Instructor:Natalia K. Nikolova COURSE ELECTRICAL ENGINEERING 2FH3
Duration of Examination: 3 hours
McMaster University Final Examination April 8, 2011 THIS EXAMINATION PAPER INCLUDES 6 PAGES AND 6 QUESTIONS. YOU ARE RESPONSIBLE FOR ENSURING THAT YOUR COPY OF THE PAPER IS COMPLETE. BRING ANY DISCREPANCY TO THE ATTENTION OF YOUR INVIGILATOR.
Instructions:
1. You can use only a standard calculator (Casio-FX991).
2. Write your name and student ID on each page, the exam booklets incl. 3. You are allowed to bring 2 sheets of letter-size paper with any writing on both sides of the sheet.
4. Answer ALL questions. Provide the solutions in the exam booklet. Problem 1 [20 points] (HL6 D3.9 HB)
Given the electric flux density
6 sin(0.5 )ρ φ ρ 1.5 cos(0.5 )ρ φ φ
= +
D a a , C/m2
find the charge Q enclosed in the volume bounded by the following surfaces: ρ =2 m,φ =0, φ π= , 0
z= , and z=5 m.
METHOD 1: Calculate flux (left side of Gauss theorem) Bounded region is half a cylinder.
D has no z-component, so
, 0 0
S z= ⋅d =
∫∫
D S
and, 5 0
S z= ⋅d =
∫∫
D S
.For the curved face:
5 2 5
2 2 0 0 0 0
5 2
0 0
1 1
6 sin 6 sin
2 2
1
6 2 2 cos 240 C 2
d dz D d dz d dz d dz
z
π π
ρ ρ
ρ ρ
π
ρ φ ρ φ ρ φ ρ φ ρ φ φ
φ = = ⋅ = = = = × − =
∫∫
D a∫∫
∫ ∫
∫ ∫
We split the remaining face into two faces, one for φ =0 and one for φ π= .
( )
5 2 5 20 0 0 0 0
2
5 2 5
2 0 0 0
0
1 1
1.5 cos 1.5cos
2 2
1 1
1.5cos 0 1.5 15 C
2 2
d dz d dz d dz
d dz z
φ
φ ρ ρ φ ρ φ ρ ρ
ρ ρ ρ
= ⋅ − = − = − = − × = − × × = −
∫∫
∫ ∫
∫ ∫
∫ ∫
D a and5 2 5 2
0 0 0 0
5 2
0 0
1 1
1.5 cos 1.5cos
2 2
1
1.5cos 0 2
d dz D d dz
d dz d dz
d dz
φ φ
φ π ρ φ π ρ
ρ φ ρ φ ρ ρ
π ρ ρ
For the total:
240 15 225 C
S ⋅d = − =
∫∫
D S
METHOD 2: Calculate charge (right side of Gauss theorem)
(
)
(
)
1 1 1 1 1 1 1
12 sin 0.75 sin 11.25sin
2 2 2
z D D D z φ ρ
ρ ρ φ ρ φ φ
ρ ρ ρ φ ρ ρ
∂ ∂ ∂ ∇ ⋅ = ⋅ + ⋅ + = + − = ∂ ∂ ∂ D 2 2
5 2 5
0
vol 0 0 0
0 0
1 1
11.25sin 11.25 2 cos
2 2 2
11.25 2 2 5 225 C
dv d d dz z
π
π ρ
φ ρ ρ φ φ
∇ ⋅ = = × × − ×
= × × × =
∫
D∫ ∫ ∫
Problem 2 [20 points] (Homework T11)
A capacitor is made of two concentric spheres. The radii of the spheres are a = 2 cm and b = 6 cm, respectively. The insulator between the two spherical electrodes has relative permittivity εr =9.
