Problems
1. Given a 60-μC point charge located at the origin, find the total electric flux
passing through: (a) that portion of the sphere r = 26 cm bounded by 0 <
θ <π/2 and 0 < φ <π/2 ; (b) the closed surface defined by ρ = 26 cm and z =
±26 cm; (c) the plane z = 26 cm.
Solution:
 =  Ds  dS
(a)
S
Where
 =
 /2  /2
 
0
0
Dat distance r =
Q
aˆr
4 r 2
dS=r 2 sin  d  d aˆr
Q
Q
aˆ  r 2 sin  d  d aˆr =
2 r
4 r
4
 /2
 /2
0
0
Q
 d  sin d  = 8 =
60
= 7.5  C
8
(b) The closed surface designed by ρ = 26 cm and z = ±26 cm, enclose the point
charge, so its flux of 60-μC pass through the surface.
  = Q = 60 C
(c) The plane z = 26 cm, intercept all flux that enters the +z half space so,
 =
Q
= 30 C
2
2. An electric field in free space is E = (5z 2 /ɛ0) a z V/m. Find the total charge
contained within a cube, centered at the origin, of 4-m side length, in
which all sides are parallel to coordinate axes (and therefore each side
intersects an axis at ±2).
Solution:
Qencl =
D
s
S
 dS
where,
D = o E = 5z 2az C/m2
Qencl =  Ds  dS =
S
 D
front
 D
 dS +
front
back
left
2 2
top
right
 dS
right
+  Dleft  dS +  Dtop  dS +
Qencl =
 D
 dS +
back

Dbottom  dS
bottom
2 2
2 2
2
2
  (5z az ) x =2  dydzax +   (5z az ) x =−2 dydz (−ax ) +   (5z az ) y =2  dxdzay
2
−2 −2
−2 −2
2 2
+   (5z 2 a
z
−2 −2
−2 −2
2 2
)
y =−2
2 2
 dxdz ( −ay ) +   (5z 2 a )  dxdyaz +   (5z 2 a )
 dxdy ( −az )
z
z
z =2
z =−2
−2 −2
−2 −2
2
Qencl = 0 + 0 + 0 + 0 + 5 (2)
2
2
−2dx −2dy
2
- 5 (-2)
2
2
−2dx −2dy =
0
3. Calculate D in rectangular coordinates at point P(2,−3, 6) produced by: (a)
a point charge Q A=55 mC at Q(−2, 3,−6); (b) a uniform line charge ρ LB=20
mC/m on the x axis; (c) a uniform surface charge density ρ SC=120 μC/m 2
on the plane z=−5 m.
Solution:
(a) a point charge Q A=55 mC at Q(−2, 3,−6)
D = o E =
Q
Q
a =
RQP
2 RQP
4 RQP
4 RQP 3
RQP = 4ax − 6ay + 12az
RQP = 16 + 36 + 144 = 14
55  10−3
D=
(4ax − 6ay + 12az ) = 6.38ax − 9.57ay + 19.14az  C/m 2
3
4 (14)
(b) a uniform line charge ρ LB=20 mC/m on the x axis;
D = o E =
 LB

aR = LB 2 R
2 R
2 R
RQP = −3ay + 6az
D=
R = 9 + 36 = 45
20  10−3
(−3ay + 6az ) = −212ay + 424az  C/m 2
2
4 ( 45)
(c) a uniform surface charge ρ SC=120 μC/m 2 on the plane z=−5 m.
D = o E =
D=
SC
2
aN
where, aN = az
120  10−6
az = 60az  C/m 2
2
4. Given the electric flux density, D=0.3r 2ar nC/m 2 in free space:(a) find E at
point P(r=2, θ=25◦, φ=90◦); (b) find the total charge within the sphere r=3;
(c) find the total electric flux leaving the sphere r=4.
Solution:
(a) E =
D
o
=
0.3r 2
o
at r = 2
ar
E = 135.7ar
V/m
(b) The total charge within the sphere r=3
Qenc =  Ds  dS =  Dr =3 .dS =
S
S
2

0
0
2 
  0.3(3) a
2
r
 (3) 2 sin  d  d ar
0 0
= 0.3(3)4  10−9  d  sin  d  = 0.3(3)4  10−9  4 = 305 nC
OR Q =  = Dr =3  (4 r ) = 0.3(3)  10  4  (3) = 305 nC
2
−9
2
2
(c) The total electric flux leaving the sphere r=4.
 = Qencl =  Ds  dS =  Dr =4 .dS =
S
S
2

