1. Four 10nC positive charges are located in the z = 0 plane at the corners of a square 8cm on a side. A fifth 10nC positive charge is located at a point 8cm distant from the other charges. Calculate the magnitude of the total force on this fifth charge for = 0 :
Arrange the charges in the xy plane at locations (4,4), (4,-4), (-4,4), and (-4,-4). Then the fifth charge will be on the z axis at location z = 4
2, which puts it at 8cm distance from the other four. By symmetry, the force on the fifth charge will be z-directed, and will be four times the z component of force produced by each of the four other charges.
F =
×
q^2 4 π 0 d^2
×
( 10 −^8 )^2
4 π( 8. 85 × 10 −^12 )( 0. 08 )^2
= 4. 0 × 10 −^4 N
2.2. A charge Q 1 = 0. 1 μC is located at the origin, while Q 2 = 0. 2 μC is at A( 0. 8 , − 0. 6 , 0 ). Find the locus of points in the z = 0 plane at which the x component of the force on a third positive charge is zero.
To solve this problem, the z coordinate of the third charge is immaterial, so we can place it in the xy plane at coordinates (x, y, 0 ). We take its magnitude to be Q 3. The vector directed from the first charge to the third is R 13 = x a x + y a y ; the vector directed from the second charge to the third is R 23 = (x − 0. 8 ) a x + (y + 0. 6 ) a y. The force on the third charge is now
F 3 =
Q 3
4 π 0
[
Q 1 R 13
| R 13 |^3
Q 2 R 23
| R 23 |^3
]
Q 3 × 10 −^6
4 π 0
[
- 1 (x a x + y a y ) (x^2 + y^2 )^1.^5
0 .2[(x − 0. 8 ) a x + (y + 0. 6 ) a y ] [(x − 0. 8 )^2 + (y + 0. 6 )^2 ] 1.^5
]
We desire the x component to be zero. Thus,
[
- 1 x a x (x^2 + y^2 )^1.^5
- 2 (x − 0. 8 ) a x [(x − 0. 8 )^2 + (y + 0. 6 )^2 ]^1.^5
]
or x[(x − 0. 8 )^2 + (y + 0. 6 )^2 ]^1.^5 = 2 ( 0. 8 − x)(x^2 + y^2 )^1.^5
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 π 0
[
R CA
| R CA|^3
R DA
| R DA|^3
R BA
| R BA|^3
]
where R CA = a x − a y , R DA = a x + a y , and R BA = 2 a x. The magnitudes are | R CA| = | R DA| =
and | R BA| = 2. Substituting these leads to
F =
( 50 × 10 −^9 )^2
4 π 0
[
]
a x = 21. 5 a x μN
where distances are in meters.
2.6. Point charges of 120 nC are located at A( 0 , 0 , 1 ) and B( 0 , 0 , − 1 ) in free space.
a) Find E at P ( 0. 5 , 0 , 0 ): This will be
E P =
120 × 10 −^9
4 π 0
[
R AP
| R AP |^3
R BP
| R BP |^3
]
where R AP = 0. 5 a x − a z and R BP = 0. 5 a x + a z. Also, | R AP | = | R BP | =
1 .25. Thus:
E P =
120 × 10 −^9 a x 4 π( 1. 25 )^1.^5 0
= 772 V/m
b) What single charge at the origin would provide the identical field strength? We require Q 0 4 π 0 ( 0. 5 )^2
from which we find Q 0 = 21 .5 nC.
2.7. A 2 μC point charge is located at A( 4 , 3 , 5 ) in free space. Find Eρ , E (^) φ , and Ez at P ( 8 , 12 , 2 ). Have
E P =
2 × 10 −^6
4 π 0
R AP
| R AP |^3
2 × 10 −^6
4 π 0
[
4 a x + 9 a y − 3 a z ( 106 )^1.^5
]
= 65. 9 a x + 148. 3 a y − 49. 4 a z
Then, at point P , ρ =
8 2 + 12 2 = 14 .4, φ = tan−^1 ( 12 / 8 ) = 56. 3 ◦, and z = z. Now,
Eρ = E p · a ρ = 65. 9 ( a x · a ρ ) + 148. 3 ( a y · a ρ ) = 65 .9 cos( 56. 3 ◦) + 148 .3 sin( 56. 3 ◦) = 159. 7
and
Eφ = E p · a φ = 65. 9 ( a x · a φ ) + 148. 3 ( a y · a φ ) = − 65 .9 sin( 56. 3 ◦) + 148 .3 cos( 56. 3 ◦) = 27. 4
Finally, Ez = − 49. 4
2.8. Given point charges of − 1 μC at P 1 ( 0 , 0 , 0. 5 ) and P 2 ( 0 , 0 , − 0. 5 ), and a charge of 2 μC at the origin, find E at P ( 0 , 2 , 1 ) in spherical components, assuming = 0.
