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Solution to the Drill problems of chapter 02
(Engineering Electromagnetics,Hayt,A.Buck 7th ed)
BEE 4A,4B & 4C
Following Exercise questions are IMPORTANT!
2.4, 2.5, 2.13, 2.14, 2.16 ,2.17, 2.18, 2.19, 2.22, 2.23, 2.27, 2.28,2.29,2.30,2.31
D2.1 (a). QA = −20µC located at A(-6,4,7) ,QB = 50µC located at B(5,8,-2)
~ AB
Find R
~ AB = (5 − (−6))âx + (8 − 4)ây + (−2 − 7)âz = 11âx + 4ây − 9âz
R
~ AB |=
(b). | R
p
(112 ) + 42 + (−9)2 = 14.76m
~ AB /4πo | R
~ AB |3 = (−20 × 10−6 × 50 × 10−6 (11âx + 4ây − 9âz ))/(4π × (10−9 /36π) | 14.76 |3 )
(c). F~AB = QA QB R
⇒ F~AB = 30.76âx + 11.184ây − 25.16âz mN
~ AB /4πo | R
~ AB |3 = (−20 × 10−6 × 50 × 10−6 (11âx + 4ây − 9âz ))/(4π × 8.85 × 10−12 | 14.76 |3 )
(d).F~AB = QA QB R
⇒ F~AB = 30.72âx + 11.169ây − 25.13âz mN
D2.2(a). QA = −0.3µC located at A(25,-30,15) in cm ,QB = 0.5µC located at B(-10,8,12)
~ at the origin O(0,0,0).
Find E
~ at the origin is denoted by E
~ o and it will be the sum of E
~A ( E
~ due to QA located at point A)
Let E
~
~
and EB ( E due to QB located at point B)
~ OA /4πo | R
~ OA |3
~ A = QA R
E
~ OA = (0 − 25))âx + (0 − (−30))ây + (0 − 15)âz = (−25âx + 30ây − 15âz )cm
R
p
~ OA |= (−25)2 + (30)2 + (−15)2 = 41.83cm
|R
~ A = (−0.3×10−6 )×(−25âx +30ây −15âz )×10−2 /4π×8.85×10−12 × | 41.83×10−2 |3 = −368.55(−25âx +30ây −15âz )
E
~ OB /4πo | R
~ OB |3
~ B = QB R
E
~ OB = (0 − (−10)))âx + (0 − 8)ây + (0 − 12)âz = (10âx − 8ây − 12âz )cm
R
p
~ OB |= (10)2 + (−8)2 + (−12)2 = 17.55cm
|R
~ B = (0.5×10−6 )×(−25âx +30ây −15âz )×10−2 /4π ×8.85×10−12 × | 17.55×10−2 |3 = 8317.36(−25âx +30ây −15âz )
E
~o = E
~A + E
~ B = (−368.55(−25âx + 30ây − 15âz )) + 8317.36(10âx − 8ây − 12âz ) = (92.3âx − 77.6ây − 94.2âz )KV /m
E
~ at the point P(15,20,50).
(b). Find E
~ P A and R
~ P B and the rest of the problem is similar to
It is the same as part(a) but this time we have to calculate R
part(a)
D2.3 (a).
