Student Name: Student ID:

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 =

∫∫



, 5 0

S z= ⋅d =

∫∫



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 = =     ⋅ = = _{ } = _{ }       = × _− _ =  

∫∫

∫∫

∫ ∫

∫ ∫

We split the remaining face into two faces, one for φ =0 and one for φ π= .

( )

0 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

=     ⋅ − = − _{ } = − _{ }         = − _ × _ = − ×_{ } × = −    

∫∫

∫ ∫

∫ ∫

∫ ∫

5 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 = − =

∫∫



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 

   

∇ ⋅ = _{ } = × × −_{  } ×

    

= × × × =

∫

∫ ∫ ∫

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

− − − − − − − − − − − − = = ⋅ = −   = ⋅ ⋅ ⋅ = ⋅ −_{ } = ⋅ − − −   − ≈ −

∫∫∫

∫ ∫ ∫

∫

12 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⋅

∫

(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 ρ_sf 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