EEC130A: Practice Problems for Midterm 2

Instructor: Xiaoguang Leo Liu (lxgliu@ucdavis.edu)

Updated: Mar. 6
th
2012
P-1. Charge is distributed along the z-axis from z = 5 m to and from z = 5 m to
with charge density
l
= 20 nC/m. Find E at (2, 0, 0) m.
5
5
y
x
z

l
(2,0,0)
Figure 1: Problem 1.
The z components of the electric eld vanishes due to symmetry. The differential electric
eld is
dE =

l
dz
4
0
(2
2
+ z
2
)
2
_
2
2
+ z
2
x
The total electric eld is
E =

l
4
0
_
_

5
2dz
(4 + z
2
)
3/2
+
_
5

2dz
(4 + z
2
)
3/2
_
x = 13 x V/m.

P-2. Charge lies on the circular disk r a, z = 0 with density

s
=
0
sin
2
. Determine
E at (0, , h)

Some problems are adapted from The Schaums Outlines on Electromagnetics and 2008+ Solved Prob-
lems in Electromagnetics.
1
5
5
y
x
z

l
(2,0,0)
dl
dE
1
dE
2
Figure 2: Solution to Problem 1.
Choose a differential surface element on the disk dS
z
. Due to symmetry, the r component
of the electric eld vanishes. The differential electric eld is
dE
z
=

0
(sin
2
)rdrd
4
0
(r
2
+ h
2
)
_
h
_
r
2
+ h
2
z
_
It follows that
E =

0
h
4
0
_
2
0
_
a
0
(sin
2
)rdrd
(r
2
+ h
2
)
3/2
z =

0
h
4
0
_
1

a
2
+ h
2
+
1
h
_
z

P-3. A point charge Q is at the origin of a spherical coordinate system. Find the ux
which crosses the portion of a spherical shell described by . What is the result
if = 0 and = /2.
The total ux that crosses a full spherical shell (total area of 4r
2
) is
t
= Q (Gausss
Law). The area of the strip is
A =
_
2
0
_

r
2
sin dd = 2r
2
(cos cos )
Then the ux through the strip is
=
A
4r
2
Q =
Q
2
(cos cos )
For = 0 and = /2 (a hemisphere), this becomes = Q/2.
P-4. Two identical uniform line charges lie along the x and y axes with charge densities

l
= 20 C/m. Obtain D at (3, 3, 3) m.
2
Figure 3: Problem 3.
The distance from the observation point to either line charge is 3

2 m.
Consider rst the line charge on the x-axis,
D
1
=

l
2r
1
_
y + z

2
_
.
Then the y-axis line charge,
D
2
=

l
2r
2
_
x + z

2
_
.
The total ux density is the vector sum of the two,
D = D
1
+ D
2
=
10
3

2
_
x + y +2 z

2
_
C/m
2
.

P-5. A charge conguration in cylindrical coordinate system is given by = 5re

2r
C/m
3
. Use Gausss Law to nd D.
Since is not a function of or z, the ux is completely radial. It is also true that, for
r constant, the ux density D must be of constant magnitude due to symmetry. Then a
proper Gaussian surface is a closed cylinder. The integrals over the end surfaces vanish
(because D is parallel to the end surfaces), so that Gausss law becomes
_
sidewall
D dS = Q
enclosed
3
D
1
y
x
z

l
(3,3,3)
D
2

l
Figure 4: Problem 4.
_
L
0
_
2
0
_
r
0
5re
2r
rdrddz = D2rL
5L
_
e
2r
_
r
2
r
1
2
_
+
1
2
_
= D2rL
Therefore,
D =
2.5
r
_
1
2
e
2r
_
r
2
+ r +
1
2
__
r.

P-6. The volume in cylindrical coordinate system between r = 2 m and r = 4 m contains

a uniform charge density (C/m
3
). Use Gausss law to nd D in all regions.
2 m
4 m
z
Figure 5: Problem 6.
4
Due to the symmetry of the problem, the electric ux density has only r component and
is independent of and z. Dene a cylindrical Gaussian surface as illustrated in Fig. 6.
The ux out of the top and bottom surfaces is 0 because the electric ux density is parallel
to these surfaces.
For 0 < r < 2 m, Gausss Law gives
D(2rL) = Q
enclosed
= 0
For 2 m r 4 m,
D(2rL) = Q
enclosed
= L(r
2
2
2
)
D =

2r
(r
2
4) r C/m
2
For r > 4 m,
(4
2
2
2
)L = D(2rL)
D =
6
r
r C/m
2
2 m
4 m
Gaussian surface
r
z
Figure 6: Solution to Problem 6.

