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ELECTROMAGNETICS HAYT 8 EDITION
DRILL SOLUTION FROM CHAPTERS 1 TO 5
CHAPTER 1
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 |. ( 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 | = (^) ( 1) 2 22 12 = 6 = 2.
( d ) a MP = 2 5 2 5 2 ( 1) ( 5) ( 5)
(^) x (^) y z
^ a^ a^ a =^5 51
^ a^ x^^ ^ a y^^ 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 | = ( 13) 2 32 ( 8)^2 = 11 2 = 15.
D1.2. A vector field S is expressed in rectangular coordinates as S = {125/[( x – 1)^2 + ( y – 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.
( a ) S = 2 125 2 2 {(2 1) (4 2) (3 1) } (2 1) (4 2) (3 1) x^ y^ z
a a a
1 2 4 x^ y^ z
a a a =^125 ( 2 4 ) 21 x^ y^ z
a a a = 5.95 a x + 11.9 a y + 23.8 a z
( b ) a S =^ 5.95^2 11.9^2 23.8 2 5.95 11.9 23.
x ^ y z
^ a^ a^ a = 5.95^ 11.9^ 23.
a x (^) a y (^) a z
= 0.218 a x + 0.436 a y + 0.873 a z ( c ) 1 = | S | = 2 2 2 2 2 2 2 2 2 2 2 2
x y z x y z x y z x y z
2 2 2 2 2 2 2 2
125 [( 1) ( 2) ( 1)
[( 1) ( 2) ( 1) ]
x y z x y z
2 2 2 2 2 2
x y z x y z
Transposing, 2 2 2 2 2 2
x y z x y z
Conjugating, 2 2 2 2 2 2 2 2 2 2 2 2
x y z x y z x y z x y z
( x 1)^2 ( y 2)^2 ( z 1)^2 = 125
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. ( 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 ( 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
θBAC = cos 1 AB^ AC AB AC
R^ R
R R
1 2 2 3 2 2 3 cos^62 ( 8) 4 ( 6) ( 9) 2 3
= 53.6° ( d ) Analyzing the projection,
| R AB (L)| = R AB a AC = R AB AC AC
R
R
R AB (L) = | R AB (L)| a AC = | R AB (L)| AC AC
R
R
(^) x (^) y z
a a a
= – 5.94 a x + 1.319 a y + 1.979 a z D1.4. 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 × R AC ; ( b ) the area of the triangle; ( c ) a unit vector perpendicular to the plane in which the triangle is located.
( a ) R AB × R AC = 8 4 6 9 2 3
x y z
a a a = [4(3) – (–6)(2)] a x + [(–6)(–9) – (–8)(3)] a y + [(–8)(2) – 4(–9)] a z
= 24 a x + 78 a y + 20 a z
( b ) A =^1 base height 2
=^1
2 AB^ BC
R R
Finding the direction of vector R BC R BC = (– 3 – (–2)) a x + (1 – 3) a y + (5 – (–4)) a z = – a x – 2 a y + 9 a z
R AB × R BC = 8 4 6
x y z
a a a = 24 a x + 78 a y + 20 a z
| R AB × R BC | = 242 782 202 = 84
A = (1/2)(84) = 42
( c ) a RAB RAC =^24 78 84
a^ x^^ ^ a y^^ a z = 0.286 a x + 0.928 a y + 0.238 a z
D1.5. ( a ) Give the rectangular coordinates of the point C ( ρ = 4.4, ϕ = – 115°, z = 2). ( b ) Give the cylindrical coordinates of the point D ( x = – 3.1, y = 2.6, z = – 3). ( c ) Specify the distance from C to D. ( a ) Converting cylindrical to rectangular coordinates Table 1 Cylindrical to rectangular coordinate systems x = ρ cos ϕ y = ρ sin ϕ z = z x = 4.4 cos – 115° = – 1. y = 4.4 sin – 115° = – 3. z = 2 C ( x = – 1.86, y = – 3.99, z = 2) (b) Converting rectangular to cylindrical coordinates Table 2 Rectangular to cylindrical coordinate systems ρ = x^2^ y^2 ϕ = tan–^1 y / x z = z ρ = ( 3.1) 2 2.6^2 = 4. ( ϕ is added with 180° which it lies on the second quadrant.) ϕ = tan–^1 y / x + 180° = tan–^1 (2.6/–3.1) + 180° = 140° z = – 3 D ( ρ = 4.05, ϕ = 140°, z = – 3) (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
x = 5 sin 20° cos – 70° = 0. y = 5 sin 20° sin – 70° = – 1. z = 5 cos 20° = 4. D ( x = 0.585, y = – 1.607, z = 4.7) ( c ) Using rectangular coordinates to find the distance
| CD | = (0.585 ( 3)) 2 ( 1.607 2)^2 (4.7 1)^2 = 6.