0
0
2 
  0.3(4) a
2
r
 (4) 2 sin  d  d ar
0 0
= 0.3(4)4  10−9  d  sin  d  = 0.3(4) 4  10−9  4 = 965.1 nC
OR Q =  = Dr =4  (4 r ) = 0.3(4) 10  4  (4) = 965.1 nC
2
2
−9
2
5. Let D=4xya x+2(x2+z2)ay+4yza z nC/m 2,evaluate surface integrals to find the
total charge enclosed in the rectangular parallelepiped 0<x<2, 0<y<3,
0<z<5 m.
6. Calculate the total electric flux leaving the cubical surface formed by the
six planes x, y, z=±5 if the charge distribution is: (a) two point charges, 0.1
μC at (1,−2, 3) and 1/7μC at (−1, 2,−2); (b) a uniform line charge of π μC/m
at x=−2, y=3;
(c) a uniform surface charge of 0.1 μC/m 2 on the plane y=3x.
Solution:
 = Qenc
(a) since both the given charges are enclosed by the cubical volume, according
to the gauss's law
1
7
 = Q1 + Q 2 = 1 + = 0.243 C
(b) It is clear that, the total length of the line enclosed by the given cubical
volume is 10 units as z = ± 5 so
 = Qenc = L  Linside =  10 = 10 = 31.4 C
(c) A uniform surface charge of 0.1 μC/m 2 on the plane y=3x.
• y=3x, 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 is given by
z
Lhor = 2 (5 / 3)2 + (5)2 = 10.54 m
x =-5/3
Lver = 10 m
•
y=5
The enclosed area inside the given
volume is
y
S = Lhor  Lver = 10.54 10 = 105.4 m2
 = Qenc = S  S inside
= 0.1 105.4 = 10.54 C
x
7. In free space, a volume charge of constant density ρ ν =ρ0 exists within the
region −∞ < x < ∞,−∞ < y < ∞, and −d/2 < z < d/2. Find D and E everywhere.
Solution:
Using Gauss's law
Qencl =  Ds  dS
z
S
The gaussian surface is chosen as cubical
surface where -1<x<1, -1<y<1 and -d/2<z<d/2
as shown in the figure
• From symmetry of the charge distribution we
find that
y
x
D = Dz aˆz , Dz is uniform with x and y
i)
Inside the charge dist.
1 1
Qencl =
 D
−1 −1
Qencl =

vol
z
1 1
1 1
−1 −1
−1 −1
(az )  dxdyaz +   D z (−az )  dxdy (−az ) = 2   D z dxdy = 8D z
z
v dv = + 
1 1

− z −1 −1
o
dxdydz ' = 8z o → (2
→ (1
From 1) and 2)
Din = z o aˆz ,
Dz = z o
ii)
and
Ein =Din /o = (o z /o )aˆz
Outside the charge dist.
1 1
1 1
1 1
−1 −1
−1 −1
−1 −1
  D z (az )  dxdyaz +   D z (−az )  dxdy (−az ) = 2   D z dxdy = 8D z
Qencl =
Qencl =