P-7. Given that D = 10r

3
/4 r in cylindrical coordinate system, evaluate both sides of the
divergence theorem for the volume enclosed by r = 1 m, r = 2 m, z = 0, and z = 10 m.
_
D dS =
_
( D)dv
Since D has no z component, D dS is zero for the top and bottom surfaces.
On the outer cylindrical surface, dS
r
is in the r direction; on the inner cylindrical surface,
dS
r
is in the r direction.
5
1 m
2 m
z
Figure 7: Problem 7.
_
D dS =
_
10
0
_
2
0
10
4
r
3
rddz |
r=1
+
_
10
0
_
2
0
10
4
r
3
rddz |
r=2
= 750 (C)
The right hand side of the divergence theorem is
D =
1
r

r
_
10r
4
4
_
= 10r
2
_
Ddv =
_
10
0
_
2
0
_
2
1
(10r
2
)rdrddz = 750 (C)

P-8. Determine the value of E in a material for which the electric susceptibility is 3.5 and
P = 2.3 10
7
C/m
2
.
E =
1

0
P = 7.42 10
3
V/m

P-9. Region 1, dened by x < 0, is free space, while region 2, x > 0, is a dielectric
material for which
r2
= 2.4. Given D
1
= 3 x 4 y +6 z C/m
2
, nd E
2
and the angles
1
and
2
.
The x components are normal to the interface: D
n
and E
t
are continuous.
D
1
= 3 x 4 y +6 z E
1
=
3

0
x
4

0
y +
6

0
z
6

2
D
1
D
2
1 2
O
x
z
Figure 8: Problem 9.
D
2
= 3 x + D
y2
y + D
z2
z E
2
= E
x2
x
4

0
y +
6

0
z
Then D
2
=
0

r2
E
2
gives
3 x + D
y2
y + D
z2
z =
0

r2
E
x2
x 4
r2
y +6
r2
z
Therefore,
E
x2
=
3

r2
=
1.25

0
D
y2
= 4
r2
= 9.6 D
z2
= 6
r2
= 14.4
To nd the angles:
D
1
x = |D
1
| cos(90

1
)

1
= 22.6

Similarly,
2
= 9.83

.
P-10. A free-space parallel-plate capacitor is charged by momentary connection to a
voltage source V, which is then removed. Determine how W
E
, D, E, C, and V change
as the plates are moved apart to a separation distance d
2
= 2d
1
without disturbing the
charge.
Since the capacitor is disconnected from the source, the total charge Q remains the same
as the distance between the plates changes.

7
Relationship Explanation
D
2
= D D = Q/A
E
2
= E
1
E = D/
0
W
E2
= 2W
E1
W
E2
=
1
2
_

0
E
2
dv, and the volume is doubled
C
2
=
1
2
C
1
C = A/d
V
2
= 2V
1
V = Q/C
P-11. A spherical conducting shell of radius a, centered at the origin, has a potential of
V = V
0
, for r a ,and V
0
a/r, for r > a
with the zero reference at innity. Find an expression for the stored energy that this
potential represents.
E = V = 0, for r a, ,and (V
0
a/r
2
) r, for r > a
The total electric energy
W
E
=
1
2
_

0
E
2
dv = 0 +

0
2
_
2
0
_

0
_

a
_
V
0
a
r
2
_
2
r
2
sin drdd = 2
0
V
2
0
a

P-12. Find H at the center of a square current loop of side L.

y
x
L/2
L/2
-L/2
-L/2
I
Figure 9: Problem 12.
8
Choose the Cartesian coordinate system because of the square shape. By symmetry, each
half side contributes the same amount to H at the center. Consider an increment source
on the half side of 0 x L/2, y = L/2, the Biot-Savart law gives the increment H
y
x
L/2
L/2
-L/2
-L/2
R
I
Figure 10: Problem 12.
dH =
1
4
Idx x
x x + (L/2) y
_
x
2
+ (L/2)
2
x
2
+ (L/2)
2
=
Idx(L/2) z
4 [x
2
+ (L/2)
2
]
3/2
Therefore,
H = 8
_
L/2
0
Idx(L/2) z
4 [x
2
+ (L/2)
2
]
3/2
=
2

2I
L
z
(1)

P-13. In the region 0 < r < 0.5 m, in cylindrical coordinate system, the current density
is
J = 4.5e
2r
z (A/m
2
)
and J = 0 elsewhere. Find H.
Because the current density is symmetrical around the z-axis, a circular path on the con-
stant z plane is chosen as the Ampere contour.
9
For r < 0.5 m,
2rH =
_
2
0
_
r
0
4.5e
2r
rdrd
H =
1.125
r
_
1 e
2r
2re
2r
_

(A/m)
For r 0.5 m, the enclosed current is the same, 0.594A. Then,
2rH =
_
2
0
_
0.5
0
4.5e
2r
rdrd = 0.594.
Therefore,
H =
0.297
r

(A/m)

P-14. In region 1 of Fig. 11, B

1
= 1.2 x +0.8 y +0.4 z (T). Find H
2
and the angles
1
and
2
.
1
2
x
z

r1
=15

r2
=1
Figure 11: Problem 14.
Magnetic boundary conditions dictates that a) the normal components of B are continu-
ous; and b) the tangential components of H are contiuous. Therefore,
B
1
= 1.2 x +0.8 y +0.4 z (T)
B
2
= B
x2
x + B
y2
y +0.4 z (T)
H
1
=
1

r1
(1.2 x +0.8 y +0.4 z) (A/m)
10
H
2
=
1

r1
(1.2 x +0.8 y +
0

r1
H
z2
z) (A/m)
It follows that,
B
x2
=
0

r2
H
x2
= 8 10
2
(T)
B
y2
= 5.33 10
2
(T)
H
z2
=
B
z2

r2
= 3.18 10
5
(A/m)
Angle
1
can be found by
cos(90

1
) =
B
1
z
|B
1
|
= 0.27
Therefore,
1
= 15.5

.
Similarly,
2
= 76.5

11