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D1.1. Given points M (–1, 2, 1), N (3, –3, 0), and P (–2, –3, –4), find: ( a ) R MN ; ( b )
R MN + R MP ; ( c )
| r M |; ( d ) a MP ; ( e ) |2 r P – 3 r N |.
CHAPTER 1
125( x 1)
125( y 2)
( x^ 1)^ (^ y^ 2)^ (^ z^ 1)
2 2 2
( x 1) ( y 2) ( z 1)
2 2 2
125( z 1)
( x 1) ( y 2) ( z 1)
2 2 2
D1.3. The three vertices of a triangle are located at A (6, –1, 2), B (–2, 3, –4), and C (–3, 1, 5).
Find:
( a ) R AB ; ( b ) R AC ; ( c ) the angle θBAC at vertex A ; ( d ) the (vector) projection of R AB on R AC.
ELECTROMAGNETICS HAYT 8 EDITION
DRILL SOLUTION FROM CHAPTERS 1 TO 5
(a) R MN = (3 – (–1)) a x + (–3 – 2) a y + (0 – 1) a z = 4 a x – 5 a y – a z
(b) R MN + R MP = R MN + (–2 – (– 1)) a x + (–3 – 2) a y + (–4 – 1) a z
= R MN – a x – 5 a y – 5 a z = 4 a x – 5 a y – a z – a x – 5 a y – 5 a z
= 3 a x – 10 a y – 6 a z
(c) | r M | = = = 2.
(d) a MP =
a x 5 a y 5 a z
=
a x 5 a y 5 a z
= –0.14 a x – 0.7 a y – 0.7 a z
(e) 2 r P – 3 r N = 2(–2 a x – 3 a y – 4 a z ) – 3(3 a x – 3 a y ) = –4 a x – 6 a y – 8 a z – 9 a x + 9 a y
= –13 a x + 3 a y – 8 a z
|2 r P – 3 r N | = = 11 = 15.
(a) S =
{(2 1) a (^) (4 2) a (3 1) a }
2 (4 2)
2 (3 1)
2 x y z
( a 2 a 4 a )
=
( a
2 a 4 a )
2 2
2 4
2 x y z x y z
= 5.95 a x + 11.9 a y + 23.8 a z
(b) a S =^ 5.95 a x 11.9 a y 23.8 a z
5.95 a x 11.9 a y 23.8 a z
= 0.218 a x + 0.436 a y + 0.873 a z
(c) 1 = | S | =
Transposing,
( x 1)
2 ( y 2)
2 ( z 1)
2
( x 1)
2 ( y 2)
2 ( z 1)
2
Conjugating,
( x 1)
2 ( y 2)
2 ( z 1)
2
(a) R AB =^ (–2^ – 6) a x +^ (3 –^ (–1)) a y +^ (–4 –^ 2) a z
= –8 a x + 4 a y – 6 a z
(b) R AC =^ (–3 –^ 6) a x +^ (1^ – (–1)) a y +^ (5^ –^ 2) a z
= –9 a x + 2 a y + 3 a z
2 2
2 1
2 6
2 (5)
2 (5)
2
2 3
2 (8)
2 2
2 11.
2 23.
2
2 [( x 1)
2 ( y 2)
2 ( z 1)
2
[( x 1)
2 ( y 2)
2 ( z 1)
2 ]
2
125 ( x 1)
2 ( y 2)
2 ( z 1)
2
( x 1)
2 ( y 2)
2 ( z 1)
2
( x 1)
2 ( y 2)
2 ( z 1)
2
( x 1)
2 ( y 2)
2 ( z 1)
2
( x 1)
2 ( y 2)
2 ( z 1)
2
( x 1)
2 ( y 2)
2 ( z 1)
2
D1.2. A vector field S is expressed in rectangular coordinates as S = {125/[( x – 1)
2
2
( z + 1)2]}{( x – 1) a x + ( y – 2) a y + ( z + 1) a z }. ( a ) Evaluate S at P (2, 4, 3). ( b ) Determine a unit
vector (^) that gives the direction of S at P. ( c ) Specify the surface f ( x , y , z ) on which | S | = 1.
(c) Locating angle θBAC ,
R AB R AC = (–8)(–9) + 4(2) + (–6)3 = 72 + 8 – 18 = 62
Using dot product,
R AB R AC = | R AB || R AC | cos θBAC
(c) CD^ =^ (–3.1 – (–1.86)) a x +^ (2.6^ – (–3.99)) a y +^ (–3 –^ 2) a z
= –1.24 a x + 6.59 a y – 5 a z
2 6.