D1.8. Transform the following vectors to spherical coordinates at the points given: ( a ) 10 a x at P ( x = – 3, y = 2, z = 4); ( b ) 10 a y at Q ( ρ = 5, ϕ = 30°, z = 4); ( c ) 10 a z at M ( r = 4, θ = 110°, ϕ = 120°). ( a ) Taking note that the dot product of the unit vectors in spherical to rectangular coordinate systems and vice versa Table 6 Dot products of the unit vectors in spherical and rectangular coordinate systems a r a θ a ϕ a x sin θ cos ϕ cos θ cos ϕ – sin ϕ a y sin θ sin ϕ cos θ sin ϕ cos ϕ a z cos θ – sin θ 0 10 a x = 10 (sin θ cos ϕ a r + cos θ cos ϕ a θ – sin ϕ a ϕ ) From Table 4, we calculate and have θ = 42.03° and ϕ = 146.31°. Substituting, = 10 sin 42.03° cos 146.31° a r + 10 cos 42.03° cos 146.31° a θ – 10 sin 146.31° a ϕ = – 5.57 a r – 6.18 a θ – 5.55 a ϕ ( b ) 10 a y = 10 (sin θ sin ϕ a r + cos θ sin ϕ a θ + cos ϕ a ϕ ) We manipulate to convert in cylindrical to spherical coordinates. From Table 1, we calculate and have x = 4.33, y = 2.5 and z = 4. Then from Table 4, we convert rectangular to spherical coordinates. We calculate and have θ = 51.34° and ϕ = 30°. Substituting, = 10 sin 51.34° sin 30° a r + 10 cos 51.34° sin 30° a θ + 10 cos 30° a ϕ = 3.9 a r – 3.12 a θ – 8.66 a ϕ ( c ) 10 a z = 10 (cos θ a r + (– sin θ ) a θ ) = 10 cos 110° a r – 10 sin 110° a θ = – 3.42 a r – 9.4 a θ
CHAPTER 2 D2.1. A charge QA = – 20 μC is located at A (–6, 4, 7), and a charge QB = 50 μC is at B (5, 8, – 2) in free space. If distances are given in meters, find: ( a ) R AB ; ( b ) RAB. Determine the vector force exerted on QA by QB if ϵ 0 = ( c ) 10–^9 /(36π) F/m; ( d ) 8.854 × 10–^12 F/m. ( a ) R AB = (5 – (–6)) a x + (8 – 4) a y + (– 2 – 7) a z m = 11 a x + 4 a y – 9 a z m
( b ) RAB = 112 42 ( 9)^2 m = 14.76 m
( c ) F AB = 0
1 2 4 π ϵ (^) AB^2 AB
Q Q
R
a =^12 0
2 4 π ϵ
AB AB AB
Q Q
R R
^ R =
0
1 2 4 π ϵ (^) AB^3 AB
Q Q
R
R
There is an ambiguous value of F AB with regards from the answer of the textbook. Trying in the higher decimal number of RAB = 14.7648, =
6 6 9 3
4 π (10 36π) (14.7648) x^ y^ z
^ ^ ^ a a a = (–2.7961 × 10–^3 )(11 a x + 4 a y – 9 a z )
= 30.76 a x + 11.184 a y – 25.16 a z mN
( d ) F AB = 0
1 2 4 π ϵ (^) AB^3 AB
Q Q
R
R =
6 6 12 3
4 π(8.854 10 )(14.7648) x^ y^ z
a a a
= (–2.7923 × 10–^3 )(11 a x + 4 a y – 9 a z ) = 30.72 a x + 11.169 a y – 25.13 a z mN D2. 2. A charge of – 0.3 μC is located at A (25, – 30, 15) (in cm), and a second charge of 0.5 μC is at B (–10, 8, 12) cm. Find E at: ( a ) the origin; ( b ) P (15, 20, 50) cm. ( a ) Solving first the vector and magnitude from charge A to the origin O , R AO = (0 – 25) a x + (0 – (–30)) a y + (0 – 15) a z cm = – 25 a x + 30 a y – 15 a z cm = – 0.25 a x + 0.3 a y – 0.15 a z m | R AO | = ( 0.25) 2 0.3^2 ( 0.15)^2 m = 0.4183 m
E A = 4 π ϵ 0 2
A AO AO
Q
R
a =^ A^ 2 AO AO AO
Q
R R
^ R =
4 π ϵ 0 3
A AO AO
Q
R
R
4 π ϵ 0
6 0 3
4 π ϵ (0.4183)
x y z
a a a 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 | = 0.1^2 ( 0.08) 2 ( 0.12)^2 m = 0.1755 m
E B = 4 π ϵ 0 3
B BO BO
Q
R
R = 3
0
4 π ϵ (0.1755) x^ y^ z
a a a V/m
= 83.13 a (^) x (^) 66.51 a (^) y (^) 99.76 a z kV/m Combining E A and E B , 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
D2. 3. Evaluate the sums: ( a )
5 0 2
m m m
( b )
4 1 2 1.
m m m
( a )
0 1 2 3 4 5 2 2 2 2 2 2
( b )
1 2 3 4 2 1.5 2 1.5 2 1.5 2 1.