The field will take the general form:
E P =
10 −^6
4 π 0
[
R 1
| R 1 |^3
2 R 2
| R 2 |^3
R 3
| R 3 |^3
]
where R 1 , R 2 , R 3 are the vectors to P from each of the charges in their original listed order. Specifically, R 1 = ( 0 , 2 , 0. 5 ), R 2 = ( 0 , 2 , 1 ), and R 3 = ( 0 , 2 , 1. 5 ). The magnitudes are | R 1 | = 2 .06, | R 2 | = 2 .24, and | R 3 | = 2 .50. Thus
E P =
10 −^6
4 π 0
[
( 2. 06 )^3
( 2. 24 )^3
( 2. 50 )^3
]
= 89. 9 a y + 179. 8 a z
Now, at P , r =
5, θ = cos−^1 ( 1 /
5 ) = 63. 4 ◦, and φ = 90 ◦. So
E (^) r = E P · a r = 89. 9 ( a y · a r ) + 179. 8 ( a z · a r ) = 89 .9 sin θ sin φ + 179 .8 cos θ = 160. 9
Eθ = E P · a θ = 89. 9 ( a y · a θ ) + 179. 8 ( a z · a θ ) = 89 .9 cos θ sin φ + 179. 8 (− sin θ) = − 120. 5 Eφ = E P · a φ = 89. 9 ( a y · a φ ) + 179. 8 ( a z · a φ ) = 89 .9 cos φ = 0
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:
E P =
100 × 10 −^9
4 π 0
R AP
| R AP |^3
where R AP = (x + 1 ) a x + (y − 1 ) a y + (z − 3 ) a z, and where | R AP | = [(x + 1 )^2 + (y − 1 )^2 + (z − 3 )^2 ]^1 /^2. The x component of the field will be
Ex =
100 × 10 −^9
4 π 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
b) Find y 1 if P (− 2 , y 1 , 3 ) lies on that locus: At point P , the condition of part a becomes
[
1 + (y 1 − 1 )^2
] 3
from which (y 1 − 1 )^2 = 0 .47, or y 1 = 1 .69 or 0. 31
2.10. Charges of 20 and -20 nC are located at ( 3 , 0 , 0 ) and (− 3 , 0 , 0 ), respectively. Let = 0. Determine | E | at P ( 0 , y, 0 ): The field will be
E P =
20 × 10 −^9
4 π 0
[
R 1
| R 1 |^3
R 2
| R 2 |^3
]
where R 1 , the vector from the positive charge to point P is (− 3 , y, 0 ), and R 2 , the vector from the negative charge to point√ P , is ( 3 , y, 0 ). The magnitudes of these vectors are | R 1 | = | R 2 | = 9 + y^2. Substituting these into the expression for E P produces
E P =
20 × 10 −^9
4 π 0
[
− 6 a x ( 9 + y^2 )^1.^5
]
from which | E P | =
( 9 + y^2 )^1.^5
V/m
2.11. A charge Q 0 located at the origin in free space produces a field for which Ez = 1 kV/m at point P (− 2 , 1 , − 1 ).
a) Find Q 0 : The field at P will be
E P =
Q 0
4 π 0
[
− 2 a x + a y − a z 61.^5
]
Since the z component is of value 1 kV/m, we find Q 0 = − 4 π 0 6 1.^5 × 10 3 = − 1. 63 μC.