2 (5)
2
D1.6. Transform to cylindrical coordinates: ( a ) F = 10 a x – 8 a y + 6 a z at point P (10, –8, 6);
( b )
G = (2 x + y ) a x
- ( y – 4 x ) a y at point Q ( ρ , ϕ , z ). ( c ) Give the rectangular components of the vector
H = 20 a ρ – 10 a ϕ + 3 a z at P ( x = 5, y = 2, z = –1).
D1.7. Given the two points, C (–3, 2, 1) and D ( r = 5, θ = 20°, ϕ = –70°), find: ( a ) the spherical
coordinates of C ; ( b ) the rectangular coordinates of D ; ( c ) the distance from C to D.
| CD | = = 8.
(a) Taking note that the dot product of the unit vectors in cylindrical to rectangular coordinate
systems and vice versa
Table 3 Dot products of the unit vectors in cylindrical and rectangular coordinate systems
a ρ a ϕ a z
a x cos ϕ – sin ϕ 0
a y sin ϕ cos ϕ 0
a z 0 0 1
F = (10 a x – 8 a y + 6 a z ) a ρ + (10 a x – 8 a y + 6 a z ) a ϕ + (10 a x – 8 a y + 6 a z ) a z
= (10 cos ϕ – 8 sin ϕ + 0) a ρ + [(10(– sin ϕ ) + (–8)(cos ϕ ) + 0] a ϕ + (0 + 0 + 6) a z
Since ϕ = (tan
= (10 cos 141.43° – 8 sin 141.43°) a ρ + [(10(– sin 141.43°) + (–8)(cos 141.43°)] a ϕ + 6 a z
= 12.81 a ρ + 6 a z
(b) G^ =^ [(2 x^ +^ y ) a x – ( y^ –^4 x ) a y ]^ a ρ +^ [(2 x^ +^ y ) a x –^ ( y^ – 4 x ) a y ]^ a ϕ
Segregating the unit vectors to avoid confusion,
Gρ = [(2 x + y ) a x – ( y – 4 x ) a y ] a ρ
Converting also the variables of rectangular to cylindrical coordinates,
= (2 ρ cos ϕ + ρ sin ϕ ) cos ϕ – ( ρ sin ϕ – 4 ρ cos ϕ ) sin ϕ
= 2 ρ cos
2 ϕ + ρ sin ϕ cos ϕ – ρ sin
2 ϕ + 4 ρ sin ϕ cos ϕ
= 2 ρ cos
2 ϕ – ρ sin
2 ϕ + 5 ρ sin ϕ cos ϕ
Gϕ = [(2 x + y ) a x – ( y – 4 x ) a y ] a ϕ
= (2 ρ cos ϕ + ρ sin ϕ )(–sin ϕ ) – ( ρ sin ϕ – 4 ρ cos ϕ ) cos ϕ
= –2 ρ sin ϕ cos ϕ – ρ sin
2 ϕ – ρ sin ϕ cos ϕ + 4 ρ cos
2 ϕ
= –2 ρ sin ϕ cos ϕ – ρ sin
2 ϕ – ρ sin ϕ cos ϕ + 4 ρ cos
2 ϕ
= 4 ρ cos
2 ϕ – ρ sin
2 ϕ – 3 ρ sin ϕ cos ϕ
Adding together,
G = (2 ρ cos2 ϕ – ρ sin2 ϕ + 5 ρ sin ϕ cos ϕ ) a ρ + (4 ρ cos2 ϕ – ρ sin2 ϕ – 3 ρ sin ϕ cos ϕ ) a ρ
(c) Applying the dot product from Table 3,
H = (20 a ρ – 10 a ϕ + 3 a z ) a x + (20 a ρ – 10 a ϕ + 3 a z ) a y + (20 a ρ – 10 a ϕ + 3 a z ) a z
Since ϕ = (tan
Hx = 20 cos ϕ + 10 sin ϕ + 0 = 20 cos 201.8° + 10 sin 201.8° = 22.
H
y = 20 sin ϕ – 10 cos ϕ + 0 = 20 sin 201.8° – 10 cos 201.8° = –1.
