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6.013/ESD.013J Electromagnetics and Applications, Fall 2005

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Massachusetts Institute of TechnologyDepartment of Electrical Engineering and Computer Science

6.013 Electromagnetics and Applications

Problem Set #1 Issued: 9/7/05

Fall Term 2005 Due: 9/14/05

Reading Assignment: Sections 1.1 – 1.5, Appendices B, C of Electromagnetics and Applications

Problem 1.1 – Coulomb Force Lawa. View combined videos 1.3.1 (Coulomb’s Force Law and 1.5.1 (Measurement of Charge)

at (http://web.mit.edu/6.013_book/www/VideoDemo.html).

b. An electroscope measures charge by the angular deflection of two identical conducting

balls suspended by an essentially weightless insulating string of length l . Each ball has

mass M in the gravity field g and when charged can be considered a point charge.

A total charge Q is deposited on the two identical balls of the electroscope when they are

touching. The balls then repel each other and the string is at an angle θ from the normal

which obeys a relation of the form

tanθ

sin2θ =const

What is the constant?c.

Conservation of charge requires that

d∫ J da + ∫ ρ dV =0i dt s V

2

Figure 1.5.5 in Electromagnetic Fields and Energy, by Hermann A. Haus and James R. Melcher, 1989.

Adapted from Problem 2.6 in Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permiss

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where i = ∫ J d i a is the terminal current and q = ∫ ρ dV is the total charge inside the s

V

volume V for the Faraday cage geometry shown above. If the instantaneous charge

within the inner spherical volume is q t ( ) , what is the voltage v across the load resistor R?

Evaluate for q t 9 ) =q0 cosω t .

Problem 1.2

a. View video 10.2.1, Edgerton’s Boomer at(http://web.mit.edu/6.013_book/www/VideoDemo.html).

b. Use Ampère’s integral law over the contour C , shown above over a radius a equal to theaverage coil radius to approximate +1 due to the N turn toroidal coil carrying a current i.

c. Estimate the coil self-inductance L.

d.

Neglecting the coil resistance, if the capacitor C is charged to voltage V , what is the coil

current I and at what frequency f is it oscillating?

e. If a metal disk of mass M is placed on the coil, when the charged capacitor at voltage V is

discharged into the coil and if all losses are negligible, what is the maximum initial diskvelocity v and to what height h will it go?

f. Evaluate the coil self-inductance L of part c, coil current I and frequency f of part d,

initial disk velocity v and height h in part e for parameters capacitance C =25μ F ,

voltage V =4000 volts, N =50 turns, average coil radius a=7 cm.

3


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Problem 1.3

A charge q of mass m with initial velocity v =v0i x is injected at x=0 into a region of uniform

electric field E E i . A screen is placed at the position x=L. At what height h does the charge= 0 z hit the screen? Neglect gravity.

Problem 1.4

a. View H/M video 1.4.1, Magnetic Field of a Line Current, at

(http://web.mit.edu/6.013_book/www/VideoDemo.html) where the magnetic fieldstrength is measured by a Hall effect probe. This probe works by the principle that when

charges flow perpendicular to a magnetic field, the transverse displacement due to theLorentz force can give rise to an electric field.

A magnetic field perpendicular to a current flow deflects the charges transversely giving rise

to an electric field and the Hall voltage. The polarity of the voltage is the same as the sign of

the charge carriers.

4

Problem 2.8 in Electromagnetic

Field Theory: A Problem Solving

Approach, by Markus Zahn, 198

Used with permission.

Figure 5.6 in Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.

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b. A uniform magnetic field B = B i = μ H i is applied to a material carrying a current ino z o o zthe y direction. For positive charges, as for holes in a p-type semiconductor, the charge

velocity v v i is in the positive y direction, while for negative charges as typically= y yoccur in metals or in n-type semiconductors, the charge velocity v y is negative. In the

steady state, the charge velocity v y does not vary with time so the net force on the charges

must be zero. What is the electric field (magnitude and direction) in terms of v y and Bo?c. What is the Hall voltage, V H = ( x d ) Φ( x = 0) xΦ = − = − E d in terms of v y , Bo and d ?

d. Can this measurement determine the polarity of the charge carriers assuming that thecurrent i is positively y-directed and B0 is positively z-directed?

