Transcript Practice Problems Lecture 7 Prof. Viviana Vladutescu
Lecture 7
Practice Problems
Prof. Viviana Vladutescu
Divergence Theorem and Gauss’s Law
Suppose
D
= 6
r
cos
φ
a
φ
C/m 2 .(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)
D
cylinder
1
D
1
v
6
C m
3 .
(b)
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 d
S
directions.
90 90
a
a
0 (c) So, 0 0
a
D
192 .
dv
a
192
C
D
dv
6 90 0 sin
d
4 0 4 0
dz
Electric Potential
E
The potential field in a material with ε r = 10.2 is
V
= 12
xy
2 (V). Find
E
,
P
and
D
.
12
xy
2
x
a
x
12
xy
2
y
a
y
12
y
2 a
x
24
xy
a
y V m
D
r o
E
-1.1
y
2
a
x
2.2
xy
a
y nC m
2
e
r
P
e o
E
12
E =
-9.8
y
2
a
x
2.00
xy
a
y nC m
2
Boundary Conditions
For
z
a 30 does ≤ 0,
r1
= 9.0 and for
z
> 0,
r2
= 4.0. If
E
2 make with a normal to the surface?
E
1 makes angle with a normal to the surface, what angle
E T
1
E
1 sin 1 ,
E T
2
E
2 sin 2 , and
E T
1
E T
2
D N
1
r
1
o E
1 also
E T
1
D N
1 1
D N
2
E T D N
2 2 , Therefore
r
2
o E
2 cos 2 , and
D N
1
D N
2
s
0 and after routine math we find 2 tan 1
r r
1 2 tan 1 Using this formula we obtain for this problem
2 = 14 ° .
1. A spherical capacitor consists of a inner conducting sphere of radius R 1 =3cm, and surface charge density ρ s =2nC/m 2 and an outer conductor with a spherical inner wall of radius R 2 =5cm . The space in between is filled with polyethylene.
a) Determine the electric field intensity from ρ s between plates(15 points) b) Determine the voltage V 12 (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
ρ s
= 12
r
nC/cm 2 (when
r
is in cm). Find the electric field intensity a point 20.0 cm above the origin on the z axis.(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
E
1 =2y a x a)Find
E
2 -3x a and y
D
+(5+z) a 2 z , at the boundary side of region 2? (20 points) b)Can we determine
E
2 and
D
2 at any point in region 2?
Explain.(5 points) 4. Two spherical conductive shells of radius
a
are separated by a material with conductivity and
b
(
b
>
a) σ
. Find an expression for the resistance between the two spheres.(20 points)
Problem 2 For a ring of charge of radius
a
,
E
2
o
L ah
a
z a
2
h
2 3 2 .
Now we have
L =
s
d and where
s
=
A
nC/cm 2 .
d
E
A d
2
o
2
h h
a
2 3
z
2 , Now the total field is given by the integral:
E
Ah
a
z
2
o
2 2
d
h
2 3 2 .
This can be solved using integration by parts, where
u
d
u
= d , = ,
v
1 2
h
2 , and
dv
2
h
2 .
This leads to
E
Ah
2
o
a a
2
h
2
a
a
2
h
2
a
z
.
h
Plugging in the appropriate values we arrive at
E
= 6.7 kV/cm
a
z .
At the z=0 plane
Problem 3
E
1 2
y a x
3
x a y
5
a z
From the boundary condition of two dielectrics we get
E
1
T
(
z
0 )
E
2
T
(
z
0 ) 2
y a x
3
x a y D
1
N
(
z
0 )
D
2
N
(
z
0 ) 1
E
1
N E
2
N
(
z
(
z
0 ) 0 ) 1 2 2
E
2
N
( 5
a z
) (
z
0 )
E
2
D
2 z=0 2
y a x
3
x a y
2
y a x
3
x a y
1 2 1 2
a z a z
r
2 0
l σ J a b
Problem 4- Hint
V ab I
I V ab
E
l
E
V ab l
s J
ds
JS
J
I S J V ab
E l
S
I
V ab S l I
where
R
l
S
( )
Resistance of a straight piece of homogeneous material of a uniform cross section for steady current
R
V ab I
s
a
b E
d l E
ds
Resistances connected in series
R sr
R
1
R
2
R
3
..........
R n
Resistances connected in parallel
1
R
//
1
R
1
1
R
2
1
R
3
..........
...
1
R n
Problem 4
First find
E
for
a < r < b
, assuming +
Q
at
r = a
and –
Q
at
r = b
. From Gauss’s law:
E
Q
4
o r
2
a
r
Now find
V ab
:
V ab
a b
a b
Q
4
o r
2
a
r dr
a
r
Q
4
o a b
dr r
2
Q
4
o
1
a
r b Q
4
o
a b
.
Now can find
I
:
I
J
Q 4
o
0 sin 2 0
d
Q 4
o
Q
o
.
1
r
2
a
r r
2 sin Finally,
R
V ab I
1 4 1
a
1
b
a
r