Hz = 0 + 0 + 3 = 3
(a) Converting rectangular to spherical coordinates,
Table 4 Rectangular to spherical coordinate systems
r = x
2 y
2 z
2
θ = cos
^1
z
r
ϕ = tan
r = (3)
2 2
2 1
2
= 3.
θ = cos
ϕ = tan
C ( r = 3.74, θ = 74.5°, ϕ = 146.3°)
(b) Converting spherical to rectangular coordinates,
Table 5 Spherical to rectangular coordinate systems
x = r sin θ cos ϕ
y = r sin θ sin ϕ
z = r cos θ
2 0.
2 (0.15)
2
AO
= –0.25 a x + 0.3 a y – 0.15 a z m
| R AO | = m = 0.4183 m
E A =
QA
a
Q
A
R
AO
=
Q
A R
4π ϵ R
2 4π ϵ R
2 R 4π ϵ R
3
0 AO 0 AO AO 0 AO
AO
y 2
z 2
ρ 4
cos 0.6
z 3
BO
π
D2.4. Calculate the total charge within each of the indicated volumes: ( a ) 0.1 ≤ | x |, | y |, | z | ≤ 0.2:
ρ v
= 1/( x 3 y 3 z 3); ( b ) 0 ≤ ρ ≤ 0.1, 0 ≤ ϕ ≤ π, 2 ≤ z ≤ 4; ρv = ρ 2 z 2sin 0.6 ϕ ; ( c ) universe: ρv = e –2 r / r 2.
^6
4π ϵ 0 (0.4183)
3 (0.25 a x 0.3 a y 0.15 a z )^ V/m
= 9.21 a x 11.05 a y 5.53 a z kV/m For
the charge B to the origin,
R BO = (0 – (–10)) a x + (0 – 8) a y + (0 – 12) a z cm = 10 a x – 8 a y – 12 a z cm
= 0.1 a x – 0.08 a y – 0.12 a z m
| R BO | =
E B =
Q
B
R
BO
=
^6
m = 0.1755 m
(0.1 a x 0.08 a y 0.12 a z ) V/m
4π ϵ 0 R
3 4π ϵ 0 (0.1755)
3
= 83.13 a x 66.51 a y 99.76 a z
Combining E A and E B ,
kV/m
E = 92.3 a x – 77.6 a y – 94.2 a z kV/m
(b) Same concept in ( a )
R AP = (15 – 25) a x + (20 – (–30)) a y + (50 – 15) a z cm = –0.1 a x + 0.5 a y + 0.35 a z m
| R
AP | = 0.6185 m
E A = 1.14 a x 5.7 a y 3.99 a z kV/m
R BP = (15 – (–10)) a x + (20 – 8) a y + (50 – 12) a z cm = 0.25 a x + 0.12 a y + 0.38 a z m
| R BP | = 0.4704 m
E B = 10.79 a x 5.18 a y 16.41 a z kV/m
E = 11.9 a x – 0.52 a y – 12.4 a z kV/m
( a
0
2 1
1
2 1
2
2 1
3
2 1
4
2 1
5
2 1
( b
1 1
2 1
3 1
4 1 = 0.
2 )
(4 2
2 )
(4 3
2 )
(4 4
2 )
(a) The differential volume of the rectangular coordinates is dv = dx dy dz. Using the formula,
Q =
vol ρ v
dv
0.2 0.2 0.
=
0.1 0.1 0.
dx dy dz x
3 y
3 z
3
x
^3 dx
y
3 dy
z
3 dz
0.2 0.2 0.
= 112.5 C
0.1 0.
Notice that the charge Q is a large number. Neglecting the rate of change of the cubical
volume, we have xyz = (0.1)
3 and xyz = (0.2)
3 which are very small; then the volume Δ v is
nearest to 0. Yielding,
Q = ρv Δ v = [1/( x 3 y 3 z 3)] 0 = 0
(b) The differential volume of the cylindrical coordinates is dv = ρ dρ dϕ dz. Using the formula,
Q =
π
4
( ρ
2 z
2
sin 0.6) ρ dρ d dz =
ρ
3 dρ
sin 0.6 d
4
z
2 dz
0 0 2 0 0 2
π ^
4 0.