Problem 1.5 y

x

1 I

2 I

d

a. Two line currents of infinite extent in the z direction are a distance d apart along the y-

axis. The current I 1 is located at y=d/2 and the current I 2 is located at y=-d/2. Find themagnetic field (magnitude and direction) at any point in the y=0 plane and for any point

the z =0 plane.

Hint: In cylindrical coordinates1 1

φ 1 = − − ⎡ d i x y ⎤ ⎡2 2 ⎤ 2

φ 2 = − + y d i x y2 y d + )2 ⎤ 2i ⎣ ( y ) + xi ⎦ / ⎣ x + ( y − d ) ⎦ ; i ⎡⎣ ( ) + xi ⎤⎦ / ⎡⎣ x + ( ⎦

b. Find the force per unit length on I 1.

c. For what values of I / I are H x y( , = 0) = 0 or H x y = 0)1 2 x y ( , = 0 ?

5

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Massachusetts Institute of Technology

Department of Electrical Engineering and Computer Science


Problem Set #1 SOLUTION

Fall Term 2005

Problem 1.1

1 2 b. T = Mg

=Q Q

where sinθ =2

s

l,

⎝

⎛ ⎜Q1 =Q =

Q ⎞

2 cosθ

4πε 0s sinθ

22 ⎠

⎟

l2

Q2 Mg4πε ( 2 sinθ ) sinθ

=1 , sin θ tanθ =Q Q0 2 1 2 =

2 2 Q Q cosθ 16πε l Mg 64πε l Mg1 2 0 0

(−c. i +

d q)=0 ⇒i =

dq, v iR R

dq= −

Rq ω sin ( )dt dt

=

dt

0ω t

Problem 1.2

Ni b. Ampere’s integral law ∫

C b

H ds = ∫S b

J da , H ≈i i 2π a

λ N Φ NBS b µ N 2a

= 0c. L = = =i i i 2

di dv= = ( = t ) ,d. v L , i C ,v t =0) =V ⇒i I sin (ω ω =

1

dt dt LC

2 ( =C

; Note :1

LI2 =

1CVAt t =0 , v t =0) =V LIω ⇒ I =

V=V

Lω L 2 2

C 1 I V

=

sin ( )

, f =

ω =ω t

L 2π 2π LC

2 Ce.

1 Mv2 =

1CV ⇒v =V

2 2 M

1

CV2 =

Mgh ⇒h =1

CV2

2 2 Mg

, / ,f. L =0.1mH I =2000 A, v =707m s h =255m

Problem 1.3

2

0 0 md z

=qE ⇒ z =qE t

2

+v t + z =qE t

2

, v t = z0

=02m

z0 0dt 2

02m

z0

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2d x

m = ⇒ x =v t ⇒t = dt

20

0v

x

0

qE x2

( =

) =

qE L2

0 0 z =2mv02

, z x L h = 2mv02

Problem 1.4

b. The Lorentz force law F q E v B)= ( + ×

In the steady state F =0 , so: E = − ×v B

⎧⎪ v i ; postive charge carriesv =

y y

, B B i= 0 z⎨⎪⎩−v i ; negative charge carries y y

⎧⎪ v B i ; postive charge carries E =

y 0 x

⎨⎪⎩−v B i ; negative charge carries y 0 x

0

c. V H

= Φ ( x =d ) − Φ ( x =0) = −d

E dx = ∫d E dx

x x∫0

⎧ v B d; positive charge carriers=

y 0 V H ⎨

⎩−v B d; negative charge carriers y 0

d. As seen in part c, different polarity charge carriers have opposite polarity voltage, so the

answer is an indubitable “Yes!”.

Problem 1.5

a. As the line currents have infinite extent in the z direction the magnetic field has no

dependence on the z coordinate.

