Download March 13, 2002

Prof. Yen-Chieh Huang
Dept of Electrical Engineering
National Tsing-Hua University
office: Delta 856, HOPE 301
ext: 62340
email:[email protected]
EE214002 Electromagnetics, Fall 2012
___________________________________________________________________________________
Nov. 11, 2012
EE214000 Electromagnetics, Fall 2012
Homework Assignment # 4 (credit points: 150)
Due in class Nov.21, 2012
1. (10 points) An infinitely long line charge along    z   has a line charge
density of l   z , where  is a constant. The line charge is placed in vacuum.
(a) (5 points) What is the electric field at ( r ,  , z  0 )? Give the magnitude and
direction of your answer.
(b) (5 points) What is the energy required to take a charge of q from
( r  a,   0, z  0 ) to ( r  b,    , z  0 )?
Ans:
(a) Refer to the following plot. Only the components in the â r direction exit
due
to
cos  
symmetry.
1
40

 z

r  z2

r
work
2
cos dz 
.
r 2  z2

2r
E  E r aˆ r  aˆ r
40
(b) The
Er 
Use

z 

0
(r  z  )
necessary
2
2 3/ 2
to
dz  
be
,
where
Therefore

aˆ r
20
done
is
independent
of
path.

b 
q
W  qVba  q  E  dl 
( a  b)
a
20
2. (20 points) An infinitesimal point charge of q locates at the x-y plane, and an
infinitely long line charge of a line charge density  l is arranged in parallel with
the z axis.
A
90o
R1
B
y
30o
l
45o
q
x
a. (4 points) Assume  l = 0 and q  0. What
is the work needed to be done for
Prof. Yen-Chieh Huang
Dept of Electrical Engineering
National Tsing-Hua University
office: Delta 856, HOPE 301
ext: 62340
email:[email protected]
EE214002 Electromagnetics, Fall 2012
___________________________________________________________________________________
moving a positive unit point charge from A to B on the x-y plane, as shown
above?
b. (8 points) Assume q = 0 and  l  0. What is the work needed to be done for
moving a positive unit point charge from A to B on the x-y plane?
c. (8 points) Assume  l  0 and q  0. What is the work needed to be done for
moving a positive unit point charge from A to B on the x-y plane?
Ans:
(a)
E
q
A
40 R 2
B
1
A
40 R
V    EdR 
(1
|R1
3
) R1
3

( 3  1)
80 R1
B
q
(b)
^
E  ar
l
20 r
A
A
B
A
A'
V    E  dR   E  dR  0 
l
2 3
ln
40
3
B
30°
,
A’
45°
ρl
q
where A and A’ are in a circle of the same radius R1
(c)
Wc  Wa  Wc  q
( 3  1) q l
2 3

ln
80 R1 40
3
3. (10 points) The far-field electric potential at a distance R from an electric dipole

p  aˆ R
with a dipole moment = p is given by V 
, where â R is the unit vector
40 R 2
pointing from the dipole to the field location. Suppose that two dipoles of dipole
Prof. Yen-Chieh Huang
Dept of Electrical Engineering
National Tsing-Hua University
office: Delta 856, HOPE 301
ext: 62340
email:[email protected]
EE214002 Electromagnetics, Fall 2012
___________________________________________________________________________________
moments  aˆ z 5 nCm and aˆ z 9 nCm are located at points (0, 0, 2 m) and (0, 0, 3
m), respectively, in a Cartesian coordinate system. (hint: 1 nC = 10-9 C,
0 
1
 10 9 F/m).
36
a. (5 points) Find the electric potential at the origin.
b. (5 points) Find the electric field intensity at the origin. Give the magnitude
and direction of your answer.

  
p  aˆ R  R
p  ( R  R )
Ans: Rewrite the dipole potential as V 
  2 
  3
40 R  R 
40 R  R 


ˆ z ; and for the
(a) For the dipole with dipole moment  aˆ z 5 , R  R  2a


ˆ z . Taking the linear sum
dipole with dipole moment aˆ z 9 , R  R  3a
of the electric potential from the two dipoles, one has -19.25 V at the origin.