4 cos 0.6(180) cos 0.6(0) (^) (^43 )
3
0
2
0.6 0.6
= 1.018 mC
(c) The differential volume of the spherical coordinates is dv = r
2 sin θ dθ dϕ dr. Assuming this
universe to be a perfect sphere, we have limits as 0 ≤ r ≤ ∞, 0 ≤ ϕ ≤ 2π, 0 ≤ θ ≤ π. Using the
formula,
D2.3. Evaluate the sums:
( a )
m 0
5 1 (1)
m 4 (0.1)
m 1
m 1
2
; ( b )
m 1 (4 m )
2 1.
2 (0.12)
2
x 2
D2.6. Three infinite uniform sheets of charge are located in free space as follows: 3 nC/m
2
at z = –4, 6 nC/m2 at z = 1, and –8 nC/m2 at z = 4. Find E at the point: ( a ) P A
(b) PB (4, 2, –3); ( c ) PC (–1, –5, 2); ( d ) PD (–2, 4, 5).
D2.7. Find the equation of that streamline that passes through the point P (1,
4, –2) in the field
E = ( a )
8 x
y
a x + a y ; ( b ) 2 e [ y (5 x + 1) a x + x a y ].
4 x
2
5 x
y 2
( a )=
dy E y
dx E x
4 x
2
y
2
8 x y
x
2 y
= 6.28 C
(a) E = ρL
2π ϵ 0 ρ
a ρ where ρ is just a radical distance R , then
ρL R
ρL
R 2π ϵ 0 R R 2π ϵ 0 R
2
Finding E x and E y at Point A since infinite uniform line charges lie along x and y axes,
R x = 4 a z
^9
E x =
2π ϵ 0
2
(4 a z
= 22.469 a z V/m
R y = 4 a z
9
E y =
2π ϵ 0
2
(4 a z
= 22.469 a z V/m
Adding together,
E = 22.469 a z + 22.469 a z V/m = 45 a z V/m
(b) Finding R and E at Point B along x and y axes,
R x = 3 a y + 4 a z
R y = 4 a z
E x =
^9
2
(3 a y 4 a z ) =^ 10.785 a y +^ 14.38 a z V/m
2π ϵ 0
2 4
2
E y = 5 10
^9
2
(4 a z )
= 22.469 a z V/m
2π ϵ 0 4
2
E = 10.785 a y + (14.38 + 22.469) a z V/m = 10.8 a y + 36.9 a z V/m
(a) Obtaining first the electric field due to charges,
E z = –4 =
ρs
a 2 ϵ 0
^9
= a z
2 ϵ 0
= 169.41 a z V/m
E z = 1 =
^9
a z
2 ϵ 0
= 338.82 a z V/m
E
z = 4
9
a z
2 ϵ 0
= –451.76 a z V/m
Since the point A is located below all the sheets of charge, all directions of electric field E are
negative a z direction with relative to normal, which is a N , of PA. Combining together,
E = – E 1 – E 2 – E 3 = –169.41 a z – 338.82 a z – (–451.76 a z ) V/m = –56.5 a z V/m
(b) The location of point B suggests that the electric field E contributed by the surface charge at
z = –4 will be in positive a z direction while others are negative. Combining together,
E = E 1 – E 2 – E 3 = 282.3 a z V/m
(c) Same arguments in ( b ), only z = 4 will be in negative a z direction. Combining together,
E = E 1 + E 2 – E 3 = 960 a z V/m
(d) We know that the point D is located above all the sheets of charge and all electric fields E are
positive a z direction. Combining together,
E = E 1 + E 2 + E 3 = 56.5 a z V/m
N
D2.5. Infinite uniform line charges of 5 nC/m lie along the (positive and negative) x and y axes
in free space. Find^ E^ at: ( a )^ PA (0, 0, 4); ( b )^ PB (0, 3, 4).