I The magnetic field of a z-directed line current at the origin is: H = iφ

2π r

Convert cylindrical coordinates to Cartesian coordinates and move the line current to

(

0,d / 2)

, the magnetic field is

I ⎛ ⎛

d ⎞ ⎞

H = ⎛ 2 ⎛ d ⎞

2 ⎞ ⎜⎝ − ⎜⎝

y −

i xi y ⎟+

2 ⎠⎟

x ⎠

2π ⎜⎜

x + ⎜ y −2 ⎠

⎟ ⎟ ⎠⎟

⎝ ⎝

Moving the line current to ( 0,−d / 2) gives the magnetic field as

I ⎛ ⎛ d ⎞ ⎞ H =

⎛ 2 ⎛ d ⎞

2 ⎞ ⎜⎝ − ⎜

⎝ y + i xi y ⎟+

2 ⎠⎟

x

⎠2π ⎜⎜

x + ⎜

y +2 ⎠

⎟ ⎟ ⎠⎟

⎝ ⎝

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The total magnetic field due to the two line currents is

I 1 ⎛ ⎛ d ⎞ ⎞ I 2 ⎛ ⎛ d ⎞ ⎞ y y H total = ⎜ − − ⎟ i x + xi y ⎟ + ⎜ − + ⎟ i x + xi y ⎟⎛ 2 ⎛

d ⎞2

⎞ ⎝ ⎝ ⎜

2 ⎠

⎠ ⎛ 2 ⎛

d ⎞2

⎞ ⎝ ⎝ ⎜

2 ⎠ ⎠ y y2π ⎜⎜

x + − 2 ⎠

⎟ ⎟ ⎠⎟ 2π ⎜⎜

x + + 2 ⎠

⎟ ⎟ ⎠⎟⎜ ⎜

⎝ ⎝ ⎝ ⎝

b. The force density on a line current (force per length) is F I B= × .

xAt (0,d / 2) the magnetic field is: H = − I

2i

2π d

0 1 2 iF I 1= × µ H = − µ I I

y0 22π d

d d I

1

c. H x y = = 2 − I

2

.⎛

2 ⎛ ⎞ 2 ⎞

x ( , 0) ⎛ ⎛ ⎞

2 ⎞

2

2 d d 2π ⎜ x + ⎜ ⎟ ⎟⎟2π ⎜ x + ⎜ ⎟ ⎟⎟⎜ 2 2⎝ ⎝ ⎠ ⎠ ⎝

⎜ ⎝ ⎠ ⎠

When I I 2 = 1 , H x y = = 01 / x ( , 0)

I x1 + 2 H x y = =

⎛ 2 ⎛ ⎞

2 ⎞ ⎛ 2 ⎛ ⎞

2 ⎞

y ( , 0) I x

,

d

d 2π ⎜ x + ⎜ ⎟ ⎟

⎟

2π ⎜ x + ⎜ ⎟ ⎟⎟⎜

2

2⎝

⎝ ⎠ ⎠

⎝ ⎜

⎝ ⎠ ⎠

When I I = −1 , H x y = = 01 / 2 y ( , 0)

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______________________________________________________________________________






Reading Assignment: Sections 1.5, 2.1, 2.4, 2.5 of Electromagnetics and Applications.6.013 Formula Sheet attached.

Problem 2.1The gradient, curl, and divergence operations have simple relationships that will be used

throughout the subject.

a. One might be tempted to apply the divergence theorem to the surface integral in Stokes’

theorem. However, the divergence theorem requires a closed surface while Stokes’

theorem is true in general for an open surface. Stokes’ theorem for a closed surfacerequires the contour to shrink to zero giving a zero result for the line integral. Use the

G

divergence theorem applied to the closed surface with vector ∇ × A to prove thatG∇ • (∇ × A) = 0 .

b. Verify (a) by direct computation in cylindrical coordinates.

(

theorem to show that

c. Integrate the normal component of the vector ∇ × ∇ f ) over a surface and use Stokes’

∫ ( f ) • dS = v∫ f d A= v∫ df∇ × ∇ ∇ • = 0S C C

where f(x, y, z) is an arbitrary scalar function.