v dv = +
vol
d /2 1 1
 
o
→ (1
dxdydz ' = 4d o → (2
−d /2 −1 −1
From 1) and 2)
D z = d o /2
and
(d o /2)aˆz
Dout = 
Eout
z d /2
z d /2
−(d o /2)aˆz
z d /2
(d  /2 )aˆ
= o o z
z d /2
−(d o /2 o )aˆz
8. A point charge of 0.25 μC is located at r=0, and uniform surface charge
densities are located as follows: 2 mC/m 2 at r=1 cm, and −0.6 mC/m 2 at r
= 1.8 cm. Calculate D at: (a) r=0.5 cm; (b) r=1.5 cm; (c) r=2.5 cm. (d) What
uniform surface charge density should be established at r = 3 cm to cause
D = 0 at r = 3.5 cm?.
Solution:
(a) At r = 0.5 cm: the virtual sphere encloses only Qo
Qo
0.25  10−6
a
=
a = 796 ar C/m2
2 r
−2 2 r
4 r
4 (0.5  10 )
(b) At r = 1.5 cm: the virtual sphere encloses Q o and ρ S1
Qo + Q1
0.25  10−6 + ( s 1  4 r 2 )
D=
ar =
ar
4 r 2
4 (1.5  10−2 ) 2
D=
D=
0.25  10−6 + (2  10−3  4 (1  10−2 ) 2 )
0.25  10−6 + 2.513  10−6
a
=
ar
r
4 (1.5  10−2 ) 2
4 (1.5  10−2 ) 2
= 977 ar  C/m 2
(c) At r = 2.5 cm: the virtual sphere encloses Qo, ρS1 and ρS2
D=
=
Qo + Q1 + Q 2
0.25  10−6 + 2.513  10−6 + ( s 2  4 r 2 )
a
=
ar
r
4 r 2
4 (2.5  10−2 ) 2
0.25  10−6 + 2.513  10−6 + (−0.6  10−3  4 (1.8  10−2 ) 2 )
ar
4 (2.5  10−2 ) 2
0.25  10−6 + 2.513  10−6 − 2.44  10−6
=
ar = 40.8 ar  C/m 2
−2 2
4 (2.5  10 )
(d) At r = 3.5 cm: the virtual sphere encloses Qo , ρS1, ρ S2 and ρ S3
D r =3.5 =
D r =3.5 = 0
→ Q 3 = −(Qo + Q1 + Q 2 ) = −0.323  C
Qo + Q1 + Q 2 + Q 3 = 0
Q 3 = s 3 (4 r32 )
Qo + Q1 + Q 2 + Q 3
ar
4 r 2
→ s 3 =
Q3
−0.323  C
=
= −28.5
2
4 r3
4 (3  10−2 ) 2
C / m 2
9. Volume charge density is located in free space as ρ ν =2e−1000r nC/m 3 for
0<r<1 mm, and ρ ν = 0 elsewhere. (a) Find the total charge enclosed by the
spherical surface r=1 mm. (b) By using Gauss’s law, calculate the value of
Dr on the surface r=1 mm.
10. In free space, let D=8xyz 4ax+4x 2z4ay+16x 2yz3az pC/m 2. (a) Find the total
electric flux passing through the rectangular surface z = 2, 0<x<2, 1<y<3,
in the a z direction. (b) Find E at P(2,−1, 3). (c) Find an approximate value
for the total charge contained in an incremental sphere located at P(2,−1,
3) and having a volume of 10 −12 m 3 .
11. In each of the following parts, find a numerical value for div D at the point
specified: (a) D = (2xyz−y 2)ax + (x 2 z−2xy)ay+x2ya z C/m 2 at P A(2,3,−1); (b)
D=2ρz 2 sin 2 φ a ρ +ρz2sin2φ a φ + 2ρ 2 z sin 2φ a z C/m 2 at P B (ρ=2, φ =110◦,
z=−1); (c) D=2r sin θ cos φ a r +r cos θ cos φ a θ − r sin φ a φ C/m 2 at PC
(r=1.5, θ=30◦,φ=50◦).
12. Determine an expression for the volume charge density associated with
each D field: (a) D=(4xy/z)a x +(2x 2 /z)ay–(2x2/yz 2)az; (b) D=z sinφ a ρ +z cosφ
aφ +ρ sin φ a z ; (c) D= sin θ sin φ a r +cos θ sin φ a θ +cos φ a φ .
Solution:
v =  D
Maxwell's first equation states that
v
(a)
  4xy 
  2x
=
+
+
=

+

x
y
z
x  z   y  z
D y
 Dx
D z
4 y 2x 2 4 y 2 + 2x 2
v =
+
=
z
yz
yz
v =
(b)
1 
 
+

z
(D  ) +
(  sin ) =
1  D
 
zsin

+
+
D z
z
=
1 
 
( − zsin )

2
2
2x
 
 +  z (− yz 2 )

C/m
3
(  zsin ) +
+0=0
1 
 
(z cos  )
1  2
1

1  D
(r D r ) +
(sin  D ) +
2
r r
r sin   
r sin   
1  2
1

1 
= 2
(r sin sin ) +
(sin  cos sin ) +
(cos  )
r r
r sin   
r sin   
v =
(c)
=
2sin sin cos2θsin sin
1
+
−
=
(2sin 2 sin + cos2θsin − sin )
r
r sinθ
r sinθ r sinθ
13. Given the field D = 6ρsin0.5φ a ρ +1.5ρ cos 0.5φ a φ C/m 2 , evaluate both sides
of the divergence theorem for the region bounded by ρ=2, φ=0, φ=π, z=0,
and z=5.
Solution:
divergence theory states that
 D  dS =    D dv
s
S
v
Over ρ=2, φ=0, φ=π, z=0, and z=5
L .H .S =  Ds  dS =
S
For
For
For
For