D2. 4. Calculate the total charge within each of the indicated volumes: ( a ) 0.1 ≤ | x |, | y |, | z | ≤ 0.2: ρv = 1/( x^3 y^3 z^3 ); ( b ) 0 ≤ ρ ≤ 0.1, 0 ≤ ϕ ≤ π, 2 ≤ z ≤ 4; ρv = ρ^2 z^2 sin 0.6 ϕ ; ( c ) universe: ρv = e –^2 r^ / r^2. ( a ) The differential volume of the rectangular coordinates is dv = dx dy dz. Using the formula,
Q = vol ρ dvv =
0.2 0.2 0. 0.1 0.1 0.1^3 3
(^1) dx dy dz
x y z
0.2 3 0.2 3 0.2 3 0.1^ x^ dx^ 0.1 y^ dy^ 0.1 z^ dz
2 0.2^2 0.2^2 0. (^2) 0.1 2 0.1 2 0.
x ^ y ^ z
= 112.5 C
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 =
0.1 π (^4 2 )
0 0 2 ( ρ z sin 0.6 ) ρ dρ d dz =
0.1 3 π (^42)
0 ρ dρ^ 0 sin 0.6^ ^ d^ 2 z^ dz
4 0.1^ π 34
0 0 2
cos 0. 4 0.6 3
ρ z
^
^
0.1^4 cos 0.6(180 ) cos 0.6(0 ) 43 23 0 4 0.6 0.6 3 3
= 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, Q =
2 π π (^2 2 ) 0 0 2 (^ )(^ sin^ )
e r r r θ dθ d dr
2 π π (^2) 0 0 sin^2
d θ dθ ^ e rdr
= ^20 π^ cos θ π 0 e^2 r 20
^ = (2π – 0)(–cos 180° – (–cos 0°))(– e –∞^ 2 – (– e^0 2))
2 ydy = – xdx ^2 ydy^ = xdx
y^2 =
2 2 2 2
^ x^ C For P ( x = 1, y = 4, z = – 2), 42 =
^ C ; C^2 = 33
x^2 + 2 y^2 = 33
( b ) dy dx
x
y x
ydy^ = 5 1
x (^) dx x Letting u = 5 x + 1 and du = 5 dx , and then x = ( u – 1)/5, 2 2
y = 1 1 5 5
u du u
(^) =
u (^) du u
(^) =
du du u
(^) =^1 ( ln ) 25
u u 2 2
y = (^1) [(5 1) ln 5 1]^2 25 2
x x ^ C For P ( x = 1, y = 4, z = – 2), 42 2
[(5(1) 1) ln 5(1) 1] 25 2
^ C ; C^2 = 15.
2 2
y = 0.04(5 1) 0.04ln 5 1 15. 2
x x y^2 = 15.7 + 0.4 x – 0.08 ln |5 x + 1|
CHAPTER 3 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.
( a ) D = 2 4 π r
Q r
a
dΨ = 2 ( 2 sin ) 4 π r^ r
Q (^) r θ dθ d r
a a = sin
4 π
Q (^) θ dθ d
π 2 π 2 4 π^0 0 sin
Q (^) θ dθ d (^) =
π 2 π 2 4 π^0 sin^0
Q θ dθ d
(^) = π/2 π/ 4 π^ (^ cos^0 ) (^0 )
Q θ
( cos90 ( cos 0 ))((π 2) 0) 4 π
= 7.5 μC
( b ) Deriving Gauss’s law of the cylinder,
Q = (^) D S (^) d S = DS (^) ρ d dz =
π 0 0
L DS ρ (^) d dz = DS ρ ^20 π z 0^ L = DS 2 π ρL
DS = Dρ = 2 π
Q
ρL
or D ρ = 2 π
Q
ρL
a ρ
Solving Ψ , dΨ = 2 π
Q
ρL
a ρ ( ρ dϕ dz ) a ρ = 2 π
Q
ρL
ρ dϕ dz =
2 π 2 π^0
Q L
ρ d dz ρL
(^) = (2π ) 2 π
Q (^) ρL ρL Ψ = Q = 60 μC ( c ) Getting the equation in ( b ) with changing the limits of ϕ as 0 ≤ ϕ ≤ π because only the flux lines pass in half through the plane, dΨ =
π 2 π^0
Q L
ρ d dz ρL
(^) = (π ) 2 π
Q (^) ρL ρL Ψ = Q 2 = 30 μC
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/m^2 on the plane z = – 5 m.
