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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 = o
E = Q
RP Q/(4π |
RP Q |
3 )
R
P Q
= (2 − (−2))ˆa x
(− 3 − 3)ˆa y
(6 − (−6))ˆa z
= 4ˆa x
− 6ˆa y
R
P Q
√
2
2
2
D = (55 × 10
− 3 (4ˆa x
− 6ˆa y
))/(4π(14)
3 ) = 6.38ˆa x
− 9 .57ˆa y
μC/m
2
(b). ρ L
= 20mC along X axis (∞ ≤ X ≤ ∞), now
E = ρ L
ˆa px
/(2π o
R
px
|) = ρ L
R
px
/(2π o
R
px
2 )
R
px
= P (2, − 3 , 6) − (x, 0 , 0) = (2 − x)ˆa x
− 3ˆa y
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
R
px
= −3ˆa y
R
px
√
2
2
D =
o
E = ρ L
R
px
/(2π |
R
px
2 ) = 20 × 10
− 3 (−3ˆa y
)/ 2 π × 45
D = −212ˆay + 424ˆaz μC/m
2
(c).
E = (ρ s
o
)ˆa N
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 =
o
E = (ρ s
/2)ˆa z
= (120/2)μˆa z
= 60μˆa z
C/m
2
D 3. 3.
D = 0. 3 r
2 ˆarnC/m
2
(a).
E =? at point P(r=2,θ = 25
o , φ = 90
o ),
D = o
E ⇒
E =
D/o = ((0. 3 r
2 ˆarnC/m
2 )/ 8. 85 × 10
− 12 F/m)r=2 =
(0. 3 × 2
2 ˆa r
nC/m
2 )/ 8. 85 × 10
− 12 F/m = 135.5ˆa r
V /m
(b).Q=? for a sphere of radius r=3, we have Q=
∮
~ D · d~s, d~s = r
2 sinθdθdφ ⇒ Q = 0. 3 r
2 × 10
− 9
∫ 2 π
0
∫ π
0
r
2 sinθdθdφ
⇒ 0. 3 r
2 (4πr
2 ) × 10
− 9 = 1. 2 πr
4 × 10
− 9 = (1. 2 πr
4 )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. 2 πr
4 × 10
− 9
(1. 2 πr
4 )r=4 × 10
− 9 = 965nC
D 3. 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 = Q 1
+ Q
2
= 0. 1 μC + (1/7)μC = 0. 243 μC
(b). ρ L
= 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 = ρ L
× 10 = πμ × 10 = 31. 4 μC
(c). ρs = 0. 1 μC 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
√
2
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. 1 μC now according to gauss’s law ψ = Q enclosed
⇒ ψ = ρs× (area of the
plane)⇒ ψ = 0. 1 μC × (105.40) = 10. 54 μC
1 This document is prepared in L
A TEX. (Email: ahmadsajjad01@ciit.net.pk)
D 3. 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/ 4 πr
2 )ˆa r
D = (0. 25 × 10
− 6 / 4 π(0. 5 × 10
− 2 )
2 )ˆa r
= 796ˆa r
μC
(b). at r = 1. 5 cm It is again clear that a sphere of this much radius will enclose the point charge 0. 25 μC which is
located at the origin and also the uniform surface charge density of 2 × 10
− 3 C/m
2 which is uniformly distributed
over a sphere of radius r = 1 now we have Q 1 = 0.25 μC and Q 2 = ρs× area of sphere= 2× 10
− 3 C × 4 π(1 × 10
− 2 )
2
2. 513 × 10
− 6 C and now
D = (Q
1
+ Q
2
/ 4 πr
2 )ˆa r