2.14. Let
ρv = 5 e−^0.^1 ρ^ (π − |φ|)
z^2 + 10
μC/m 3
in the region 0 ≤ ρ ≤ 10, −π < φ < π, all z, and ρv = 0 elsewhere.
a) Determine the total charge present: This will be the integral of ρ (^) v over the region where it exists; specifically,
Q =
−∞
∫ (^) π
−π
0
5 e−^0.^1 ρ^ (π − |φ|)
z^2 + 10
ρ dρ dφ dz
which becomes
Q = 5
[
e−^0.^1 ρ ( 0. 1 )^2
] 10
0
−∞
∫ (^) π
0
(π − φ)
z^2 + 10
dφ dz
or Q = 5 × 26. 4
−∞
π^2
z^2 + 10
dz
Finally,
Q = 5 × 26. 4 × π^2
[
tan−^1
z √ 10
)]∞
−∞
5 ( 26. 4 )π^3 √ 10
= 1. 29 × 10 3 μC = 1 .29 mC
b) Calculate the charge within the region 0 ≤ ρ ≤ 4, −π/ 2 < φ < π/2, − 10 < z < 10: With the limits thus changed, the integral for the charge becomes:
Q′^ =
− 10
∫ (^) π/ 2
0
0
5 e−^0.^1 ρ^ (π − φ)
z^2 + 10
ρ dρ dφ dz
Following the same evaulation procedure as in part a, we obtain Q′^ = 0 .182 mC.
2.15. A spherical volume having a 2 μm radius contains a uniform volume charge density of 10 15 C/m 3.
a) What total charge is enclosed in the spherical volume? This will be Q = ( 4 / 3 )π( 2 × 10 −^6 )^3 × 10 15 = 3. 35 × 10 −^2 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
ρ (^) v,avg =
3. 35 × 10 −^2
( 0. 003 )^3
= 1. 24 × 10 6 C/m 3
2.16. The region in which 4 < r < 5, 0 < θ < 25 ◦, and 0. 9 π < φ < 1. 1 π contains the volume charge density of ρv = 10 (r − 4 )(r − 5 ) sin θ sin(φ/ 2 ). Outside the region, ρv = 0. Find the charge within the region: The integral that gives the charge will be
Q = 10
∫ (^1). 1 π
. 9 π
0
4
(r − 4 )(r − 5 ) sin θ sin(φ/ 2 ) r^2 sin θ dr dθ dφ
2.16. (continued) Carrying out the integral, we obtain
Q = 10
[
r^5 5
r^4 4
r^3 3
] 5
4
[
θ −
sin( 2 θ)
] 25 ◦
0
[
−2 cos
θ 2
)] 1. 1 π
. 9 π = 10 (− 3. 39 )(. 0266 )(. 626 ) = 0 .57 C
2.17. A uniform line charge of 16 nC/m is located along the line defined by y = −2, z = 5. If = 0 :
a) Find E at P ( 1 , 2 , 3 ): This will be E P =
ρ (^) l 2 π 0
R P
| R P |^2
where R P = ( 1 , 2 , 3 ) − ( 1 , − 2 , 5 ) = ( 0 , 4 , − 2 ), and | R P |^2 = 20. So
E P =
16 × 10 −^9
2 π 0
[
4 a y − 2 a z 20
]
= 57. 5 a y − 28. 8 a z V/m
b) Find E at that point in the z = 0 plane where the direction of E is given by ( 1 / 3 ) a y − ( 2 / 3 ) a z: With z = 0, the general field will be
E z= 0 =
ρ (^) l 2 π 0
[
(y + 2 ) a y − 5 a z (y + 2 )^2 + 25
]
We require |E (^) z| = −| 2 E (^) y |, so 2(y + 2 ) = 5. Thus y = 1 /2, and the field becomes:
E z= 0 =
ρ (^) l 2 π 0
[
- 5 a y − 5 a z ( 2. 5 )^2 + 25
]
= 23 a y − 46 a z
2.18. Uniform line charges of 0. 4 μC/m and − 0. 4 μC/m are located in the x = 0 plane at y = − 0 .6 and y = 0 .6 m respectively. Let = 0.
a) Find E at P (x, 0 , z): In general, we have
E P =
ρ (^) l 2 π 0
[
R +P
| R +P |
R −P
| R −P |
]
where R +P and R −P are, respectively, the vectors directed from the positive and negative line charges to the point P , and these are normal to the z axis. We thus have R +P = (x, 0 , z) − ( 0 , −. 6 , z) = (x,. 6 , 0 ), and R −P = (x, 0 , z) − ( 0 ,. 6 , z) = (x, −. 6 , 0 ). So
E P =
ρ (^) l 2 π 0
[
x a x + 0. 6 a y x^2 + ( 0. 6 )^2
x a x − 0. 6 a y x^2 + ( 0. 6 )^2
]
0. 4 × 10 −^6
2 π 0
[
- 2 a y x^2 + 0. 36
]
- 63 a y x^2 + 0. 36
kV/m
