Electromagnatic

1. 1 Solution to the Drill problems of chapter 03
(Engineering Electromagnetics,Hayt,A.Buck 7th ed)
BEE 4A,4B & 4C
D3.2 (a). D =? at point P(2,-3,6)
QA = 55mC at point Q(-2,3,-6) now D = oE = QRPQ/(4π | RPQ |3)
RPQ = (2 − (−2))ˆax + (−3 − 3)ˆay + (6 − (−6))ˆaz = 4ˆax − 6ˆay + 12ˆaz, | RPQ |= 42 + (−6)2 + 122 =
√
196 = 14
⇒ D = (55 × 10−3(4ˆax − 6ˆay + 12ˆaz))/(4π(14)3) = 6.38ˆax − 9.57ˆay + 19.14ˆazµC/m2
(b). ρL = 20mC along X axis (∞ ≤ X ≤ ∞), now E = ρLˆapx/(2π o | Rpx |) = ρLRpx/(2π o | Rpx |2)
Rpx = P(2, −3, 6) − (x, 0, 0) = (2 − x)ˆax − 3ˆay + 6ˆaz since the infinite line charge is along X axis so the E filel
at point P is having only Y and Z components present and the X component is cancelled due to symmetry so
Rpx = −3ˆay + 6ˆaz, | Rpx |= (−3)2 + 62 =
√
45, D = oE = ρLRpx/(2π | Rpx |2) = 20 × 10−3(−3ˆay + 6ˆaz)/2π × 45
⇒ D = −212ˆay + 424ˆazµC/m2
(c). E = (ρs/2 o)ˆaN for infinite surface charge density also z=-5 is an infinite x-y plane located at z=-5 and the
charge is spread on this plane,so D = oE = (ρs/2)ˆaz = (120/2)µˆaz = 60µˆazC/m2
D3.3. D = 0.3r2ˆarnC/m2
(a). E =? at point P(r=2,θ = 25o, φ = 90o), D = oE ⇒ E = D/ o = ((0.3r2ˆarnC/m2)/8.85 × 10−12F/m)r=2 =
(0.3 × 22ˆarnC/m2)/8.85 × 10−12F/m = 135.5ˆarV/m
(b).Q= ? for a sphere of radius r=3, we have Q= D · ds, ds = r2sinθdθdφ ⇒ Q = 0.3r2 × 10−9 2π
0
π
0 r2sinθdθdφ
⇒ 0.3r2(4πr2) × 10−9 = 1.2πr4 × 10−9 = (1.2πr4)r=3 × 10−9 = 305nC
(c). ψ = Q =? for a sphere of radius r=4 follow the same procedure as in part (b) we will get ψ = 1.2πr4 × 10−9 =
(1.2πr4)r=4 × 10−9 = 965nC
D3.4. Find total electric flux leaving the cubical surface formed by the six planes at x,y,z=±5
(a). since both the given charges are enclosed by the cubical volume ( as evident by their given locations) according
to the gauss’s law ψ = Q1 + Q2 = 0.1µC + (1/7)µC = 0.243µC
(b). ρL = πµC at (-2,3,z), clearly this line charge distribution is passing through point x=-2 and y=3 and is parallel
to z axis, the total length of this charge distribution enclosed by the given cubical volume is 10 units as z = ±5 so
ψ = Q = ρL × 10 = πµ × 10 = 31.4µC
(c). ρs = 0.1µC on the plane y=3x, now this is a straight line equation in xy plane which passes through the
origin,we need to find the length of this line which is enclosed by the given volume, and once we find it, this length
is moving up and down along z axis between z = ±5 to form a plane, by putting y=5 we get x=5/3 and hence the