(b) E  V . Since both the dipole moments are aligned along the z axis, what
we are concerned with is E z or ER (  0) for the dipole at (0, 0, 2 m)
and E R (  180  ) for the dipole at (0, 0, 3 m). E R ( )  
V
p cos 

.
R 40 R 3
p1
p
where
aˆ z  23 (aˆ z )) ,
3
40 R1
R2

p1  9, R1  2, p2  5, R2  3 . Therefore, at the origin, E  E z aˆ z  203 / 24
E1, R (  0)aˆ1, R  E2, R (  180  )aˆ 2, R 
1
(
V/m.
4. (35 points) A sphere consists of four regions, (1) an inner conductor region in R < a,
(2) a vacuum region between a < R < 2a, (3) a dielectric region with a dielectric
constant of r = 1.5 between 2a < R < 4a, (4) an outer conductor region in R > 4a.
The inner conductor is grounded and the outer conductor is maintained at a voltage
V0.
a. (5 points) What is the capacitance of this sphere?
b. (5 points) What is the electric field intensity in region (2)?
c. (5 points) What is the electric field intensity in region (3)?
d. (5 points) What is the surface charge density on the inner conductor?
e. (5 points) What is the surface polarization charge density at R = 2a?
f. (5 points) What is the surface polarization charge density at R = 4a on the
Prof. Yen-Chieh Huang
Dept of Electrical Engineering
National Tsing-Hua University
office: Delta 856, HOPE 301
ext: 62340
email:[email protected]
EE214002 Electromagnetics, Fall 2012
___________________________________________________________________________________
dielectric?
g. (5 points) What is the surface charge density at R = 4a on the outer
conductor?
(hint: you can verify your answers from conservation of charges)
(4)
(2)
(3)
(1)
Ans:
(a) Assume the inner conductor has a charge Q. The electric field intensity in

region (2) is E 2 
Q
40 R
2

aˆ R and that in region (3) is E3 
Q
aˆ R .
40  1.5R 2
The potential between the inner and outer conductor is given by the integration
 a 

2a 
Q
1
1
Q 1 1
Q
V  V0    E3  dR   E 2  dR 
(  )
(  )
4a
2a
60 2a 4a 40 a 2a
60 a
The capacitance is C 
Q
 60 a .
V
(b) From (a), the charge in region (1) can be calculated to be Q  60 aV0 .

E2 
Q
40 R
2
aˆ R 
 3aV0
aˆ R .
2R 2

 aV0
(c) Substitute the charge calculated in (b), E3 
aˆ R
R2
(d) The surface charge density on the inner conductor is equal to

 s  aˆ R  D2 
 3 0V0
2a
 

 a 0V0
1 
(e) P  D3   0 E3   0 E3 
aˆ R . The surface polarization charge
2
2R 2

density at R = 2a is given by  ps   aˆ R  P
R 2 a

a 0V0
2R 2

R 2 a
 0V0
8a
Prof. Yen-Chieh Huang
Dept of Electrical Engineering
National Tsing-Hua University
office: Delta 856, HOPE 301
ext: 62340
email:[email protected]
EE214002 Electromagnetics, Fall 2012
___________________________________________________________________________________
(f) Continued from (e), the surface polarization charge density at R = 4a is given

by  ps  aˆ R  P
R 4a

a 0V0
2R 2

R 4 a
 0V0
32a
(g) Similar to (d) The surface charge density on the outer conductor is given by

 s  aˆ R  D3 
1.5 0 aV0 3 0V0
. The result in (c) is used.