∂ f ∂ f ∂ f Hint: df = dx + dy + dz = f ∇ • d A

∂ x ∂ y ∂ z

Since the equality is true for any surface dS conclude that ( f ) = 0∇ × ∇ .

d. Verify the results of part (c) that ∇ × ∇( f ) = 0 by direct computation in spherical

coordinates.

2

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Problem 2.2

A cylinder of radius R1 has a volume current distribution

( ) = ( / R 2 < R2 . There is no surface current on the = 1r R surface.

a. What is the total z directed current flowing through the cylinder?

b. What is the magnetic field H for 0 < < 2 ?r R

c. What is the surface current density on the cylinder of radius R2?

d. What is the total z directed current flowing on the = 2r R cylinder and how is it related to

your answer in (a)?

Problem 2.3

A sphere of radius R1 and free space permittivity ε 0 has a volume charge distribution

ρ ( ) = ρ ( / R ) 0 r R f r 0 r 14 < <

1

The sphere is surrounded by free space and a perfectly conducting sphere of radius R2 so that

E = 0 for > 2 . r R1 surface.r R There is no surface charge on the =

a. What is the total charge on the sphere?

b. What is the electric field E for 0 < < 2 ?r R

c. What is the surface charge density on the perfectly conducting sphere of radius R2?d. What is the total charge on the = 2r R spherical surface and how is it related to your

answer in (a)?

3

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Problem 2.4

The N turn rectangular coil of height h and length l shown above is used to measure the magnetic

field intensity H due to the current i I 0 sinω t in the infinitely long wire of height R= above the

coil. The N turn coil is open circuited and thus its current is zero.

a. What is the total magnetic flux

z +A

R h+

f = 0 ∫ ∫ φ dr λ μ N dz H z R

linked by the N turn coil?

b. With wire current i I 0 sinω t what is v t across the terminals of the N turn coil?= ( )

Evaluate for N =20 turns, h=8 cm, l =20cm, I = 6 amp peak, and ω = 120π radians (600

Hertz).c. How should the N turn coil be positioned with respect to the line current i so that

v t ( ) = 0 ?

4


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Problem Set #2 SOLUTION

Fall Term 2005

Problem 2.1

a. By the divergence theorem;

∇ ∇ × A dV =∫ (∇ × A da where S encloses V.∫ i( ) )iV S

By Stokes’ theorem:

)i i∫ (∇ × A da =∫ A dlS ' C

Suppose S’=S is a closed surface

S S’

C

Or S’ is an open surface with boundary contour C. i.e., S’ is the same as S, except for the

curve C, which makes S’ not a closed surface.

Now consider the limit as CÆ0;

SS’ S’

So that S’=S.

If C is 0 then ∫ (∇ × A da =∫ A dl =0 and ∫∇i(∇ × A dV =∫ (∇ × A da =0)i i ) )iS ' C V S

Since V can be any volume, the argument of the integral must be identically 0:

∇i(∇ × A) =0

b. A = A i + A i + A i in the cylindrical geometry.r r φ φ

z z

∇ × A =⎜⎛ 1 ∂ A

z − ∂ A

φ ⎞ ⎛ ∂ Ar ∂ A z ⎞ 1⎛ ∂ (rAφ )