length of this line on the plane formed by the +ive x and +ive y axis is given by 52 + (5/3)2 = 5.270, the same
length we will get on the plane formed by -ive x and -ive y axix and the sum of these two lengths is 10.540,now
this straight line is moving between z = ±5 to form a plane whose area is given by 10×10.540 = 105.40, and this
plane has a surface charge density ρs = 0.1µC now according to gauss’s law ψ = Qenclosed ⇒ ψ = ρs× (area of the
plane)⇒ ψ = 0.1µC × (105.40) = 10.54µC
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This document is prepared in LATEX. (Email: ahmadsajjad01@ciit.net.pk)
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2. D3.5.(a).at r=0.5cm It is very much clear that a sphere of this much radius will enclose only the point charge
0.25 µC which is located at the origin now D = (Q/4πr2)ˆar ⇒ D = (0.25 × 10−6/4π(0.5 × 10−2)2)ˆar = 796ˆarµC
(b). at r = 1.5cm It is again clear that a sphere of this much radius will enclose the point charge 0.25µC which is
located at the origin and also the uniform surface charge density of 2 × 10−3C/m2 which is uniformly distributed
over a sphere of radius r = 1 now we have Q1 = 0.25 µC and Q2 = ρs× area of sphere= 2×10−3C × 4π(1 × 10−2)2 =
2.513 × 10−6C and now D = (Q1 + Q2/4πr2)ˆar ⇒ D = ((0.25 × 10−6 + 2.513 × 10−6)/4π(1.5 × 10−2)2)ˆar
⇒ D = 977ˆarµC/m2
(c). It is the same as part(b) the only difference is that now we have r = 2.5 which not only encloses the charges Q1
and Q2 (same as in part(b)) but also encloses Q3 and we can calculate the value of Q3 just like we calculated Q2 but
for Q3 calculation we are going to use r = 1.8cm and ρs = −0.6mC/m2
Q3 = ρs× area of sphere= -0.6×10−3C × 4π(1.8 × 10−2)2 = −2.5 × 10−6C and then the rest of the problem is same
as part(b)
(d). find ρs at r = 3 to make D = 0 at r = 3.5cm
now the point is that the uniform surface charge density established at r = 3 should give us a charge Q4 equal to the
sum of Q1, Q2 and Q3 but opposite in sign, so that when we calculate D = (Q1+Q2+Q3+Q4/4πr2)ˆar it should give us
a 0 value, now Q1 +Q2 +Q3 = 0.25×10−6C and Q4 = ρs× (area of sphere)r=3 ⇒ −0.25×10−6C = ρs ×4π(3×10−2)2
⇒ ρs = 0.25 × 10−6C/4π(3 × 10−2)2 = −28.3µC/m2
D3.6(a). Since we are going to find the total electric flux passing through the given rectangular surface in the ˆaz
direction at z = 2, so we get ds = dxdyˆaz and now ψ = D·ds ⇒ ψ = 3
1
2
0 (8xyz4ˆax +4x2z4ˆay +16x2yz3ˆaz)·dxdyˆaz
⇒ ψ = 3
1
2
0 16x2yz3dxdy = 16z3 3
1 ydy 2
0 x2dx = (16z3 × 4 × 8/3)z=2 = 1365pC