D S dS = D
C
π L
2
2
2
r r
2 ydy = – xdx
2 ydy =^
xdx
2
x
2 C
2
y =
For P ( x = 1, y = 4, z = –2),
2
2 C
2
2 4 = ; C
x
2
2 = 33
( b )
dy
dx
x
y (5 x 1)
ydy
x
dx 5 x 1
Letting u = 5 x + 1 and du = 5 dx , and then x = ( u – 1)/5,
y 2 u (^1) (^1) du (^) u 1 1 1 1
5
u
25 u
du
25
du
25 u
du
( u ln u ) 25
y
[(5 x 1) ln 5 x 1]
C
For P ( x = 1, y = 4, z = –2),
[(5(1) 1) ln 5(1) 1] ; C 2
y
= 0.04(5 x 1) 0.04 ln 5 x 1
y
2 = 15.7 + 0.4 x – 0.08 ln |5 x + 1|
(a) D =
dΨ
Q
4π r
2
Q
a r
a ( r
2
sin θ dθ d ) a =
Q
sin θ dθ d
4π r
2 4π
Q
π 2
π 2
sin θ dθ
d
Q
π 2
sin θ dθ
π 2
d
Q
(cos θ
π/
π/ )
4π
0 0 4π
0 0 4π
0 0
^6
4π
(cos 90 (cos 0 ))((π 2) 0) (^) = 7.5 μC
(b) Deriving Gauss’s law of the cylinder,
Q =
ρ d dz = D ρ
d
dz = D ρ
2π z
L
= D 2π ρL
DS = Dρ =
Solving Ψ ,
Q
2π ρL
S
or D ρ =
S
0 0
Q
a ρ
2π ρL
S 0 0
S
dΨ =
Q
a ρ ( ρ dϕ dz ) a ρ =
Q
ρ dϕ dz
Q
ρ
2π
d
L
dz =
Q
(2π ρL )
2π ρL
Ψ = Q = 60 μC
2π ρL 2π ρL
0 0 2π ρL
(c) Getting the equation in ( b ) with changing the limits of ϕ as 0 ≤ ϕ ≤ π because only the flux
D3.1. Given a 60-μC point charge located at the origin, find the total electric
flux passing through:
( a ) that portion of the sphere r = 26 cm bounded by 0 < θ < π/2 and 0 < ϕ <
π/2; ( b ) the closed surface defined by ρ = 26 cm and z = ±26 cm; ( c ) the
plane z = 26 cm.
CHAPTER 3
2
x
0.3 10 r a
r sin θ dθ d a
R
(a) D = ϵ 0 E = Q
4π R
2
a R =
Q
R
4π R
3
R QP = (2 – (–2)) a x + (–3 – 3) a y + (6 – (–6)) a z = 4 a x – 6 a y + 12 a z
RQP = 14
D QP =
Q
4π R
R
QP
^3
4π
3 (4 a x – 6 a y + 12 a z ) = 6.38 a x – 9.57 a y + 19.14 a z μC/m
QP
(b) D = ϵ 0 E =
ρ L
a =
ρ L
R
2π R 2π R
2
R x = (2 – x ) a x – 3 a y + 6 a z
Since the infinite line charge density is along x axis, the electric field E at point P is having
only y and z components present. Canceling x component due to symmetry,
R x = –3 a y + 6 a z
Rx =
D =
ρ L
R
^3 (–3 a + 6 a ) = –212 a + 424 a μC/m
2
x
x
2π R
2
2π
y z y z
(c) D = ϵ 0 E =
ρS
a 2
The infinite surface change density is an infinite x-y plane located at z = –5 and the charge is
spread on that plane. Going back the formula,
D = (
ρS
- a z = (120 2) μ a z = 60 a z μC/m
(a) E = (^) D ϵ 0 =^ (0.3^ ^10
2 a r ) ϵ 0 = (0.3 10
2 ) a r ) ϵ 0 = 135.5 a r V/m
(b) Q =
D
d S =
9 2
2π π 2 r r
0 0
^9 r
2 (4π r
2 )
^9 (4π r
4 ) = 0.3 10
^9 (4π(
4 )) (^) = 305 nC
(c) Same procedure in ( b ),
Ψ = Q =
^9 π r
4 = 1.2 10
^9 π(4)
4 = 965 nC
(a) Both given charges are enclosed by the cubical volume according to Gauss’s law. Adding,
Ψ = Q 1 + Q 2 = 0.1 μC + (1 7) μC = 0.243 μC
(b) The line charge distribution passes through x = –2 and y = 3; it is parallel to z axis. The total
length of that charge distribution enclosed by the given cubical volume is 10 units as z = ±5.
Applying Gauss’s law of line charge,
Ψ = ρL L = π(10) μC = 31.4 μC
(c) A straight line equation, y = 3 x , passes through the origin. We need to find the length of that
line which is enclosed by the given volume. The length is moving up and down along z axis
as z = ±5 to a form a plane. Putting y = 5,
5 = 3; x = 5
Finding the length of that line on the plane formed by positive x and y axes,
3
2
R
D3.2. Calculate D in rectangular coordinates at point P (2, –3, 6) produced by: ( a ) a point charge
QA = 55 mC at Q (–2, 3, –6); ( b ) a uniform line charge ρLB = 20 mC/m on the axis; ( c ) a uniform
surface charge density ρSC =120 μC/m2 on the plane z = –5 m.