Σ50 ((1 + (−1)m )/(m2 + 1)) = (1 + (−1)0 )/(02 + 1) + (1 + (−1)1 )/(12 + 1) + (1 + (−1)2 )/(22 + 1) + (1 + (−1)3 )/(32 +
1) + (1 + (−1)4 )/(42 + 1) + (1 + (−1)5 )/(52 + 1) = 2 + 0 + 2/5 + 0 + 2/17 + 0 = 2.52
(b). Similar to the part(a)
D2.4 (a). 0.1 ≤ (| x |, |R y |, | z |) ≤ 0.2 , given ranges of x,y and z co-ordinates doesnot constitute a cubical
volume so dv = 0 ⇒ Q = vol ρv dv = 0
R
(b). Differential volume in cylindrical co-ordinates is given by dv = ρdρdφdz , we have Q = vol ρv dv
R
R
R R
R
R
R
⇒ Q = vol (ρ2 z 2 sin(0.6)φ)ρdρdφdz = 00.1 0π 24 (ρ2 z 2 sin(0.6)φ)ρdρdφdz = 00.1 ρ3 dρ 0π (sin(0.6)φ)dφ 24 dz
π
4
4
0
⇒ Q =| ρ4 /4 |0.1
0 × | (−cos(0.6φ))/0.6 |0 × | z |2 =| (0.1) /4 | × | (−cos(108 ) − (−cos(0)))/0.6 | × | (64 − 8)/3 |
⇒ Q =| (0.1)4 /4 | × | (1.31)/0.6 | × | 56/3 |= 1.018mC
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This document is prepared in LATEX. (Email: ahmadsajjad01@ciit.net.pk)
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(C). Assuming this universe to be a perfect sphere we have limits as 0 ≤ r ≤ ∞, 0 ≤ φ ≤ 2π, 0 ≤ θ ≤ π
and the Rdifferential Rvolume for spherical co-ordinates
system
is gien
by dv = (r sin θdφ)(rdθ)(dr) = r2 sin θdθdφdr
R 2π
Rπ
R ∞ −2r
π
−2r /−2 |∞
−2r
2
2
⇒ Q = vol ρv dv = vol (e /r )r sin θdθdφdr = 0 dφ 0 sin θdθ 0 e dr =| φ |2π
0 × | (−cosθ) |0 × | e
0
⇒ Q = 2π × 2 × 1/2 = 2πC = 6.28C
~ is given by E
~ = ρL âR /2πo | R
~ |= ρL R/2π
~
~ 2
~
D2.5 (a). For infinite uniform line charge E
o | R | where R is
the perpendicular distance vector between the line charge and the point under consideration, in this case the point
~ at PA is
is PA (0,0,4), since we have two infinite line charges,one along x-axis and one along y-axis so the value of E
~ 1 (E
~ due to infinite line charge along x-axis ) and E
~ 2 (E
~ due to infinite line charge along y-axis )
the sum of E
2
~
~
~
~
so E1 = ρL Rpx /2πo | Rpx | , Rpx =( Perpendicular distance vector between point PA and the line charge along
x-axis)
~ px = 4âz , now E
~ 1 = 5 × 10−9 × 4âz /(2π × 8.85 × 10−12 × 16) = 22.479âz V /m
⇒R
using similar arguments we can find
~ py /2πo | R
~ py |2 = 5 × 10−9 × 4âz /(2π × 8.85 × 10−12 × 16) = 22.479âz V /m
~ py = 4âz so E
~ 2 = ρL R
R
~ =E
~1 + E
~ 2 = 22.479âz V /m + 22.479âz V /m = 45âz V /m
⇒E
~ px = 3ây + 4âz , since the point is PB (0, 3, 4)
(b). Using the same arguments as in part(a) we have R
~ py = +4âz
and R
~ px /2πo | R
~ px |2 = 5 × 10−9 × (3ây + 4âz )/(2π × 8.85 × 10−12 × 25) = 10.8ây + 14.38âz
~ 1 = ρL R
E
~ 2 = ρL R
~ py /2πo | R
~ py |2 = 5 × 10−9 × 4âz /(2π × 8.85 × 10−12 × 16) = 22.479âz
E
~ 2 = 10.8ây + 14.38âz + 22.479âz = 10.8ây + 36.9âz
~ =E
~1 + E
⇒E
D2.6 (a). PA = (2, 5, −5) , since the point PA is located below all the given surfaces or shet of charges so the