(b). D = oE ⇒ E = D/ o = ((8xyz4ˆax + 4x2z4ˆay + 16x2yz3ˆaz) × 10−12/8.85 × 10−12)x=2,y=−1,z=3
⇒ E = 146.4ˆax + 146.4ˆay − 195.2ˆazV/m
(c). Find Q, ∆v = 10−12m3
now ∂Dx/∂x = ∂(8xyz4)/∂x = (8yz4)x=2,y=−1,z=3 = −648, ∂Dy/∂y = ∂(4x2z4)/∂y = 0, ∂Dz/∂z = ∂(16x2yz3)/∂z
= (48x2yz2)x=2,y=−1,z=3 = −1728, now charge enclosed in volume ∆v = (∂Dx/∂x + ∂Dy/∂y + ∂Dz/∂z) × ∆v =
(−648 − 1728) × 10−12 × 10−12 = −2.38 × 10−24C
D3.7. (a). Find divD, atPA(2, 3, −1), D = (2xyz − y2)ˆax + (x2z − 2xy)ˆay + x2yˆaz
div D = (∂Dx/∂x + ∂Dy/∂y + ∂Dz/∂z) = (2yz − 2x)x=2,y=3,z=−1 = −10
(b). Find divD, atPB(ρ = 2, φ = 110o, z = −1), D = 2ρz2 sin2
φˆaρ + ρz2 sin 2φaφ + 2ρ2z sin2
φˆaz
div D = (1/ρ)∂(ρDρ)/∂ρ + (1/ρ)∂Dφ/∂φ + ∂Dz/∂z
⇒(1/ρ)∂(ρDρ)/∂ρ = (1/ρ)∂(2ρ2z2 sin2
φ)/∂ρ = (1/ρ)(4ρz2 sin2
φ) = (4z2 sin2
φ)ρ=2,φ=110o,z=−1 = 3.532
⇒(1/ρ)∂Dφ/∂φ = (1/ρ)∂(ρz2 sin 2φ)/∂φ = (1/ρ)(2ρz2 cos 2φ) = (2z2 cos 2φ)ρ=2,φ=110o,z=−1 = −1.532
⇒ ∂Dz/∂z = ∂(2ρ2z sin2
φ)/∂z = (2ρ2 sin2
φ)ρ=2,φ=110o,z=−1 = 7.064
⇒ div D = (1/ρ)∂(ρDρ)/∂ρ + (1/ρ)∂Dφ/∂φ + ∂Dz/∂z = 3.532 − 1.532 + 7.064 = 9.06
(c). Part (c) is similar to part (b) but we have to use the formula for div D in spherical coordinates as given
in the book.
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3. D3.8. Find volume charge density ρv
(a).D = (4xy/z)ˆax + (2x2/z)ˆay − (2x2y/z2)ˆaz
ρv = (∂Dx/∂x + ∂Dy/∂y + ∂Dz/∂z)
ρv = 4y/z + 0 + 4x2y/z3 = (4y/z3)(x2 + z2)
(b). Use the equation of ρv for cylindrical co ordinates,rest is the same as part(a)
(c). Use the equation of ρv for spherical co ordinates,rest is the same as part(a)
D3.9. D = (6ρ sin(φ/2))ˆaρ + (1.5ρ cos(φ/2))ˆaφ
s D · ds = vol ∆· dv
⇒ s D · ds = s((6ρ sin(φ/2))ˆaρ + (1.5ρ cos(φ/2))ˆaφ) · (ρdφdzˆaρ − dρdzˆaφ)
⇒ s(6ρ sin(φ/2)ρdφdz − s(1.5ρ cos(φ/2)dρdz
⇒ π
0
5
0 (6ρ sin(φ/2)ρdφdz − 2
0
5
0 (1.5ρ cos(φ/2)dρdz (this second surface lies at φ = 0o)
⇒ 6ρ2 π
0 (sin(φ/2)dφ 5
0 dz − (1.5 cos(φ/2)) 2
0 ρdρ 5
0 dz
⇒24 × | −2 cos(φ/2) |π
0 × | z |5
0 −(1.5 cos(φ/2))× | ρ2/2 |2
0 × | z |5
0
⇒ 24 × 2 × 5 − ((1.5 cos(φ/2)) × 2 × 10)φ=0o = 240 − 15 = 225
now ∆ · D=(1/ρ)∂(ρDρ)/∂ρ + (1/ρ)∂Dφ/∂φ + ∂Dz/∂z
⇒ (1/ρ)∂(ρDρ)/∂ρ = (1/ρ)∂(ρ(6ρ sin(φ/2))/∂ρ = (1/ρ)∂(6ρ2 sin(φ/2)/∂ρ = 12 sin φ/2
⇒ (1/ρ)∂(Dφ)/∂φ = (1/ρ)∂((1.5ρ cos(φ/2))/∂φ = (−1.5/2) sin φ/2
⇒ ∆ · D=(1/ρ)∂(ρDρ)/∂ρ + (1/ρ)∂Dφ/∂φ + ∂Dz/∂z = 12 sin φ/2 + (−1.5/2) sin φ/2 = 11.25 sin φ/2
vol ∆ · D dv= π
0
2
0
5
0 (11.25 sin φ/2)ρdφdρdz = 11.25 π
0 (sin φ/2)dφ 2
0 ρdρ 5
0 dz
⇒ 11.25× | −2cos(φ/2) |π
0 × | ρ2/2 |2
0 × | z |5
0= 11.25 × 2 × 2 × 5 = 225
THE END
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