curved
Ds  dS +  Ds  dS +
top

Ds  dS +
bottom
curved dS= ρdφdz a ρ ,
ρ=2
top dS= ρdρdφ a z ,
z=5
bottom dS= -ρdρdφ a z ,
z=0
inside
(1) at +ve x-direction dS= dρdz (-a φ ), φ=0
(b) at -ve x-direction dS= dρdz a φ, φ= π/2
 D  dS
s
inside
5
 2
00
 2
00
L .H .S =  D  =2  d dz a +  Dz =5  d d  az
52
52
+  D z =0  d d  ( - az ) +  D =0  d dz ( - a ) +  D =  d dz a
00
5
00
00


52
L .H .S =  6  sin ( )  =2 d dz + 0 + 0 - 1.5 cos ( ) =0 d dz
2
2
00
00

52
5


5
2
+ 1.5 cos ( ) = d dz = 6(2)2  dz  sin ( )d  +  dz 1.5(−  )d  + 0
2
2
00
0
0
0
0

5

5
2
L .H .S = 6(2)2  dz  sin ( )d  +  dz 1.5(−  )d  = 24  5  2 -1.5  5  2 = 255 C
2
0
0
0
0
R .H .S =    D dv
dv = d d dz
v
D =
=
1 
 
( D  ) +
1  D
 
+
D z
z
 1 


(6  2 sin ) +
(1.5  cos ) +
(0)
 
2  
2 z
 3  45 
  D = 12 sin − sin = sin
C/m3
2 4
2 4
2
1 
5 2
R .H .S =    D dv =   
v
0 0 0
45

sin d d dz
4
2
45  2 5
45
(
)0 (z)0 (-2cos( / 2))0 = * 2*5* 2 = 225
4 2
4
2
=
  D  dS =    D dv = 225 C
S
v
C
14. An infinitely long cylindrical dielectric of radius b contains charge within
its volume of density ρ v = aρ 2 , where a is a constant. Find the electric field
strength, E, both inside and outside the cylinder.
Solution:
Qencl =  Ds  dS
Using Gauss's law
z
b
S
•
From symmetry of the charge distribution we find that
D = D aˆ ,
ρ
D  is uniform with  and z
i)
For inside field:
The gaussian surface is chosen as cylinder of radius ρ and
unit length in z where ρ < b as shown in the figure.
 Ds  dS =
S

curved
 D  dS = 
s
S
curved
+
Ds  dS +  Ds  dS +
top

bottom
top
D  aˆ  (−aˆz ) d d 
bottom
1 2
1
2
 D  dS =   D  d dz =  D   dz  d  = 2 D 
s
S
x
D  aˆ  aˆ d dz +  D aˆ  aˆz d d 

y
Ds  dS
0 0
0
→ (1)
0
The total enclosed charge:
1 2 
4
Qenc =  v dv =    (a ' )  'd  'd dz = 2 a
→ (2)
4
vol
0 0 0
4
a 3
From (1) and (2)
2 D  = 2 a
→ D =
2
 D=D  aˆ =
a
aˆ
4
4
3
and
E=
D a
=
aˆ
o 4o 
3
4
for   b
ii)
For outside field:
The gaussian surface is chosen as cylinder of radius ρ and unit length in z where
ρ > b as shown in the figure.
 D  dS = 
s
S
D
S
curved
s
Ds  dS +  Ds  dS +
top