32a
( 4a ) 2
5. (15 points) A large-area parallel-plate capacitor is charged to store an amount of
charge Q and then disconnected from the power supply, as shown in (a) in the
following. The parallel-plate capacitor has an area A and an electrode separation d
in vacuum. If one inserts a perfect dielectric of area A, thickness d/2, and relative
permittivity r = 2 into the capacitor, as shown in (b),
(1) what is the ratio Vb / Va , where Va, Vb are the voltages across the conducting
electrodes before and after inserting the perfect dielectric, respectively. (5
points)
Ans: C a 
0 A
d
, Cb 
4 A 4
1
 0  C a independent of
y0
d / 2  y0
d /2
3d
3


0 A
0 A
 0 r A
y0. Since V  Q C with a constant Q,
Vb / Va  C a C b 
3
.
4
(2) what is the ratio Fb / Fa , where Fa , Fb are the forces holding the
conducting electrodes before and after inserting the perfect dielectric,
respectively. (5 points)
Ans: F 
1 / Cb ( y) 
dWe
dy
Q

y
1 Q2
1 2 d (1 / C )
 1 / C a ( y) 
, Fa 
;
Q
0 A
2 dC a
2
dy
C
3y
1 Q2
3
, Fb 
 Fb / Fa  a  .
4 0 A
2 dCb
Cb 4
(3) what is the polarization charge density induced on the surface of the dielectric
facing electrode 1? (5 points)
Ans: The voltage across the two conducting plates for capacitor (b) is
Q
3Qd
D d D d
Q
Vb 



.  D  . According to
A
C b 4 0 A 2 0 2  0 2
Prof. Yen-Chieh Huang
Dept of Electrical Engineering
National Tsing-Hua University
office: Delta 856, HOPE 301
ext: 62340
email:[email protected]
EE214002 Electromagnetics, Fall 2012
___________________________________________________________________________________
P  D   0 E  (1 
1
r
)D 
1
Q
D
. The surface polarization charge density
2
2A

Q
on the dielectric facing electrode 1 is there P  aˆ s 
.
2A
d
d
d/2
vacuum
1
2
Q
-Q
r =2
1
2
-Q
Q
y0
(a)
(b)
6. (20 points) Find the work to assemble a sphere of charges with its charge density
varying with   
b
between the radius distance 0  R  R0 , where  0 and b
0 R
are constants. The sphere of charges is in a vacuum. Present at least TWO methods
to calculate and cross check the answer.
Ans:
Method 1

2
b 2
R sin dRd d  2bR 2  0
0 0 0
R
Rb  0
QR
VR 

40 R
2 0
QR  
R

0
dWe  VR dQR
R0
We   VR dQR  
0
Method 2
R0
0
202 b 2 R 2
0
202 bR0
dR 
3 0
3
Prof. Yen-Chieh Huang
Dept of Electrical Engineering
National Tsing-Hua University
office: Delta 856, HOPE 301
ext: 62340
email:[email protected]
EE214002 Electromagnetics, Fall 2012
___________________________________________________________________________________
R  R0
 0bR02
40 R 2 2 0 R 2
b
QR
ER 
 0
2
40 R
2 0
QR0
ER 
0  R  R0
R
R0


V    EdR   

R  b
 0 bR02
 bR  bR
dR   0 dR  0 0  0
2
R
2 0 R
2 0
0
2 0
0
We  20 b  V  20 b 
b 0
0

R0
0
20 b 2 R03 R03
202 bR0
R
)dR 
(
 )
2
0
2
6
3 0
2
( R0 
3
Method 3
R  R0
 0bR02
40 R 2 2 0 R 2
b
QR
ER 
 0
2
40 R
2 0
ER 
0  R  R0
We 
QR0

R0  b
 0  2   0bR02 2 2
1
2

E

(  ( (
) R dR   ( 0 ) 2 R 2 dR ) sin dd )
0
2

0
2V
2 0 0 R0 2 0 R
2 0
2
3
4
 R
R
   b
 2 b
R3
202 bR0
 0  0  4 (  02 dR   R 2 dR )  0 ( R03  0 ) 
R R
0
2  2 0 
2 0
3
3 0
0
0
7. (40 points) Visit a few electronics stores to find out various kinds of capacitors. Use
your cell phone camera to take photographs for those capacitors with your photo ID in
photographs. Explain the functioning principles of those capacitors that you found in
the stores.