Electromagnatic

1. 1Solution to the Drill problems of chapter 04
(Engineering Electromagnetics,Hayt,A.Buck 7th ed)
BEE 4A,4B & 4C
D4.1 (a). E = (1/z2)(8xyzˆax + 4x2zˆay − 4x2yˆaz)V/m, Q = 6nC, | dL |= 2µm, P(2, −2, 3)
ˆaL = (−6/7)ˆax + (3/7)ˆay + (2/7)ˆaz, Find dW
dL = ˆaL | dL |= 2 × 10−6((−6/7)ˆax + (3/7)ˆay + (2/7)ˆaz) = ((−12/7)ˆax + (6/7)ˆay + (4/7)ˆaz) × 10−6
dW= −QE · dL⇒ dW=-6 ×10−9((1/z2)(8xyzˆax + 4x2zˆay − 4x2yˆaz)) · (((−12/7)ˆax + (6/7)ˆay + (4/7)ˆaz) × 10−6)
= −6 × 10−15((1/z2)((−96/7)xyz + (24/7)x2z − (16/7)x2y)
⇒ dWP(2,−2,3) = −6 × 10−15((1/32)((−96/7)(2)(−2)(3) + (24/7)(2)2(3) − (16/7)(2)2(−2))
= −6 × 10−15((1/32)((1152/7) + (288/7) + (128/7)) = −149.3fJ
(b). Similar to part(a)
(c). Similar to part(a)
D4.2 (a). Find the work done W, Q = 4C, from B(1, 0, 0) to A(0, 2, 0) along the path y = 2 − 2x, z = 1,
E = 5ˆaxV/m
we have W = −Q E · dL, since the path of integration is a straight line so we have dL = dxˆax + dyˆay + dzˆaz
⇒ W = −4 (5ˆax + 0ˆay + 0ˆaz) · (dxˆax + dyˆay + dzˆaz) = −4 0
1 5dx = 20J
(b). Follow the same procedure as in part(a) and we get W = −4 (5xˆax + 0ˆay + 0ˆaz) · (dxˆax + dyˆay + dzˆaz)
⇒ W = −4 0
1 5xdx = −20× | x2/2 |0
1= 10J
(C). Follow the same procedure as in part(a) and we get W = −4 (5xˆax + 5yˆay + 0ˆaz) · (dxˆax + dyˆay + dzˆaz)
⇒ W = −4( 0
1 5xdx + 2
0 5ydy) = −20 × (| x2/2 |0
1 + | y2/2 |2
0) = −20 × (−(1/2) + 2) = −30J
D4.3 (a). E = yˆax, Q = 3C, along the straight line segments joining (1,3,5) to (2,3,5) to (2,0,5) to (2,0,3)
we have W = −Q E · dL
for (1,3,5) to (2,3,5)
W1 = −3 (yˆax + 0ˆay + 0ˆaz) · (dxˆax) (dy and dz are zero for this line segment)
⇒ W1 = −3 2
1 ydx = −3y | x |2
1= (−3y)y=3 = −9J
for (2,3,5) to (2,0,5)
W2 = −3 (yˆax + 0ˆay + 0ˆaz) · (dyˆay) (dx and dz are zero for this line segment)
⇒ W2 = 0
for (2,0,5) to (2,0,3)
W3 = −3 (yˆax + 0ˆay + 0ˆaz) · (dzˆaz) (dx and dy are zero for this line segment)
⇒ W3 = 0
⇒ W = W1 + W2 + W3 = −9 + 0 + 0 = −9J
(b). Similar to part(a)
D4.4 (a). E = 6x2ˆax + 6yˆay + 4ˆazV/m, dL = dxˆax + dyˆay + dzˆaz find VMN , M(2, 6, −1), N(−3, −3, 2)
we have VAB = − A
B E · dL ⇒ VMN = − M
N E · dL = − M
N (6x2ˆax + 6yˆay + 4ˆaz) · (dxˆax + dyˆay + dzˆaz)
= −(6 2
−3 x2dx + 6 6
−3 ydy + 4 −1
2 dz) = −(6× | x3/3 |2
−3 +6× | y2/2 |6
−3 +4× | z |−1
2 ) = −(70 + 81 − 12) = −139V
(b). we have VAB = VA − VB ⇒ VMQ = VM − VQ = (− M
0 E · dL) − VQ
(VM and VQ are the potential differences with respect to the origin (0, 0, 0))
⇒ VMQ = − M
0 (6x2ˆax + 6yˆay + 4ˆaz) · (dxˆax + dyˆay + dzˆaz) − VQ
⇒ VMQ = −(6 2
0 x2dx + 6 6
0 ydy + 4 −1
0 dz) − VQ = −(6× | x3/3 |2
0 +6× | y2/2 |6
0 +4× | z |−1
0 ) − VQ = −120 − 0
= −120 V (since VQ = 0)⇒VM = −120V
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2. (c). we have VAB = VA − VB ⇒ VNP = VN − VP = (− N
0 E · dL) − VP
(VN and VP are the potential differences with respect to the origin (0, 0, 0))
⇒ VNP = − N
0 (6x2ˆax + 6yˆay + 4ˆaz) · (dxˆax + dyˆay + dzˆaz) − VP
⇒ VNP = −(6 −3
0 x2dx+6 −3
0 ydy+4 2
0 dz)−VP = −(6× | x3/3 |−3
0 +6× | y2/2 |−3
0 +4× | z |2
0)−VP = 19−2 = 17V
⇒ VN = 19V
D4.5 (a). Q = 15nC at the origin, calculate V1 at P1(−2, 3, −1), V = 0 at (6, 5, 4)
we have V = (Q/4π 0r) + C1, V = 0 at (6, 5, 4), r =
√
62 + 52 + 42 = 8.77 ⇒ C1 = −(Q/4π 0r)
⇒ C1 = −15 × 10−9/(4π × 8.85 × 10−12 × 8.77) = −15.37
now again we have V = (Q/4π 0r) + C1, but now we know the value of C1 and we will calculate the value of r this
times with the point P1(−2, 3, −1) ⇒ r = (−2)2 + 32 + (−1)2 = 3.74
⇒ V=(Q/4 π 0r) + C1 = (15 × 10−9/4π × 8.85 × 10−12 × 3.74) − 15.37 = 20.7V
(b). Follow the same procedure to find C1 as in part (a),by putting V = 0 and r = ∞ ⇒ C1 = 0
Now V=(Q/4 π 0r) + C1 = (15 × 10−9/4π × 8.85 × 10−12 × 3.74) − 0 = 36V
(C). Similar to part(a).
