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Lecture 7
Practice Problems
Prof. Viviana Vladutescu
Divergence Theorem and Gauss’s Law
Suppose D = 6rcos φ aφ C/m2.(a) Determine the charge density at
the point (3m, 90, -2m). Find the total flux through the surface
of a quartered-cylinder defined by 0 ≤ r ≤ 4m, 0 ≤ φ ≤ 90, and
-4m ≤ z ≤ 0 by evaluating (b) the left side of the divergence
theorem and (c) the right side of the divergence theorem.
(a)
1 D 1   6 cos  

 6sin .
  Dcylinder 
  

C
v  3,90 , 2   6 3 .
m
(b)
 D dS  
 0

   
 90
top
bottom


outside
,
note that the top, bottom and outside integrals yield zero since
there is no component of D in the these dS directions.

 90

 0
So,
(c)
  6  cos 
  6  cos 
 90
 0
a  d  dza   0
a  d  dza   192C
 D dS  192C.
 D  6sin  , dv   d  d dz
90
4
0
0
0
4
  Ddv  6  sin  d   d   dz  192C.
Electric Potential
The potential field in a material with εr= 10.2
is V = 12 xy2 (V). Find E, P and D.
E  V  
 12 xy 2 
x
ax 
 12 xy 2 
y
V
a y  12 y a x  24 xya y
m
2
D   r  o E  -1.1y a x  2.2 xya y
2
nC
m2
e   r  1  9.2
P   e o E   9.2   8.854 x10
12
 E = -9.8 y a
2
x
 2.00 xya y
nC
m2
Boundary Conditions
For z ≤ 0, r1 = 9.0 and for z > 0, r2 = 4.0. If E1 makes
a 30 angle with a normal to the surface, what angle
does E2 make with a normal to the surface?
ET 1  E1 sin 1 , ET 2  E2 sin 2 , and ET 1  ET 2
also
ET 1 ET 2

,
DN 1 DN 2
Therefore
DN1   r1 o E1 cos1 , DN 2   r 2 o E2 cos2 , and DN1  DN 2  since s  0
and after routine math we find
 r2

2  tan  tan 1 
  r1

1
Using this formula we obtain for this problem 2 = 14°.
1. A spherical capacitor consists of a inner conducting
sphere of radius R1=3cm, and surface charge density
ρs=2nC/m2 and an outer conductor with a spherical inner
wall of radius R2=5cm . The space in between is filled with
polyethylene.
a) Determine the electric field intensity from ρs between
plates(15 points)
b) Determine the voltage V12 (10 points)
c) Determine the capacitance C (10 points)
2. Consider a circular disk in the x-y plane of radius 5.0 cm.
Suppose the charge density is a function of radius such
that ρs = 12r nC/cm2 (when r is in cm). Find the electric
field intensity a point 20.0 cm above the origin on the zaxis.(25 points)
3. Assume the z=0 plane separates two lossless dielectric
regions, where the first one is air and the second one is
glass. If we know that the electric field intensity in region 1
is E1 =2y ax -3x ay+(5+z) az,
a)Find E2 and D2 at the boundary side of region 2? (20
points)
b)Can we determine E2 and D2 at any point in region 2?
Explain.(5 points)
4. Two spherical conductive shells of radius a and b (b > a)
are separated by a material with conductivity σ. Find an
expression for the resistance between the two spheres.(20
points)
Problem 2
For a ring of charge of radius a,
E
 L aha z
2 o  a 2  h 2 
Now we have L=sd and
3
dE 
where s = A nC/cm2.
.
2
A d   ha z
2 o   2  h 2 
Now the total field is given by the integral:
Aha z
E
2 o

 2d 

2
 h2 
3
.
2
3
,
2
This can be solved using integration by parts, where u = ,
du = d,
v
1
 2  h2
, and dv 
d 
 2  h2
.
This leads to
 a  a 2  h2
Ah  a

E
 ln 

2 o  a 2  h2
h



 a z .


Plugging in the appropriate values we arrive at E = 6.7 kV/cm az.
Problem 3
At the z=0 plane
E1  2 y ax  3xa y  5az
From the boundary condition of two dielectrics we get
E1T ( z  0)  E2T ( z  0)  2 y a x  3x a y
D1N ( z  0)  D2 N ( z  0)
1 E1N ( z  0)   2 E2 N ( z  0)
1
E2 N ( z  0)  (5a z )
2
z=0
1
E2  2 y a x  3 x a y  a z
2

1 
D2   2 y a x  3x a y  a z  r 2 0
2 

Problem 4- Hint
I
b
l
σ
Vab
J
a
Vab
Vab  E  l  E 
l
I
I   J  ds  JS  J 
S
s
Vab
I
J  E   
S
l
l
 l 
Vab    I where R 
()
S
 S 
Resistance of a straight piece of homogeneous
material of a uniform cross section for steady
current
b
  E  dl
Vab
R
 a
I

E

ds

s
Resistances connected in series
Rsr  R1  R2  R3  ..........  Rn
Resistances connected in parallel
1
1
1
1
1
 

 .............
R// R1 R2 R3
Rn
Problem 4
First find E for a < r < b, assuming +Q at r = a and
–Q at r = b. From Gauss’s law:
Q
E
a
2 r
4 o r
Now find Vab:
a
a
Q
Vab    E dL   
a dra r
2 r
4 o r
b
b
Q dr Q 1
Q 1 1



  .
2

4 o b r 4 o r b 4 o  a b 
a
a
Now can find I:
Q
1
2
I   J dS =   E dS =  
a r r sin  d da r
2
4 o r

2
Q
Q

sin  d  d 
.

4 o 0
o
0
Finally,
Vab
1 1 1
R

  
I
4  a b 