Ds  dS
bottom
1 2
1
2
0 0
0
0
 dS =   D  d dz =  D   dz  d  = 2 D  → (1)
The total enclosed charge:
z
1 2 b
Qenc =
2
 v dv =    (a ' )  'd  'd dz = 2 a
vol
0 0 0
4
b
4
b
→ (2)
ρ
From (1) and (2)
2 D  = 2 a
b4
4
→
D =
ab 4
4
ab 4
 D=D  aˆ =
aˆ
4
and
D ab 4
E= =
aˆ
 o 4 o  
y
x
for   b
15. In cylindrical coordinates, let ρ ν =0 for ρ<1 mm, ρ ν =2 sin(2000πρ) nC/m 3
for 1 mm<ρ<1.5 mm, and ρ ν =0 for ρ>1.5 mm. Find D everywhere.
16. Within the spherical shell, 3<r<4 m, the electric flux density is given as D
= 5(r − 3) 3 a r C/m 2 . (a) What is the volume charge density at r=4? (b) What
is the electric flux density at r=4? (c) How much electric flux leaves the
sphere r=4? (d) How much charge is contained within the sphere r=4?
17. Let D=5.00r 2ar mC/m 2 for r≤0.08 m and D=0.205a r /r2 μC/m 2 for r ≥ 0.08
m. (a) Find ρ ν for r = 0.06 m. (b) Find ρν for r = 0.1 m. (c) What surface
charge density could be located at r = 0.08 m to cause D = 0 for r > 0.08
m? (d) Repeat, but use ∇·D=ρ ν and take an appropriate volume integral.
18. In the region of free space that includes the volume 2<x, y, z<3, D =
(2/z 2 )(yz a x +xz a y −2xy a z ) C/m 2 . (a) Evaluate the volume integral side of
the divergence theorem for the volume defined here. (b) Evaluate the
surface integral side for the closed surface.
19. Given the flux density D=(16cos(2θ)/r) a θ C/m 2 , use two different methods
to find the total charge within the region 1< r <2 m, 1<θ<2 rad,1<φ<2 rad.
Solution:
1) Using ∇·D = ρ ν and volume integral
  D = v =
=
Qenc =

vol
1

1

16cos(2 )
(sin  D ) =
(sin 
)
r sin   
r sin   
r
8

8
(sin 3 − sin  )= 2
(3cos3 − cos )
r sin   
r sin 
v dv =
2
2rad 2rad 2
8
1 1 1 ( r 2 sin  (3cos3 − cos )) r
2
sin  drd  d 
2rad 2rad 2
=8
1 1
 (3cos3 − cos ) drd  d  = 8  ( r )1 ( )1
2
2rad
1
2rad
1
(3cos3 − cos )d 

2  180
2  180 
180
180  
= 8 sin 3
− sin
−  sin 3
− sin
= −3.91 C



  


2) Using D and surface integral
Qenc =  Ds  dS
= 8sin 3 − sin  1
2rad
S
D
s
S
 dS =  D r =1  (−aˆr )r 2 sin  d  d  +  D r =2  aˆr r 2 sin  d  d 
r =1
r =2
+  D =1  (−aˆ )r sin  drd  +
 D

+  D =1  (−aˆ )rdrd  +
=2
 =1
 =1
 D
 aˆ rdrd 
 =2
D=
2 rad 2
+
 16cos(2  2 
1rad 1
Qenc =
D
s
180

 dS = −16cos(2 
S
+ 16cos(2  2 
180

)sin(2 
180

)sin(2 
180

)drd 
180
)sin(
)(1)(1)

180

 aˆ r sin  drd 
=2
16cos(2 )
aˆ
r
2 rad 2
180
180
D

dS
=
−
16cos(2 
)sin(
)drd 
S s




1rad 1
Since
=2
)(1)(1) = −3.91 C
20. Calculate ∇·D at the point specified if (a) D=(1/z 2)[10xyz a x +5x2z
ay +(2z3−5x2y)az] at P(−2, 3, 5); (b) D=5z 2 a ρ +10ρz a z at P(3,−45◦, 5); (c)
D=2r sin θ sin φ a r +rcosθ sin φ a θ +r cos φ a φ at P(3, 45◦, −45◦).
21. A radial electric field distribution in free space is given in spherical
coordinates as:
where ρ0, a, and b are constants. (a) Determine the volume charge density in
the entire region (0 ≤ r ≤∞) by the appropriate use of ∇・D = ρv . (b) In terms
of given parameters, find the total charge, Q, within a sphere of radius r
where r > b.
Solution:
(a)
v =   D
D =  oE
and
1  2
1   r 3 o 
(
r
D
)
=

 = o
r1
r2 r
r2 r 
3 
1  2
1   (2a 3 − r 3 ) o
v 2 =   D2 = 2
(r D r 2 ) = 2

r r
r r 
3
v 1 =   D1 =
(r  a )

 = − o

3
3
1  2
1   (2a − b ) o 
v 3 =   D3 = 2
(r D r 3 ) = 2

=0
r r
r r 
3

(b) Q =
  dv
and
v
(a  r  b )
(r  b )
dv = r 2 sin  drd  d 
vol
2  a
Q total =
  r
o
2
sin  drd  d  +
2  a
0 0 0
a
   (− 
o
)r 2 sin  drd  d  + 0
0 0 0
b
r3 
r3 
4
= o   (4 ) − o   (4 ) =
o 2a 3 − b 3 
3
 3 0
 3 a