( a ) D = ϵ 0 E = 2 4 π R
Q
R
a = 3 4 π
Q
R
R
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 = 3 4 π (^) QP QP
Q
R
R =
3 3
4 π
^
(4 a x – 6 a y + 12 a z ) = 6.38 a x – 9.57 a y + 19.14 a z μC/m^2
( b ) D = ϵ 0 E = 2 π
ρ L R R
a = 2 2 π
ρ L R
R
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 = 45
D x = 2 2 π
L x x
ρ R
R =
3 2
2 π 45
^
(– 3 a y + 6 a z ) = – 212 a y + 424 a z μC/m^2
( c ) D = ϵ 0 E = 2
ρ S (^) a R
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 2) a z = (120 2) μ a z = 60 a z μC/m^2
D3. 3. Given the electric flux density, D = 0.3 r^2 a r nC/m^2 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. ( a ) E = D 0 = (0.3 10 –9 r^2 a r ) 0 = ( 0.3 10 –9 (2 )^2 a r ) 0 = 135.5 a r V/m
( b ) Q = (^) D d S = 9 2 2 π^ π 2
0.3 10 r r 0 0 r sin θ dθ d r
a (^) a = 0.3 10 9 r^2^ (4π r 2 ) = 0.3 10 9 (4π r^4 ) = 0.3 10 9 (4π(3 ))^4 = 305 nC ( c ) Same procedure in ( b ), Ψ = Q = 1.2 10 9 π r^4 = 1.2 10 ^9 π(4)^4 = 965 nC 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. ( 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 3
Finding the length of that line on the plane formed by positive x and y axes, 52 (5 3)^2 = 5. The same length we get on the plane formed by negative x and y axes. Adding two lengths, 5.27 + 5.27 = 10. That straight line moves between z = ±5 to form a plane. Knowing the area,
ϵ ϵ ϵ
D (^) y y = ( x z^2 2 xy ) y = – 2 x D (^) z z = ( x y^2 ) z = 0 div D = (2 yz – 2 x ) x = 2, y = 3, z = – 1 = – 10
( b ) div D =^1 (^ ρDρ^ )^1 D^ Dz ρ ρ ρ z
Finding the partial derivatives of each term, (^2 2 22 ) 2, 110 , 1
1 ( (^) ρ ) (^) = 1 (2 sin )= (4 sin ) = ρ z
ρD ρ z (^) z ρ ρ ρ ρ^
(^22) 2, 110 , 1
(^1) = 1 ( sin 2 )= (2 cos 2 ) = ρ z
D ρz (^) z ρ ρ
^
^ ^ ^ ^
(^2 22 ) D z (^) = (2 ρ z sin ) = (2 ρ sin ) (^) ρ 2, 110 , z 1 = z ρ
div D = 3.532 + (–1.532) + 7.064 = 9.
( c ) div D =
2 2
1 ( ) 1 (sin ) 1 sin sin
r D r θDθ D r r r θ θ r θ
2 3 2 2 1.5,^ 30 ,^50
1 ( (^) r ) (^) = 1 (2 sin cos ) = (6sin cos ) = r θ
r D r θ (^) θ r r r r^
1 (sin ) (^) = 1 ( sin cos cos ) sin sin
θD θ r θ θ Dθ r θ θ r θ θ
Applying product rule of the derivative of sin θ cos θ , =^ cos^ (cos^2 sin^2 ) (^) 1.5, 30 , 50 = sin r^ θ
θ θ θ^
1.5, 30 , 50
(^1) = 1 ( sin ) = ( cos sin ) = sin sin r^ θ
D r (^) θ r θ r θ
^ ^
^ ^ ^ ^ ^
div D = 1.928 + 0.643 – 1.286 = 1. D3. 8. Determine an expression for the volume charge density associated with each D field: ( a )
D =^4 xy z
a x + 2 x^2 z
a y –
2 2
2 x y z
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. Note: The volume charge density ρv is equal to div D. ( a ) D (^) x x = (4 xy z ) x = 4 y z
D (^) y y = (2 x^2 z ) y = 0 D (^) z z = ( 2 x y z^2 2 ) z = 4 x y z^2 ρv = 4 y z + 4 x y z^2 3 = (4 y z^3^ )( x^2^ z^2 )
( b )(1 ρ ) ( ρDρ ) ρ = (1 ρ ) ( ρz sin ) ρ = ( sin z ) ρ
(1 ρ ) D = (1 ρ ) ( cos z ) = ( z sin ) ρ
D z z = ( ρ sin ) z = 0
ρv = ( sin z ) ρ + ( z sin ) ρ = 0
( c ) (1 r^2^ ) ( r D^2 r ) 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 ( sin r θ )
ρv = (2sin sin ) r + (sin r sin )(cos^2 sin^2 )+( sin ( sin r ))
2sin^2 sin cos^2 sin sin^2 sin sin r sin
^ ^ =^ sin^2 sin^ cos^2 sin^ sin r sin
Since sin^2 cos^2 = 1,
sin (sin^2 cos^2 ) sin r sin
^ =^0
r sin
D3. 9. Given the field D = 6 ρ sin (1/2) ϕ a ρ + 1.5 ρ cos (1/2) ϕ a ϕ C/m^2 , evaluate both sides of the divergence theorem for the region bounded by ρ = 2, ϕ = 0, ϕ = π, z = 0, and z = 5.