(a) At what distance r from the center of the spherical capacitor is the magnitude of the E field maximum? [2]
(b) What is the breakdown voltage of the capacitor if the dielectric strength of the insulator is 100
ds
E = MV/m? [10]
(c) What is the electric energy stored in the capacitor when the applied voltage is equal to the breakdown voltage? [5]
(d) What is the capacitance C of the structure? [3]
(a) Using Gauss law, the E-field inside the capacitor is
2 0 ( ) 4 r r Q r r πε ε = E a
Therefore, the field is maximum when r is minimum, i.e., at the surface of the inner electrode where 2
r= =a cm.
(b) We have to express the field in terms of the voltage between the electrodes.
2
0 0 0
1 1 1
4 4 4
b
b b
r
a r a r r
a
Q dr Q Q
V E dr
r r a b
πε ε πε ε πε ε
= = = ⋅ − = ⋅ −
∫
∫
1 1 2
0
1 1
/ ( )
4 r r( )
Q V
V r
a b a b r
πε ε − −
⇒ = − ⇒ =
−
E a
The maximum E-field occurs at r=a:
max ( ) 1 1 2
( )
r a
V
E E
a b a
= − −
= =
−
We now set Emax =Eds when breakdown will start in the dielectric at the surface of the inner
electrode. We calculate the respective voltage, which is the breakdown voltage:
2 8 2 8 8 2 6 6
1 1 4 4
10 10 10 2 10 10 1.33 10 V
6 3
bd ds
b a b a
V E a a a
a b ab b
−
− −
= − = = = × = × ≈ ×
(c)
2 2
2 2
0 0 1 1 2 4
vol 0 0
2 2 2
0 0 1 1
0 1 1 2 2 1 1 2 1 1 2
2 0
1 1
1 1
| | sin
2 2 ( )
1 2 1 2
2 2 ( )
2 ( ) ( ) ( )
2 2
=
( )
b
bd
e r r
a
b b
r r
bd bd bd
r
a a
r bd
V
W dv r drd d
a b r
V dr V V
a b
a b r a b r a b
V
a b
π π
ε ε ε ε θ φ θ
πε ε πε ε
ε ε π
πε ε π
− − − − − − − − − − − − = = ⋅ = − = ⋅ ⋅ ⋅ = ⋅ − = ⋅ − − − − ≈ −
∫∫∫
∫ ∫ ∫
∫
E12 2 12
4 2
8.854187 10 9 1.33 10
2 6 10 26.7 J (6 2) 10
− − − ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ≈ − ⋅ (d) 12
2 2 12
2 2 26.7
30 10 F 1.33 10 e W C V − ⋅ = ≈ ≈ × × Alternatively:
from L14, we know that
4
12 12
0 2
2 6 10
4 4 8.854187 10 9 30 10 F
( ) (6 2) 10
r
ab C
b a
πε ε π − − −
−
⋅ ⋅
= = ⋅ ⋅ ⋅ ⋅ ≈ ⋅
− − ⋅
Alternatively, from (b) we see that
0
0 0
1 1 4
/ 4
1 1
4 ( )
r
r r
Q Q ab
V C
a b V b a
a b
πε ε πε ε πε ε = − ⇒ = = − == −
which is the same as above.
Problem 3 [20 points] (T09, 5.6 HB)
The current density is given in spherical coordinates as
6
1
exp( 10 ) 10r t r
= −
J a , A/m2
where t denotes time.