−∂ A ⎞

r − + ⎜ ⎟ i z⎝ r ∂φ ∂ z

⎟ ir + ⎜⎝ ∂ z ∂r

⎟ ⎠

iφ r ⎜⎝ ∂r ∂φ

⎟ ⎠ ⎠

∂ Aφ ⎞⎤ 1 ∂ ⎛ ∂ A ∂ A z ⎞ +∂

⎡1 ⎛ ∂ (rAφ ) ∂ A ⎞⎤∇ ∇ × A) =

1 ∂

⎢⎡

r ⎜⎛ 1 ∂ A z − ⎟⎥ + ⎟ ⎢ ⎜ −

r ⎟⎥∂ ⎢

i(

r r ⎣ ⎝ r ∂φ ∂ z ⎠⎦ r ∂φ ⎝ ⎜ ∂ z

r −∂r ⎠ ∂ z r ⎜ ∂r ∂φ ⎠

⎟⎦⎥⎣ ⎝

2 2 21 ∂2 A 1 ∂ Aφ ∂ Aφ 1 ∂2 Ar −

1 ∂ A 1 ∂ Aφ ∂ Aφ 1 ∂

2 Ar z += z − − + + − =0

r r r z r ∂ ∂ z r r r z r ∂ ∂ z∂ ∂φ r ∂ z ∂ ∂ φ ∂ ∂φ r ∂ z ∂ ∂ φ

c. From stokes’ theorem

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C

( i f∂ f ∂ f

∫ (∇ × ∇ f ))ida = ∫ ∇ f dl Here ∇ = ∂ f

i +∂ y

i y + i , x z∂ x ∂ z

S C

dl = dxi + dy y + dzi

S

dl x z

A

∂ f ∂ f ∂ f ∇ f idl = dx + dy + dz = df

∂ x ∂ y ∂ z

ASo ∫ (∇ × ∇ f ))ida = ∫ ∇ f dl = ∫ df = f = 0( i A

S C C

Since S can be any surface, the argument of the integral must be identically 0:

∇ × ∇ f ) = 0(

∂ f 1 ∂ f 1 ∂ f d. ∇ = ir + iθ +

∂iφ in spherical coordinate. f

∂r r ∂θ r sinθ φ

⎡ ⎛ 1 ∂ f ⎞ ⎛ 1 ∂ f ⎞ ⎤ ⎡ ⎛ ∂ f ⎞ ⎛ 1 ∂ f ⎞ ⎤

∂ ∂ ⎜ ⎢ ∂ ⎜ r ∂ ⎟

⎥r sinθ φ ⎠ ⎥ iθ ∇ × ∇ f ) =

1⎢⎢

∂ ⎜⎝ sinθ

r sinθ φ ⎠⎟

− ⎝ r ∂θ ⎠

⎟⎥⎥

i +1 ⎢ 1 ⎝ ∂r ⎠

⎟−

∂ ⎜⎝ (

r sinθ ⎢ ∂θ ∂φ ⎥r

r ⎢sinθ ∂φ ∂φ ⎥⎢

⎥ ⎢ ⎥⎣

⎦ ⎣ ⎦

⎡ ⎛ 1 ∂ f ⎞ ⎛ ∂ f ⎞ ⎤ ∂ ⎜

⎟ ⎥ 1

⎢∂ ⎜

⎝

r

r∂

θ ⎠

⎟

−

⎝ ∂

r ⎠ ⎥ iφ + ⎢r ⎢ ∂r ∂θ ⎥

⎢ ⎥⎦

⎣

2 ∂2 2 ∂2 2 ∂21 ⎛ ∂ f f ⎞ 1 ⎛ ∂ f f ⎞ 1 ⎛ ∂ f f ⎞= − ⎟ ir + ⎜ − ⎟ iθ + ⎜ − ⎟ iφ = 02 ⎜

⎝ ∂ ∂ θ φ ⎠ r ⎝ ∂ ∂φ ∂ ∂φ ⎠ r ⎝ ∂ ∂θ ∂ ∂θ ⎠r sinθ θ φ ∂ ∂ r r r r

i

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Problem 2.2

a. The total z directed current on the cylinder is:

2

∫0 R1 J 0 2π 4 R1 J R1

2π = 0 I 0 = J ( )2π rdr = ∫0

R1 J 0

⎛

⎝ ⎜ R

r

1

⎟ ⎞

⎠2π rdr =

4 R1

2r r z

0 2

b. By Ampere’s integral law H ds = ∫S J dai i∫C⎧ r

z ( ) '2π 4

( ) 2π r = ⎪⎨∫0 J r ' 2π r dr ' =

J 0 r '4

r=

J 02π

r , r

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i

3

0 1d. The total charge at r R2

is: q r R ) = σ π R = −4πρ R

= −q= ( = 2 s 4 22

7

Problem 2.4

z l R h 0 ⎛

h ⎞⎤+ +a. H φ = ,λ = µ N ∫ z dz∫ R H dr =⎡⎢µ Nl ln ⎜1+ ⎟⎥ i0 φ 2π r

f

⎣ 2π ⎝ R ⎠⎦

( ) =d λ f ⎡ µ Nl ⎛ h ⎞⎤0 ln 1

ω t b. v tdt

=⎣⎢ 2π ⎝

⎜ + ⎟⎥ω I 0 cos ( ) R ⎠⎦

with the values given in the problem statement v t (( ) =d λ f = 1.35 cos 120π t )

mVdt

c. The coil should be placed vertically centered on the wire so that half the coil flux is positive

and half is negative so that no net flux links the coil.