Using the divergence theorem with the formula: (^) (^) S D d S ,
(^) S D^ d S^ = (^) S (6^ ρ^ sin (^ ^ 2)^ a ρ^^ ^ 1.5^ ρ^ cos (^ ^ 2) a^ ^ (^ ρ d^ dz^ a ρ^ dρ dz a ) = S 6 ρ sin ( 2) ρ d dz (^) S 1.5 ρ cos ( 2) dρdz
=
π 5 2 5 0 0 6 ρ^ sin (^ ^ 2)^ ρ d^ dz^ 0 0 1.5^ ρ^ cos (^ 2) dρ dz = 6 ρ^2 (^) 0^ π sin ( 2) d 0 5 dz (1.5cos ( 2)) 02 ρdρ (^) 05 dz
= (^24) 2cos 2 0 π^ z^50 (1.5cos 2) (^) ρ^2 (^220) z^50
Since the second surface lies at ϕ = 0°,
= 24(2)(5) (1.5cos 2)(2)(5) = 240 – 15 = 225
Using another divergence theorem with the formula: (^) vol D dv but solving first the D ,
D = div D = (1 ρ ) ( ρDρ ) ρ + (1 ρ ) D + D z z
= (1 ρ ) (6 ρ^2 sin 2) ρ + (1 ρ ) (1.5 ρ cos 2) + 0
= 12sin 2 + ( 1.5 2)sin 2 = 11.25sin 2
vol D dv^ =
2 π 5 (^) 0 0 0 (11.25sin 2) ρ dρ d dz =
2 π 5 11.25 (^) 0 ρ dρ (^) 0 (sin 2) d 0 dz
= 11.25 (^) ρ^2 (^2 20) 2cos 2 0 π z^50 = 11.25(2)(2)(5) = 225
CHAPTER 4
D4. 1. Given the electric field E = (1/ z^2 )( 8 xyz a x + 4 x^2 z a y – 4 x^2 y a z ) V/m, find the differential amount of work done in moving a 6-nC charge a distance of 2 μm, starting at P (2, – 2, 3) and proceeding in the direction a L = ( a ) – 6/7 a x + 3/7 a y + 2/7 a z ; ( b ) 6/7 a x – 3/7 a y – 2/7 a z ; ( c ) 3/7 a x + 6/7 a y. ( a ) dW = – Q E d L Finding first the differential length d L , d L = a L dL = (–6/7 a x + 3/7 a y + 2/7 a z )(2 × 10–^6 ) = (–12/7 a x + 6/7 a y + 4/7 a z )(1 × 10–^6 ) dW = – (6 × 10–^9 )(1/ z^2 )(8 xyz a x + 4 x^2 z a y – 4 x^2 y a z ) (–12/7 a x + 6/7 a y + 4/7 a z )(1 × 10–^6 ) = – 6 × 10–^15 [(1/ z^2 )((–96/7) xyz + (24/7) x^2 z – (16/7) x^2 y )] x = 2, y = – 2, z = 3 = – 6 × 10–^15 (224/9) = – 149.3 fJ ( b ) Same procedure in ( a ), d L = (12/7 a x – 6/7 a y + 4/7 a z )(1 × 10–^6 ) dW = – (6 × 10–^9 )(1/ z^2 )(8 xyz a x + 4 x^2 z a y – 4 x^2 y a z ) (12/7 a x – 6/7 a y + 4/7 a z )(1 × 10–^6 ) = – 6 × 10–^15 [(1/ z^2 )((96/7) xyz – (24/7) x^2 z – (16/7) x^2 y )] x = 2, y = – 2, z = 3 = – 6 × 10–^15 (–224/9) = 149.3 fJ ( c ) Same procedure in ( a ) and ( b ), d L = (6/7 a x + 12/7 a y + 0 a z )(1 × 10–^6 ) dW = – (6 × 10–^9 )(1/ z^2 )(8 xyz a x + 4 x^2 z a y – 4 x^2 y a z ) (6/7 a x + 12/7 a y + 0 a z )(1 × 10–^6 ) = – 6 × 10–^15 [(1/ z^2 )((48/7) xyz + (48/7) x^2 z ] x = 2, y = – 2, z = 3 = – 6 × 10–^15 (0) = 0 D4. 2. Calculate the work done in moving a 4-C charge from B (1, 0, 0) to A (0, 2, 0) along the path y = 2 – 2 x , z = 0 in the field E = ( a ) 5 a x V/m; ( b ) 5 x a x V/m; ( c ) 5 x a x + 5 y a y V/m.
( a ) W =
A
B
Q (^) E d L where d L = dx a x + dy a y + dz a z
=
0 (^4) 1 (5 a (^) x (^) 0 a (^) y (^) 0 a z (^) ) ( dx a x (^) dy a y (^) dz a z ) =
0 (^4) 15 dx = 0 20 x 1 = – 20(–1) = 20 J
( b ) Same procedure in ( a ),
W =
0 (^4) 15 x dx = 2 0 20( x 2) 1 = – 20(–0.5) = 10 J
( c ) Same procedure in ( a ) and ( b ),
VP = VPM + VM = 36 V + 0 = 36 V
( c ) Same procedure in ( a ) and ( b ),
RP = 14 RM = 22 42 = 20
VPM = 0
4 π ϵ (^) 14 20
= 5.89 V
VP = VPM + VM = 5.89 V + 5 V = 10.89 V
D4. 6. If we take the zero reference for potential at infinity, find the potential at (0, 0, 2) caused by this charge configuration in free space ( a ) 12 nC/m on the line ρ = 2.5 m, z = 0; ( b ) point charge of 18 nC at (1, 2, – 1); ( c ) 12 nC/m on the line y = 2.5, z = 0, – 1.0 < x <1.0.