D3.3. Given the electric flux density, D = 0.3 r 2 a r nC/m2 in free space: ( a ) find E at point P ( r =
2, (^) θ = 25°, ϕ = 90°); ( b ) find the total charge within the sphere r = 3; ( c ) find the total electric
flux leaving the sphere r = 4.
D3.4. Calculate the total electric flux leaving the cubical surface formed by the six planes x ,
y , z = ±5 if the charge distribution is: ( a ) two point charges, 0.1 μC at (1, –2, 3) and 1/7 μC
at (–1, 2, –2); ( b ) a uniform line charge of π μC/m at x = –2, y = 3; ( c ) a uniform surface charge
of
0.1 μC/m
2 on the plane y = 3 x.
The same length we get on the plane formed by negative x and y axes. Adding two lengths,
That straight line moves between z = ±5 to form a plane. Knowing the area,
2 (5 3)
2
(cos θ sin θ ) =
D3.8. Determine an expression for the volume charge density associated
with each D field: ( a )
D =
4 xy
z
a x + a y –
2 x
2
z
2 x
2 y
z 2 a z ; ( b ) D = z sin ϕ a ρ + z cos ϕ a ϕ + ρ sin ϕ a z ; ( c ) D = sin θ sin ϕ a r +
cos θ sin ϕ a θ + cos ϕ
a ϕ.
Dy y = ( x
2 z 2 xy ) y (^) = –2 x
Dz z = ( x
2 y ) z = 0
div D = (2 yz – 2 x ) x = 2, y = 3, z = –1 = –
(b) div D =
1 ( ρDρ )
1 D
Dz
ρ ρ ρ z
Finding the partial derivatives of each term,
1 ( ρDρ ) 1 (2 ρ
2 z
2 sin
2
= = (4 z
2 sin
2
) ρ 2, 110 , z 1 =^ 3.
ρ ρ ρ ρ
1 D
1 ( ρz
2
sin 2 )
= = (2 z
2
cos 2 )
ρ 2, 110 , z 1
D
z (2 ρ
2 z sin
2
2 2
z
ρ
= (2 ρ sin ) ρ 2, 110 , z 1 =^ 7.
div D = 3.532 + (–1.532) + 7.064 = 9.
1 ( r
2 Dr ) 1 (sin θDθ ) 1 D
(c) div D =
r
2
r r sin θ
θ r sin θ
1 ( r
2 Dr ) 1 (2 r
3
sin θ cos)
= = (6sin θ cos) r 1.5, θ 30 , 50 =
r
2 r r
2 r
1 (sin θD θ
1 ( r sin θ cos θ cos D
θ
r sin θ θ r sin θ θ
Applying product rule of the derivative of sin θ cos θ ,
cos 2
r 1.5, θ 30 , 50 0.
sin θ
1 D
1 ( r sin)
= (cos
sin θ ) r 1.5, θ 30 , 50
r sin θ r sin θ
div D = 1.928 + 0.643 – 1.286 = 1.
Note: The volume charge density ρv is equal to div D.