•

I

×

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Reading Assignment: 2.2, 2.7

Problem 3.1

The superposition integral in free space for the electric scalar potential is

Φ(r ) = ∫V ′ ρ (r ′)dV ′

(1)4πε r −r ′o

The electric field is related to the potential as

E (r ) =−∇Φ(r ) (2)

An elementary volume of charge dq = ρ ( )′ ′r dV at r ′

gives rise to a potential at the observer position r .

The vector distance between a source point at Q and a field point at P is:

r −r ′ =( x − x′)i x +( y − y′)i y +( z − z ′)i zG G

r − r ′a. By differentiating in Cartesian coordinates with respect to the unprimedcoordinates at P show that

∇⎛ ⎜

G

1G

⎞⎟⎟ =

−G

(r −G

r ′)=

G

−ir G

′r

⎜ 3 2r −r ′ r −r ′ r −r ′

where ir ′r is the unit vector pointing from Q to P .

⎝ ⎠

b. Using the results of (a) show that

E (r ) =−∇Φ(r ) = − ρ (r ′)

∇⎜⎛ 1 ⎟

⎞⎟dV

′ = ∫V ′ ρ (r ′)ir ′r dV ′ (3)

2∫V ′ 4πε o ⎜⎝ r −r ′ ⎠ 4πε o r −r ′

2

Figure 4.5.1 from Electromagnetic Fields and Energy by Hermann A. Haus and James R. Melcher.

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c.

A circular hoop of line charge λ 0 coulombs/meter with radius a is centered about theorigin in the z=0 plane. Find the electric scalar potential along the z -axis for z < 0 and z

> 0 using Eq. (1) with ρ (r ′)dV ′ = λ oad φ . Then find the electric field magnitude and

direction using symmetry and Eq. (2). Verify that using Eq. (3) gives the same electric

field. What do the electric scalar potential and electric field approach as z → ∞ and howdo these results relate to the potential and electric field of a point charge?

d. Use the results of (c) to find the electric scalar potential and electric field along the z axis

for a uniformly surface charged circular disk of radius a with uniform surface charge

density σ 0 coulombs/m2. Consider z>0 and z

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a. Start with the electric potential of a point charge, and determine Φ for the electric dipole.

b. Define the dipole moment as p=qd and show that in the limit where d →0 (while premains finite), the electric potential is

p cosθ Φ =

4πε r 2o

c. What is the electric field for the dipole of part (b) with d →0 with p remaining finite?

d. The electric field lines are lines that are tangent to the electric field:

dr E r =rd θ

E θ

Using the result of (c), integrate this equation to find the field line that passes through the

radial point r 0 when θ =π/2. This analytical equation can be used to precisely plot theelectric field lines.

Hint: ∫ cotθ d θ =ln(sinθ ) +constante.

Use your favorite computer plotting routine to plot equipotential and electric field lines

for4πε

p

0

= 0.01 volt-m2. Draw electric field lines for r 0 =0.25, 0.5,1 , and 2 meters and

draw equipotential lines for Φ = 0, ±0.0025, ±0.01, ±0.04, ± 0.16, and ±0.64 volts.

Problem 3.3

When a bird perches on a dc high-voltage power line and then flies away, it does so carrying anet charge.

(a) Why?(b) For the purpose of measuring this net charge Q carried by the bird, we have the apparatus

pictured above. Flush with the ground, a strip electrode having width w and length l is

mounted so that it is insulated from ground. The resistance, R, connecting the electrodeto ground is small enough that the potential of the electrode (like that of the surrounding

ground) can be approximated as zero. The bird flies in the x direction at a height h above

4

Figure P4.7.3 in Electromagnetic Fields and Energy, by Hermann A. Haus and James R. Melcher, 1989.