( a ) V ( r ) = 0
4 π ϵ
ρ L ' dL' (^) '
r r r
where dL ʹ = ρ dϕ
Since the distance is converted in rectangular coordinates, we can mentally solve it where ϕ = 0°. | r – r ʹ| = (0 2.5)^2 (2 0)^2 = 3.
V ( r ) =
2 9 (^0 )
π (^12 ) 4 π ϵ (3.20)
^ ρ d
(^) =
(^9 2) π 0 (3.20)^0
4 π ϵ
ρ d
(^) =
(^9 2) π 0 (3.20)^0 2.
4 π ϵ ρ
ρ
= 529 V
( b ) V ( r ) = 0
1 4 π ϵ 1
Q
r r | r – r ʹ| = (0 1) 2 (0 2)^2 (2 ( 1))^2 = 14
V ( r ) =
9 0
4 π ϵ 14
^
= 43.2 V
( c ) Same formula in ( a ) but now dL ʹ = dx. Solving the distance | r – r ʹ| where x is symmetric,
| r – r ʹ| = (0 x )^2 (0 2.5)^2 (2 0)^2 = x^2^ 10.
V ( r ) = 2
1 9 (^1 )
4 π ϵ x 10.
dx
(^)
(^) = 2
(^9 ) 0 1
4 π ϵ (^) 25
dx x
Integrating 1 x^2 10.25 is solved by trigonometric substitution and yielding,
9 2 1 (^01)
(^12 10) ln 10. 4 π ϵ (^) 10.
x x
^
=
9 0
4 π
^
= 66.3 V
D4. 7. A portion of a two-dimensional ( Ez = 0) potential field is shown the figure below. The grid lines are 1mm apart in the actual field. Determine approximate values for E in rectangular coordinates at: ( a ) a ; ( b ) b ; ( c ) c. Requires graphical solution.
D4. 8. Given the potential field in cylindrical coordinates, V = [100/( z^2 + 1)] ρ cos ϕ V, and point P at ρ = 3 m, ϕ = 60°, z = 2 m, find values at P for ( a ) V ; ( b ) E ; ( c ) E ; ( d ) dV / dN ; ( e ) a N ; ( f ) ρv in free space. ( a ) V = [100/( z^2 + 1)] ρ cos ϕ = [100/(2^2 + 1)]3 cos 60° = 30 V
ϵ
( b ) E = V =^ V^ ρ^1 V^ V z ρ ρ ρ
^ ^ ^
a a a
=^1002 cos^1002 sin^2002 2 cos 1^ ρ^ 1 ( 1) z
z (^) ρ z z^ z
^
^ ^
^ a^ a^ a = – (10 a ρ – 17.32 a ϕ – 24 a z )
= – 10 a ρ + 17.32 a ϕ + 24 a z V/m
(c) E = ( 10) 2 17.32^2 242 = 31.2 V/m
( d ) dV dN = E = 31.2 V/m
( e ) E = ( dV dN ) a (^) N
a N = dV dN
E = –10^ 17.
^ ρ^^ ^32 ^24 z ^ a^ a^ a = 0.32 a ρ –^ 0.55 a ϕ –^ 0.77 a z
( f ) D = ϵ 0 E = ϵ 0 1002 cos 1002 sin 2002 2 cos 1^ ρ^ 1 ( 1) z
z (^) ρ z z^ z
^ ^
a a a
ρv = D = div D = (1 ρ ) ( ρDρ ) ρ + (1 ρ ) D + D z z
[ ( 2 1) ]^2
200 ϵ ρ cos z^ z z
^
Applying product rule of the derivative of z ( z^2^ 1)^2 and yielding,
= 0 3,
2 2 60
2 2 3 , 2
200 ϵ ρ cos 1 4 ( 1) ( 1) (^) ρ z
z z z
(^) (^) = – 234 pC/m^3
D4. 9. An electric dipole located at the origin in free space has a moment p = 3 a x – 2 a y + a z nC ∙ m. ( a ) Find V at PA (2, 3, 4). ( b ) Find V at r = 2.5, θ = 30°, ϕ = 40°.
( a ) V = 0 2
4 π ϵ
p r^ r r r r^ r
0
4 π ϵ '
p r r r r r – r ʹ = (2 – 0) a x + (3 – 0) a y + (4 – 0) a z = 2 a x + 3 a y + 4 a z | r – r ʹ| = 22 32 42 = 29
V =
9 3 0
4 π ϵ 29
a (^) x a y (^) a z (^) ^ a x (^) a (^) y a z =
9 3 0
4 π ϵ 29
^
= 0.23 V
( b ) Using Table 5 from Chapter 1 to convert the point PA to rectangular coordinates, x = 0.958, y = 0.803, z = 2. r – r ʹ = (0.958 – 0) a x + (0.803 – 0) a y + (2.165 – 0) a z = 0.958 a x + 0.803 a y + 2.165 a z | r – r ʹ| = 0.958^2 0.803^2 2.165^2 = 2.