~ will be in the -ive âz direction i.e −âz
unit normal vector âN to these surfaces ,which also shows the diretion of E
~
~
E at point PA will be the sum of E’s caused by each sheet of charge
~ =E
~ z=−4 + E
~ z=1 + E
~ z=4 , now we have E
~ = (ρs /2 × o )âN
⇒E
−9
−12
~
⇒ Ez=−4 = −(3 × 10 /2 × 8.85 × 10 )âz
~ z=1 = −(6 × 10−9 /2 × 8.85 × 10−12 )âz
⇒E
~ z=4 = −(−8 × 10−9 /2 × 8.85 × 10−12 )âz
⇒E
~ = −(3 × 10−9 /2 × 8.85 × 10−12 )âz − (6 × 10−9 /2 × 8.85 × 10−12 )âz + (8 × 10−9 /2 × 8.85 × 10−12 )âz
⇒E
~ = −(1 × 10−9 /2 × 8.85 × 10−12 )âz = −56.5âz V /m
⇒E
~ contributed by the surface charge at
(b). PB (4, 2, −3) ,location of point PB (4, 2, −3) is suggesting that the E
~
z = −4 will be in âz direction and all other E’s contributed by the other surfaces will be in -âz direction
~ = (3 × 10−9 /2 × 8.85 × 10−12 )âz − (6 × 10−9 /2 × 8.85 × 10−12 )âz + (8 × 10−9 /2 × 8.85 × 10−12 )âz
⇒E
~ = (5 × 10−9 /2 × 8.85 × 10−12 )âz = 282.4âz V /m
⇒E
(c). PC (−1, −5, 2), using the similar arguments as in part(b) about the location of the given point we can
~ in the −iveâz direction and all other surfaces will produce E
~
notice easily that the surface at z = 4 will produce E
in the +iveâz direction
~ = (3 × 10−9 /2 × 8.85 × 10−12 )âz + (6 × 10−9 /2 × 8.85 × 10−12 )âz + (8 × 10−9 /2 × 8.85 × 10−12 )âz =
⇒E
~ = (17 × 10−9 /2 × 8.85 × 10−12 )âz 960.45âz V /m
⇒E
(d). PD (−2, 4, 5), using the similar arguments as in part(b) about the location of the given point we can
~ in the +iveâz direction
notice easily that all the surfaces will produce E
−9
−12
−9
~ = (3 × 10 /2 × 8.85 × 10 )âz + (6 × 10 /2 × 8.85 × 10−12 )âz − (8 × 10−9 /2 × 8.85 × 10−12 )âz = 56.5âz V /m
⇒E
~ = (−8x/y)âx + (4x2 /y 2 )ây , P (1, 4, −2), we have dy/dx = E
~ y /E
~ x ⇒ dy/dx = (4x2 /y 2 )/(−8x/y) ⇒
D2.7(a). E
R
R
2
2
dy/dx = (−x/2y) ⇒ 2ydy = −xdx ⇒ 2 ydy = − xdx ⇒ 2 × y /2 = −x /2 + c2 /2 ⇒ y 2 = −x2 /2 + c2 /2
⇒ (y 2 = −x2 /2 + c2 /2)P (1,4,−2) ⇒ c2 = 33
⇒ x2 + 2y 2 = 33
2
~ y /E
~ x = x/y(5x + 1) ⇒ ydy = xdx/(5x + 1) ⇒ ydy = xdx/(5x + 1) ⇒ y 2 /2 = xdx/(5x + 1) + c2 /2
(b). dy/dx = E
putR (5x + 1) = t ⇒ 5dx
= dt also (5x + 1) = t ⇒R x = (t − 1)/5
R
R
R
⇒ xdx/(5x + 1) = ((t − 1)/5)dt/5t = (1/25) ((t − 1)/t)dt = (1/25)(
dt
−
dt/t) = (1/25)(t − ln | t |)
R
= (0.04t − 0.04 ln | t |)(t=5x+1) = 0.04(5x + 1) − 0.04 ln | (5x + 1) |⇒ xdx/(5x + 1) = 0.04(5x + 1) − 0.04 ln | (5x + 1) |
now we have y 2 /2 = 0.04(5x + 1) − 0.04 ln | (5x + 1) | +c2 /2
⇒ y 2 /2 = 0.04(5x + 1) − 0.04 ln | (5x + 1) | +c2 /2)P (1,4,−2) ⇒ c2 = 15.66
⇒ y 2 = 0.04(5x + 1) − 0.04 ln | (5x + 1) | +(15.66)/2
⇒ y 2 = 15.74 + 0.4x − 0.08 ln | (5x + 1) |
R
THE END
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R
R