(a) Find the current I t( ) through the sphere of radius 5 cm as a function of time. [4]
(b) Find the volume charge density ρv( , )r t as a function of position and time assuming that 0
v
ρ → as t→ ∞. [4]
(c) Find the velocity of charge v( , )r t as a vector function of position and time. [4]
(d) Find the power P t( ) as a function of time dissipated in the spherical region r≤5 cm if the conductivity of the medium is σ =20 S/m. [4]
(e) Find the energy dissipated in the spherical region r≤5 cm in the interval 0≤ ≤ ∞t if the conductivity of the medium is σ =20 S/m. [4]
(a) 2 2 6 2
sphere 0 0
1
( ) ( , ) sin exp( 10 ) sin
10 r
I t J r t r d d t r d d
r
π π
θ θ φ θ θ φ
=
∫∫
⋅ =∫ ∫
−
6 6 6
0
0.05 2
( ) exp( 10 ) 2 sin 0.4 exp( 10 ) 0.0628exp( 10 ) A 10
r
I t = ⋅ − t ⋅ π⋅
∫
π θ θd = π r − t ≈ − t(c)
(d)
r r
p=E J where r r
J E
σ
=
6 6
2 2 10 2 10
3
2 2
( , )
( , ) , W/m
20 100 2000
t t
r
J r t e e
p r t
r r
σ
− ⋅ − ⋅
⇒ = = =
⋅
6
2 10
2 0.05
2 2
0 0 0
vol
( ) ( , ) sin
2000
t e
P t p r t dv r drd d
r
π π
θ θ φ − ⋅
=
∫∫∫
=∫ ∫ ∫
⋅6
6
2 10
4 2 10
( ) 0.05 2 2 3.14 10 , W 2000
t
t
e
P t π e
− ⋅
− − ⋅
⇒ = ⋅ ⋅ ⋅ ≈ ⋅
(e)
6
6
4 2 10
0 0
4
2 10 10
6 0
1
( ) 3.14 10
3.14 10
1.57 10 J 2 10
t
t
W P t dt e dt
W e
∞ ∞
− − ⋅
− ∞
− ⋅ −
−
= ≈ ⋅
⋅
⇒ ≈ ⋅ ≈ ⋅
− ⋅
∫
∫
Problem 4 [15 points] (T10)
Two parallel plates are biased at 100 V. The plate at x=5 mm is at potential V = 100 V. The plate at x = 0 is grounded. The plates are identical and have an area of 2 5× cm2 each. Consider two cases: (1) insulator is vacuum, ε ε= 0, and (2) insulator is dielectric with ε =4ε0. In both cases
calculate:
(a) the vectors E, D, and P in the insulator; [3]
(b) the free surface charge density ρsf on each electrode (give magnitudes only); [3]
(c) the bound surface charge density ρsb at the electrode-insulator interface (magnitudes only); [3]
(a)
(b)
(d)
(e)
6 3
2
2 4 3
0
Vol 510 m
(1) 12 8 6 9
(2) 12 8 6 9
Vol 0.5 2 5 10 5 10
0.5 8.854187 10 4 10 5 10 8.85 10 J
0.5 4 8.854187 10 4 10 5 10 35.4 10 J
E E r
E
E
E
W w E
W
W
ε ε
−
− −
= ⋅
− − −
− − −
= ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅
⇒ ≈ ⋅ × ⋅ ⋅ ⋅ ⋅ ≈ ⋅
⇒ ≈ ⋅ ⋅ × ⋅ ⋅ ⋅ ⋅ ≈ ⋅
Problem 5 [15 points] (HL15/example, 8.6 HB)
A charged disk of radius a=10 cm lies in the xy plane with the z axis being the disk’s axis of rotation. The rotation is in the +aφ direction and its angular speed is Ω = ×2 103 rad/s. The surface charge density is 2 10 12
s
ρ = × − C/m2
Since Hz is an even function of z, it is the same at z=3 cm and at z= −3 cm.
2 2
12 3 2 4
2 7
2 9
10
1 1 3.48
3
2 10 2 10 10 2 9 10 (1 3.48) (0.03) ( 0.03)
2 3 10 3.48
2 10 1 0.4464
10 0.106 10 A/m
3 3.48
z z
a b
z
H H
− − −
− −
− −
= + = + ≈
⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ −
= − ≈
⋅ ⋅
⋅ −
≈ ⋅ ≈ ⋅
Problem 6 [10 points] (L16 example)
Surface current sheet of current density K= 4ax A/m lies in the plane y = 5. Find the magnetic field
vector H at the origin (0,0,0).
4 ( )
2 A/m
2 2
x y
n
z
× − ×
= K a = a a = −
H a
END OF QUESTION SHEET