V =
9 0 3
4 π ϵ 2.
^ a^ x^^ a y^^ ^ a z^^ ^ a x (^) a y (^) a z =^9 0 3
4 π ϵ 2.
^
= 1.97 V
D4. 10. A dipole of moment p = 6 a z nC ∙ m is located at the origin in free space. ( a ) Find V at P ( r = 4, θ = 20°, ϕ = 0°). ( b ) Find E at P. ( a ) Same formula in D4.9( a ). Converting the point P to rectangular coordinates, x = 1.368, y = 0, z = 3. Since the procedure is very similar with D4.9( a ) and ( b ) where p is at the origin, r – r ʹ = 1.386 a x + 3.759 a z and | r – r ʹ| = 4 V =
9 0 3
4 π ϵ 4
x^ 3.759 z z
a a ^ a =
9 (^4 )
4 π ϵ
^
= 3.167 V = 3.17 V
( b ) E = 3 0
(2cos sin ) 4 π ϵ r
Qd r^
a a
Since there is no charge given, we have to derive with the help of the voltage V since we have the answer in (a). Deriving, V = 0 2
cos 4 π ϵ
Qd r
; Qd =^4 π^02
cos
r V
Substituting,
ϵ
ρv = [ –10 (0^6. 1 1.5 )] (2 106 ) = – 15.8 mC/m^3 ( c ) Same formula in ( b ), vz = J (^) z ρv = [–1 06 (0.151.5 )] ( 2000 ) = 29.0 m/s D5. 3. Find the magnitude of the current density in a sample of silver for which σ = 6.17 × 107 S/m and μe = 0.0056 m^2 /V ∙ s if ( a ) the drift velocity is 1.5 μm/s; ( b ) the electric field intensity is 1 mV/m; ( c ) the sample is a cube 2.5 mm on a side having a voltage of 0.4 mV between opposite faces; ( d ) the sample is a cube 2.5 mm on a side carrying a total current of 0.5 A. ( a ) J = σ E Since v d = – μe E , then J = σ v d μe. Since we only find the value of the magnitude of the current density J , which we know that vectors are denoted by bold letters, we consider J and other vectors as the same dot unit vector. The negative is canceled because they refer to the direction. J = σvd μe = [(6.17 10 )(1.5 7 10 6 )] 0.0056 = 16.5 kA/m^2 ( b ) J = σE = (6.17 × 10^7 )(1 × 10–^3 ) = 61.7 kA/m^2 ( c ) J = σV L = [(6.17 10 )(0.4 7 10 ^3 )] (2.5 10 3 ) = 9.9 MA/m^2
( d ) J = I S = 0.5 (2.5 10 ^3 )^2 = 80 kA/m^2 D5. 4. A copper conductor has a diameter of 0.6 in. and it is 1200 ft long. Assume that it carries a total dc current of 50 A. ( a ) Find the total resistance of the conductor. ( b ) What current density exists in it? ( c ) What is the dc voltage between the conductor ends? ( d ) How much power is dissipated in the wire? ( a ) We must convert the English units to metric systems. A 1200 feet is about 365.76 m. A 0.6 in is about 0.01524 m. The radius is half of the diameter; so r = 7.62 × 10–^3 m. The conductivity of the copper is 5.8 × 10^7 S/m. The area of the circle is π r^2. R = L σS = 365.76 [(5.8 10 )^7 π ( 7.62 10 –^3 ) ]^2 = 0.035 Ω
( b ) J = I S = 50 [π(7.62 10 3 ) ]^2 = 2.74 × 10^5 A/m^2 ( c ) V = IR = 50(0.035) = 1.75 V With regards from the answer of the textbook, V = 50(0.03457) = 1.73 V ( d ) P = VI = 1.75(50) = 87.5 W With regards from the answer of the textbook, P = 1.7285(50) = 86.4 W D5. 5. Given the potential field in free space, V = 100 sinh 5 x sin 5 y V, and a point P (0.1, 0.2, 0.3), find at P : ( a ) V ; ( b ) E ; ( c ) | E |; ( d ) | ρs | if it is known that P lies on a conductor surface. ( a ) The function sinh( x ) is kind of a hyperbolic trigonometry where sinh( x ) = ( ex e x ) 2. Be careful that the points are in radians, not degrees. V = 100 sinh 5(0.1) sin 5(0.2) = 43.8 V ( b ) Using back the formula in Chapter 4, E = V = (^) ( V x ) a x (^) ( V y ) a (^) y (^) ( V z ) a z = ( (100sinh 5 sin 5 ) x y x ) a x (^) ( (100sinh 5 sin 5 ) x y y ) a y (^) 0 a z The derivative of sinh x is cosh x or ( ex e x ) 2. But sinh 5 x dx is 5 cosh 5 x. = 500cosh 5 sin 5 x y a x (^) 500sinh 5 cos5 x y a y = 500cosh 5(0.1)sin 5(0.2) a x (^) 500sinh 5(0.1)cos5(0.2) a y = – 474 a x – 140.8 a y V/m
( c ) E = ( 474) 2 ( 140.8)^2 = 495 V/m
