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)(8xyzax + 4x2zay − 4x2yaz)V/m,Q = 6nC, | ~dL |= 2µm,P (2,−2, 3)aL = (−6/7)ax + (3/7)ay + (2/7)az, Find dW~dL = aL | ~dL |= 2× 10−6((−6/7)ax + (3/7)ay + (2/7)az) = ((−12/7)ax + (6/7)ay + (4/7)az)× 10−6

dW= −Q~E · ~dL⇒ dW=-6 ×10−9((1/z2)(8xyzax + 4x2zay − 4x2yaz)) · (((−12/7)ax + (6/7)ay + (4/7)az)× 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 = 5axV/mwe have W = −Q

∫~E · ~dL, since the path of integration is a straight line so we have ~dL = dxax + dyay + dzaz

⇒W = −4∫

(5ax + 0ay + 0az) · (dxax + dyay + dzaz) = −4∫ 01 5dx = 20J

(b). Follow the same procedure as in part(a) and we get W = −4∫

(5xax + 0ay + 0az) · (dxax + dyay + dzaz)⇒W = −4

∫ 01 5xdx = −20× | x2/2 |01= 10J

(C). Follow the same procedure as in part(a) and we get W = −4∫

(5xax + 5yay + 0az) · (dxax + dyay + dzaz)⇒W = −4(

∫ 01 5xdx+

∫ 20 5ydy) = −20× (| x2/2 |01 + | y2/2 |20) = −20× (−(1/2) + 2) = −30J

D4.3 (a). ~E = yax, 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

∫(yax + 0ay + 0az) · (dxax) (dy and dz are zero for this line segment)

⇒ W1 = −3∫ 21 ydx = −3y | x |21= (−3y)y=3 = −9J

for (2,3,5) to (2,0,5)W2 = −3

∫(yax + 0ay + 0az) · (dyay) (dx and dz are zero for this line segment)

⇒ W2 = 0for (2,0,5) to (2,0,3)W3 = −3

∫(yax + 0ay + 0az) · (dzaz) (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 = 6x2ax + 6yay + 4azV/m, ~dL = dxax + dyay + dzaz find VMN ,M(2, 6,−1), N(−3,−3, 2)we have VAB = −

∫ AB~E · ~dL⇒ VMN = −

∫MN

~E · ~dL = −∫MN (6x2ax + 6yay + 4az) · (dxax + dyay + dzaz)

= −(6∫ 2−3 x

2dx+ 6∫ 6−3 ydy + 4

∫−12 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 = (−∫M0

~E · ~dL)− VQ(VM and VQ are the potential differences with respect to the origin (0, 0, 0))⇒ VMQ = −

∫M0 (6x2ax + 6yay + 4az) · (dxax + dyay + dzaz)− VQ

⇒ VMQ = −(6∫ 20 x

2dx+ 6∫ 60 ydy + 4

∫−10 dz)− VQ = −(6× | x3/3 |20 +6× | y2/2 |60 +4× | z |−1

0 )− VQ = −120− 0= −120 V (since VQ = 0)⇒VM = −120V

1This document is prepared in LATEX. (Email: ahmadsajjad01@ciit.net.pk)

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(c). we have VAB = VA − VB ⇒ VNP = VN − VP = (−∫N0~E · ~dL)− VP

(VN and VP are the potential differences with respect to the origin (0, 0, 0))⇒ VNP = −

∫N0 (6x2ax + 6yay + 4az) · (dxax + dyay + dzaz)− VP

⇒ VNP = −(6∫−30 x2dx+6

∫−30 ydy+4

∫ 20 dz)−VP = −(6× | x3/3 |−3

0 +6× | y2/2 |−30 +4× | z |20)−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/4πε0r) + C1, V = 0 at (6, 5, 4), r =

√62 + 52 + 42 = 8.77⇒ C1 = −(Q/4πε0r)