D4.6 (a). ρL = 12nC/m on the line ρ = 2.5m (we have a uniform line charge in the form of a circular ring in
the z = 0 plane or x-y plane), we have V (r) = ρL(r )dL /4π 0 | r − r |, dL = ρdφ, | r − r |= 22 + (2.5)2 = 3.20
⇒ V (r) = (12 × 10−9) × ρ 2π
0 dφ/(4π 0 × 3.20) = (12 × 10−9) × 2.5× | φ |2π
0 /(4π 0 × 3.20) = 529V
(b). Q = 18nC, P(1, 2, −1) for point charges we have V (r) = Q/4π 0 | r − r |
| r − r |= (0 − 1)2 + (0 − 2)2 + (2 − (−1))2 = 3.74
⇒ V (r) = 18 × 10−9/(4π × 8.85 × 10−12 × 3.74) = 43.2V
(c). Similar kind of problem.
D4.7. Graphical problem.
D4.8(a). V = (100/(z2 +1))ρ cos φV, P(ρ = 3m, φ = 600, z = 2m) ⇒ V(ρ=3m,φ=600,z=2m) = (100/(22 +1))×3×cos 600
⇒ V(ρ=3m,φ=600,z=2m) = 30V
(b). we have E = − V , now V = (∂V/∂ρ)ˆaρ + (1/ρ)(∂V/∂φ)ˆaφ + (∂V/∂z)ˆaz
⇒ (∂V/∂ρ)ˆaρ = (100/(z2 + 1)) cos φˆaρ = 10.0ˆaρ (put the values from point P)
⇒ (1/ρ)(∂V/∂φ)ˆaφ = −(100/(z2 + 1)) sin φˆaφ = −17.32ˆaφ (put the values from point P)
⇒ (∂V/∂z)ˆaz = −((100ρ cos φ)(2z))/(z2 + 1)2ˆaz = −24.0ˆaz (put the values from point P)
⇒ E = −10.0ˆaρ + 17.32ˆaφ + 24.0ˆaz
(c). | E |= (−10.0)2 + (17.32)2 + (24.0)2 = 31.24
(d). We have dV/dN = (dV/dL) |max=| E |= 31.24
(e). ˆaN = (−10.0ˆaρ + 17.32ˆaφ + 24.0ˆaz)/31.24 = −0.320ˆaρ + 0.554ˆaφ + 0.768ˆaz
(f). We have ρv = · D, now D = 0E (since the value of E is known to us so we can find D and then can
apply the divergence theorem to find ρv)
D4.9(a). p =3ˆax − 2ˆay + ˆaz, PA(2, 3, 4), We have V = p · r/4π 0 | r |3
r = (2 − 0)ˆax + (3 − 0)ˆay + (4 − 0)ˆaz = 2ˆax + 3ˆay + 4ˆaz, | r |=
√
22 + 32 + 42 = 5.38
⇒ V = (3ˆax − 2ˆay + ˆaz) · (2ˆax + 3ˆay + 4ˆaz) × 10−9/4π × 8.85 × 10−12× | 5.38 |3= 0.230V
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3. (b). r, θ & φ are given so we can find x ,y and z by using the relations from chapter 01
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
rest of the problem is similar to the part(a)
D4.10(a). This problem is similar to the problem D4.9(b).
(b). E = Qd(2cosθˆar + sin θˆaθ)/4π 0 | r |3 also V = Qd cos θ/4π 0 | r |2⇒ Qd = 4π 0 | r |2 V/ cos θ
⇒ E = V (2ˆar + tan θˆaθ)/ | r |⇒ E(r=4,θ=200) = 3.27 × (2ˆar + tan 200ˆaθ)/4 = 1.584ˆar + 0.288ˆaθV/m
we are using V = 3.17 as calculated in part(a )
D4.11(a). We have E = − V, V = 200/r
⇒ V = (∂V/∂r)ˆar +(1/r)(∂V/∂θ)ˆaθ +(1/r sin θ)(∂V/∂φ)ˆaφ ⇒ V = −200/r2 (since we are going to have a partial
derivative with respect to r only)
⇒ E = (200/r2)ˆar ⇒| E |2= 40, 000/r4 also dv = r2 sin θdθdφdr (for spherical coordinates).
now WE = (1/2) vol 0 | E |2 dv = ( 0/2) vol(40, 000/r4)r2 sin θdθdφdr = (20, 000 0) 3
2 dr/r2 π/2
0 sin θdθ
π/2
0 dφ
⇒ WE = (20, 000 0)× | −1/r |3
2 × | − cos θ |
π/2
0 × | φ |
π/2
0 = 2.3 × 10−4J
(b). Similar to part(a).
THE END
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