( d ) ρS = ϵ 0 EN = ϵ 0 (495) = 4.38 nC/m^2 D5. 6. A perfectly conducting plane is located in free space at x = 4, and a uniform infinite line charge of 40 nC/m lies along the line x = 6, y = 3. Let V = 0 at the conducting plane. At P (7, – 1, 5) find: ( a ) V ; ( b ) E. It is really good if we refer to answer first in ( b ) because it is difficult to solve a voltage with lack of given variables like the charge Q and the electric field E. In ( b ), a method of image is required. We are going to mirror the infinite line charge ρL including the distance (or radial vector) R without
the conducting plane. For more details, see Section 5.5. The book is explained that the radial vector from positive (or negative if ever) line charge to point P is symmetric where the plane is located. For example, if the line charge (not the plane) is at z = 3 and the plane is at z = 0, then we put an image line charge at z = – 3 which the plane is removed. ( b ) Locating the radial vector R from the given location of line charge and P , R + = a x – 4 a y and R – = – 5 a x – 4 a y How did we obtain them? Well, draw a two-dimensional rectangular coordinates. Locate P at x = 7 and y = – 1. Notice that no z axis is included because the conducting plane activates on x axis and the line charge activates on x and y axes. The plane is at x = 4 while the line charge is x = 6 and y = 3. Therefore, the image line charge is symmetrically at x = 2 and y = 3. We draw the plane, line charge and image line charge with horizontal line. We can now find the radial vector R. For R +, the line charge is one unit away from P ( x = 7), so 7 – 6 = 1; the line charge is four units away from P ( y = 3), so – 1 – 3 = – 4. The image line charge reacts on negative distance because we deal with directions no matter where plane is placed. For R – , the image line charge is five units away from P ( x = 7), so – (7 – 2) = – 5; the image line charge is four units away from P ( y = 3) is – (– 1 – 3) = 4. Finding their magnitudes,
R + = 17 = 4.123 and R – = 41 = 6.403.
Knowing each electric fields, E + = 2 π ϵ 0
ρ L R R
a =
9 2 0
2 π ϵ 17
x y
a a = 42.294 a x – 169.177 a y
E – =
2 π ϵ 0
ρ L R R
a =
9 2 0
2 π ϵ 41
x y
(^) a a = – 87.683 a x + 70.147 a y
We cannot get the exact answer from the textbook. Adding together, E = E + + E – = – 45.39 a x – 99.03 a y V/m ( a ) We know that the plane stands at x axis. Finding the voltage V where only the electric field E and length for P are extracted at x direction, V = ExLx = ( 45.39) (7)^2 = 317.73 V
D5. 7. Using the values given in this section for the electron and hole mobilities in silicon at 300 K, and assuming hole and electron charge densities are 0.0029 C/m^3 and – 0.0029 C/m^3 , respectively, find: ( a ) the component of the conductivity due to holes; ( b ) the component of the conductivity due to electrons; ( c ) the conductivity. The electron and hole mobilities of pure silicon is 0.12 and 0.025. ( a ) σ h = ρhμh = (0.0029)0.025 = 72.5 μS/m ( b ) σ e = – ρeμe = – (–0.0029)0.12 = 348 μS/m ( c ) σ = σ h + σ e = 421 μS/m D5. 8. A slab of dielectric material has a relative dielectric constant of 3.8 and contains a uniform electric flux density of 8 nC/m^2. If the material is lossless, find: ( a ) E ; ( b ) P ; ( c ) the average number of dipoles per cubic meter if the average dipole moment is 10–^29 C ∙ m. ( a ) D = ϵ E where ϵ = ϵ 0 ϵr E = D ϵ 0 (^) ϵr = (8 10 ^9 ) 3.8 0 = 238 V/m ( b ) D = ϵ 0 E + P ; P = D – ϵ 0 E P = (8 × 10–^9 ) – 238 ϵ 0 = 5.89 nC/m^2
( c ) P = 0 1
lim^1^ n^ v v i vi
(^) p We have one dipole moment p and no limit of the average volume Δ v. Generalizing the formula, P = p v ; 1 v = P p = 5.89 10 9 10 ^29 = 5.89 × 10^20 m–^3 D5. 9. Let Region 1 ( z < 0) be composed of a uniform dielectric material for which ϵr = 3.2, while Region 2 ( z > 0) is characterized by ϵr = 2. Let D 1 = – 30 a x + 50 a y + 70 a z nC/m^2 and find: ( a ) DN 1 ; (b) D t 1 ; ( c ) Dt 1 ; ( d ) D 1 ; ( e ) θ 1 ; ( f ) P 1. ( a ) Since the Region 1 affects the z axis, only Dz is referred.
ϵ