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the ground with a velocity U . Thus, its position is taken as y=h and x=Ut . At time t ,

what is the effective charge distribution that will allow easy calculation of the electric

scalar potential?(c) Given that the bird has flown at an altitude sufficient to make it appear as a point charge,

what is the potential distribution as a function of time and position ( x, y, z )?

(d) Determine the surface charge density σ s ( , y = 0, z t, ) on the ground plane at y=0 as a

function of time.(e) At time t , what is the net charge, q, on the electrode? (Assume that the width w is small

compared to h so that in an integration over the electrode surface, the integration in the zdirection is simply a multiplication by w.)

Hint: Let ′ x Ut = − dx x

Hint: ∫ = [a 2 + x 2 ]3/ 2 a 2[a 2 + x 2 ]1/ 2 (f) The current through the resistor is dq/dt . Find an expression for the voltage, v, that would

be measured across the resistance, R.

Problem 3.4

A line current I of infinite extent in the z-direction is at a distance d above a perfectly conducting

plane.

(a)

Use the method of images to satisfy boundary conditions and find the magnetic field for

y > 0.

Hint: i = (− + xi ) / x2 + y2φ yi x y

(b) What is the surface current that flows on the y= 0 surface?

5

From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.

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(c) What is the total current flowing on the y=0 surface?

dx 1 x−1 ⎛ ⎞Hint: ∫ 2 2 = tan ⎜ ⎟ + d d d ⎝ ⎠

(d) What is the force per unit length on the line current at y = d ?

6

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�

�

∇ =

∇ = =�)

6.013/ESD.013J — Electromagnetics and Applications Fall 2005

Problem

Set

3

- Solutions

Prof. Markus Zahn MIT OpenCourseWare

Problem 3.1

A

The idea here is similar to applying the chain rule in a 1D problem:

d

1

d

1

df

f �(x)

dx f (x)=

df f (x) dx= −

f 2(x),

where f (x) corresponds to r� .

|r

− |So, by differentiating f (x) we get part of the answer to the derivative of 1/f (x). But, we can just do itdirectly:

|r

− r

� | = (x − x�)2 + (y − y�)2 + (z − z�)2 1

∂

1

∂

1

∂

1

∇ = êx + êy + êz|r − r� | ∂x |r − r� | ∂y |r − r� | ∂z |r − r� |

So, we can apply the trick above by just considering x, y, and z components separately.

∂ ∂ ∂x

|r − � |∂x

(x − x�)2 + (y − y�)2 + (z − z�)2r =x − x

=(x − x�)2 + (y − y�)2 + (z − z�)2

x − x= |r − r� |Similarly:

∂�

y − y�∂y

|r − r | =r�|r

− |∂

�z − z�

∂z|r − r | = |r

− r� |We have

|r − r � |2 = (x − x�)2 + (y − y�)2 + (z − z�)2 ,

so: 1

−[(x − x�) êx + (y − y�) êy + (z − z�) êz]|r

− r� | [(x − x�)2 + (y − y�)2 + (z − z�)2]3/2

The denominators are clearly |r − r� |3 , thus1

(r − r 1 (r − r�)

r� r� 3 r� 2 r�|r

− | − |r

− | − |r

− | |r

− |êr�r

=r� 2

− |r − |

1

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�

C

Problem Set 3 6.013, Fall 2005

B

This follows from part A immediately by substitution. Remember∇ is derivatives in terms of the unprimed coordinates x, y, and z; ∇ does not operate on x�, y�, or z�.

ρ(r�) dV �

λ0a dφ

�

Φ(r) = =V 4πε0|r − r� | 4πε0(a2 + z2)1/2

where we consider the infinitesimal charges dq = (a dφ)λ0 around the ring.

y

ad φ

a

d φ x

Figure 1: Diagram for Problem 3.1 Part C. Differential length adφ in a circular hoop of line charge. (Imageby MIT OpenCourseWare.)