D = ((0. 25 × 10
− 6
− 6 )/ 4 π(1. 5 × 10
− 2 )
2 )ˆa r
D = 977ˆarμC/m
2
(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 Q 1
and Q 2
(same as in part(b)) but also encloses Q 3
and we can calculate the value of Q 3
just like we calculated Q 2
but
for Q 3
calculation we are going to use r = 1. 8 cm and ρ s
= − 0. 6 mC/m
2
Q
3
= ρ s
× area of sphere= -0.6× 10
− 3 C × 4 π(1. 8 × 10
− 2 )
2 = − 2. 5 × 10
− 6 C and then the rest of the problem is same
as part(b)
(d). find ρ s
at r = 3 to make
D = 0 at r = 3. 5 cm
now the point is that the uniform surface charge density established at r = 3 should give us a charge Q 4 equal to the
sum of Q 1
, Q
2
and Q 3
but opposite in sign, so that when we calculate
D = (Q
1
+Q
2
+Q
3
+Q
4
/ 4 πr
2 )ˆa r
it should give us
a 0 value, now Q 1 +Q 2 +Q 3 = 0. 25 × 10
− 6 C and Q 4 = ρs× (area of sphere)r=3 ⇒ − 0. 25 × 10
− 6 C = ρs × 4 π(3× 10
− 2 )
2
⇒ ρ s
= 0. 25 × 10
− 6 C/ 4 π(3 × 10
− 2 )
2 = − 28. 3 μC/m
2
D3.6(a). Since we are going to find the total electric flux passing through the given rectangular surface in the ˆa z
direction at z = 2, so we get
ds = dxdyˆa z
and now ψ =
∮
~ D ·
ds ⇒ ψ =
∫ 3
1
∫ 2
0
(8xyz
4 aˆ x
+4x
2 z
4 ˆa y
+16x
2 yz
3 ˆa z
)·dxdyˆa z
⇒ ψ =
∫ 3
1
∫ 2
0
16 x
2 yz
3 dxdy = 16z
3
∫ 3
1
ydy
∫ 2
0
x
2 dx = (16z
3 × 4 × 8 /3) z=
= 1365pC
(b).
D = o
E ⇒
E =
D/o = ((8xyz
4 ˆax + 4x
2 z
4 ˆay + 16x
2 yz
3 ˆaz ) × 10
− 12 / 8. 85 × 10
− 12 )x=2,y=− 1 ,z=
E = 146.4ˆa x
− 195 .2ˆa z
V /m
(c). Find Q, ∆v = 10
− 12 m
3
now ∂
D
x
/∂x = ∂(8xyz
4 )/∂x = (8yz
4 ) x=2,y=− 1 ,z=
D
y
/∂y = ∂(4x
2 z
4 )/∂y = 0, ∂
D
z
/∂z = ∂(16x
2 yz
3 )/∂z
= (48x
2 yz
2 )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
− 24 C
D3.7. (a). Find div
D, atP A
D = (2xyz − y
2 )ˆa x
2 z − 2 xy)ˆa y
2 yˆa z
div
D = (∂
D
x
/∂x + ∂
D
y
/∂y + ∂
D
z
/∂z) = (2yz − 2 x) x=2,y=3,z=− 1
(b). Find div
D, atP B
(ρ = 2, φ = 110
o , z = −1),
D = 2ρz
2 sin
2 φaˆ ρ
2 sin 2φa φ
2 z sin
2 φaˆ z
div
D = (1/ρ)∂(ρ
D
ρ
)/∂ρ + (1/ρ)∂
D
φ
/∂φ + ∂
D
z
/∂z
⇒(1/ρ)∂(ρ
D
ρ
)/∂ρ = (1/ρ)∂(2ρ
2 z
2 sin
2 φ)/∂ρ = (1/ρ)(4ρz
2 sin
2 φ) = (4z
2 sin
2 φ) ρ=2,φ=
o ,z=− 1
⇒(1/ρ)∂
D
φ
/∂φ = (1/ρ)∂(ρz
2 sin 2φ)/∂φ = (1/ρ)(2ρz
2 cos 2φ) = (2z
2 cos 2φ) ρ=2,φ=
o ,z=− 1
Dz /∂z = ∂(2ρ
2 z sin
2 φ)/∂z = (2ρ
2 sin
2 φ)ρ=2,φ=110o,z=− 1 = 7. 064
⇒ div
D = (1/ρ)∂(ρ
Dρ)/∂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.