⇒ C1 = −15× 10−9/(4π × 8.85× 10−12 × 8.77) = −15.37now again we have V = (Q/4πε0r) + C1, but now we know the value of C1 and we will calculate the value of r thistimes with the point P1(−2, 3,−1)⇒ r =

√(−2)2 + 32 + (−1)2 = 3.74

⇒ V=(Q/4 πε0r) + 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 = 0Now V=(Q/4 πε0r) + 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 inthe z = 0 plane or x-y plane), we have V (r) =

∫ρL(r

′)dL

′/4πε0 | r − r

′ |, dL′= ρdφ, | r − r′ |=

√22 + (2.5)2 = 3.20

⇒ V (r) = (12× 10−9)× ρ∫ 2π0 dφ/(4πε0 × 3.20) = (12× 10−9)× 2.5× | φ |2π0 /(4πε0 × 3.20) = 529V

(b). Q = 18nC, P (1, 2,−1) for point charges we have ~V (r) = Q/4πε0 | 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 cosφV, 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)aρ + (1/ρ)(∂V/∂φ)aφ + (∂V/∂z)az⇒ (∂V/∂ρ)aρ = (100/(z2 + 1)) cosφaρ = 10.0aρ (put the values from point P)⇒ (1/ρ)(∂V/∂φ)aφ = −(100/(z2 + 1)) sinφaφ = −17.32aφ (put the values from point P)⇒ (∂V/∂z)az = −((100ρ cosφ)(2z))/(z2 + 1)2az = −24.0az (put the values from point P)⇒ ~E = −10.0aρ + 17.32aφ + 24.0az

(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). aN = (−10.0aρ + 17.32aφ + 24.0az)/31.24 = −0.320aρ + 0.554aφ + 0.768az

(f). We have ρv = ∇ · ~D, now ~D = ε0 ~E (since the value of ~E is known to us so we can find ~D and then canapply the divergence theorem to find ρv)

D4.9(a). ~p =3ax − 2ay + az, PA(2, 3, 4), We have V = ~p · ~r/4πε0 | ~r |3~r = (2− 0)ax + (3− 0)ay + (4− 0)az = 2ax + 3ay + 4az, | ~r |=

√22 + 32 + 42 = 5.38

⇒ V = (3ax − 2ay + az) · (2ax + 3ay + 4az)× 10−9/4π × 8.85× 10−12× | 5.38 |3= 0.230V

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(b). r, θ & φ are given so we can find x ,y and z by using the relations from chapter 01x = r sin θ cosφy = r sin θ sinφz = r cos θrest 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(2cosθar + sin θaθ)/4πε0 | ~r |3 also V = Qd cos θ/4πε0 | ~r |2⇒ Qd = 4πε0 | ~r |2 V/ cos θ⇒ ~E = V (2ar + tan θaθ)/ | ~r |⇒ ~E(r=4,θ=200) = 3.27× (2ar + tan 200aθ)/4 = 1.584ar + 0.288aθV/mwe are using V = 3.17 as calculated in part(a )

D4.11(a). We have ~E = −∇V, V = 200/r⇒ ∇V = (∂V/∂r)ar+(1/r)(∂V/∂θ)aθ+(1/r sin θ)(∂V/∂φ)aφ ⇒ ∇V = −200/r2 (since we are going to have a partialderivative with respect to r only)⇒ ~E = (200/r2)ar ⇒| ~E |2= 40, 000/r4 also dv = r2 sin θdθdφdr (for spherical coordinates).now WE = (1/2)

∫vol ε0 | ~E |2 dv = (ε0/2)

∫vol(40, 000/r4)r2 sin θdθdφdr = (20, 000ε0)

∫ 32 dr/r

2∫ π/20 sin θdθ

∫ π/20 dφ

⇒WE = (20, 000ε0)× | −1/r |32 × | − cos θ |π/20 × | φ |π/20 = 2.3× 10−4J

(b). Similar to part(a).

THE END

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