(a) Dx x = (4 xy z ) x = 4 y z
D
y y = (2 x
2 z ) y = 0
Dz z = ( 2 x
2 y z
2 ) z (^) = 4 x
2 y z
3
ρv = 4 y z + 4 x 2 y z 3 = (4 y z 3)( x 2 z 2 )
(b) (
ρ ) ( ρDρ
ρ = (1 ρ )^ (^ ρz^ sin
ρ = ( z sin ) ρ
(1 ρ ) D
= (1 ρ ) ( z cos) = ( z sin) ρ
Dz
z = ( ρ sin) z = 0
ρv = ( z sin) ρ + ( z sin) ρ = 0
(c) (1 r
2 ) ( r
2 Dr ) r = (1 r
2 ) ( r
2
sin θ sin ) r = (2sin θ sin ) r
(1 r sin θ ) (sin θDθ ) θ = (1 r sin θ )(sin θ cos θ sin) θ
= (sin r sin θ )(cos
2 θ sin
2 θ )
(1 r sin θ ) D
= (1 r sin θ ) (cos) = sin ( r sin θ )
ρ v
= (2sinsin) r + (sin
r sin)(cos
2
sin
2
) + (sin ( r sin))
2sin
2
sincos
2
sinsin
2
sinsin
r sin
sin
2
sincos
2
sinsin
r sin
Since sin
2
cos
2
sin(sin
2
cos
2
) sin
r sin
r sin
0
0
4 (5 a x y z x y z
1
(b) Same procedure in ( a ),
4 5 dx
1 1
W = 4
1
5 x dx = 20 ( x
2
1
= –20(–0.5) = 10 J
(c) Same procedure in ( a ) and ( b ),
0
dy a y
dz a z
2 3
2 (1)
2
2 5
2 4
2
2 3
1
1
M
M
N
D4.5. A 15-nC point charge is at the origin in free space. Calculate V 1 if point P 1 is located
at
P 1 (–2, 3, –1) and (a) V = 0 at (6, 5, 4); ( b ) V = 0 at infinity; ( c ) V = 5 V at (2, 0, 4).
(0,2)
W = 4
(1,0)
(5 x dx 5 y
dy )
x
2
(0,2)
= –20(1.5) = –30 J
(a) Use the formula in D4.2 ( a ). One differential unit vector of each line segments is used. For
segment (1, 3, 5) to (2, 3, 5),
W =
0 a 0 a )
3 y x
2 = –9 J
1 3 (^ y^ a x y z
1
3 y dx
1 1 y 3
For segment (2, 3, 5) to (2, 0, 5),
2
W 2 = 3
1
( y a x 0 a y 0 a z ) 0
For segment (2, 0, 5) to (2, 0, 3),
2
W 3 = 3
1
( y a x 0 a y 0 a z ) 0
Adding the total work, W = W 1
+ W
2
+ W
3
= –9 J
(b) Same procedure in ( a ). For segment (1, 2, 5) to (1, 3, 3),
W 1 =
3 y 1
dx
= 3 y x
2
= 0
y 3
Both W 2 and W 3 are zeroes no matter what the line segments are. So adding the total work, W = 0.
(a) VMN = N
E d L =
(2,6,1)
(6 x
2 dx 6 y dy 4 dz ) =
(3,3,2)
(2 x
2 3 y
2 4 z )
(2,6,1)
(3,3,2)
= (70 81 12) = –139 V
(b) VMQ =
Q
E d L^ =^
(2 x
2 3 y
2 4 z )
(2,6,
(4,2,35)
= ( 112 96 136) = –120 V
VMQ = VM – VQ
VM = VMQ + VQ = –120 V + 0 = –120 V
(c) VNP =
P
E d L =
(2 x
2 3 y
2 4 z )
(3,3,2)
(1,2,4)
= ( 56 15 24) = 17 V
VNP = VN – VQ
VN = VNP + VP = 17 V + 2 V = 19 V
(a) Assume that second point is M. Applying the potential difference VAB , or VPM in the problem,
VPM =
Q 1
4π ϵ 0
RP RM
RP = =
R
M
9 1
VPM =
4π ϵ 0
= 20.67 V
VPM = VP – VM
VP = VPM + VM = 20.67 V + 0 = 20.67 V
(b) Same procedure in ( a ),
RP =
D4.3. We will see later that a time-varying E field need not be
conservative.conservative, the (Ifwork it (^) expressed by equation:is not W = – Q ∫ ifinal E nitial d L^ may^ be^ a function^ of the^ path
used.) Let E = y a x V/m at a certain instant of time, and calculate the work required to move a 3-C
charge from (1, 3, 5) to (2, 0, 3) along the straight-line segments joining: ( a ) (1, 3, 5) to (2, 3, 5)
to (2, 0, 5) to (2, 0, 3); ( b ) (1, 2, 5) to (1, 3, 3) to (1, 0, 3) to (2, 0, 3).
y 2
dx a x
D4.4. An electric field is expressed in rectangular coordinates by E = 6 x 2 a x + 6 y a y + 4 a z V/m.
Find: ( a ) VMN if points M and N are specified by M (2, 6, –1) and N (–3, –3, 2); ( b ) VM if V = 0 at
Q (4, –2, –35); ( c ) VN if V = 2 at P (1, 2, –4).