We only care about the z-axis in the problem, so, by symmetry, there is no field in the x and y directions. 2π λ0(a dφ)Φ(r) = ,

4πε0(a2 + z2)1/2

where (a2 + z2)1/2 is the distance from the charge λ0a dφ to the point z on the z-axis.

λ0a

0

Φ(r) = on the z-axis2ε0(a2 + z2)1/2

Check the limit as z → ∞λ0a q 2

Φ(z → ∞) =2ε0 z

=4πε0 z

(same form as point charge where q 2

= λ02πa) � | | | |Now,

0 0E

= −∇Φ(r) = ∂ � Φ � + êy ∂ � Φ � + êz ∂ Φ ∂

λ0a

−(êx � ∂x � ∂y ∂z

) = −êz∂z 2ε0(a2 + z2)1/2

aλ0zE

= êz2ε0(a2 + z2)3/2

Again, we check the limit as z → ∞ :êz

λ0a ; z > 0

êzq2 ; z > 0

= 2ε0 z2

= 4πε0z2

(same form as point charge)−λ0a −q2 E(z → ∞) êz 2ε0 z2 ; z < 0 êz 4πε0z2 ; z < 0

2

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D

From part C

λ0rΦ = 2ε0(r2 + z2)1/2

for a ring of radius r. But now we have σ0, not λ0. How do we express λ0 in terms of σ0?

dr

a

r

Figure 2: Diagram for Problem 3.1 Part D. Finding the scalar electric potential and electric field of a chargedcircular disk by adding up contributions from charged hoops of differential radial thickness. (Image by MITOpenCourseWare.)

Take a ring of width dr in the disk (see figure). We have

Total charge = (r)(2π)(dr)σ0

circum.

total chargeLine charge density = λ0 = = σ0 dr

length

So, λ0

= σ0

dr and

σ0r drdΦ =

2ε0(r2 + z2)1/2

Integrating gives a σ0r dr σ0

a r dr σ0

r=ar2 + z2=Φtotal = =

0 2ε0(r2 + z2)1/2 2ε0 0 (r2 + z2)1/2 2ε0 r=0

σ0 = 2ε0 a2 + z2 − |z|σ0z

1 1

E

= −∇Φtotal =2ε0 z

− √ a2 + z2

êz| |

As a → ∞, z in √ a2 + z2 can be neglected, so:Φtotal(a → ∞) = − 2σε00 (z − a)

z > 0, just like sheet charge

E(a → ∞) = −∇Φ = êz 2σε00

3

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Problem 3.2

A

r-r

r+

x

(x,y,z)z

+qd2

d2 -q

Figure 3: Diagram for Problem 3.2 Part A. (Image by MIT OpenCourseWare.)

We can simply add the potential contributions of each point charge:

q q Φ = ,

4πε0r+ −

4πε0r− 2d

r+

= x2 + y2 + z −2 2d

r− = x2 + y2 + z +2

q 1 1Φ = 4πε0 � d2 − � d2 x2 + y2 + z − 2

x2 + y2 + z +2

B

r-r

r+

x

z

+qd2

d

2 -q

r-

r

r+

x

z

θ

a= cosθd2

Figure 4: Diagrams for Problem 3.1 Part B. (Image by MIT OpenCourseWare.)

p = qd, where p is the dipole moment. We must make some approximations. As r → ∞, r+, r−, and r

4

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C


become nearly parallel. Thus:

dr+

≈ r − a = r −2

cos θ

drr+ ≈ 1 −2r

cos θ .

Similarly,

d1 + cosθr− ≈ r

2r

By part A,

q

1 1Φ =

4πε0

r+

− .r−

If |x| � 1, then 1/(1 + x) ≈ 1 − x. In addition,d

cos θ2r �

1,

so

1 1 1 1 d1 + cosθ≈

r d cos θ≈

2r1 −r+ r2r

1 1 1 1 d

r−≈

r 1 +2

dr cos θ

≈r

1 −2r

cos θ

1 1 1 d d=⇒

r+ −

r−≈

r rcos θ = cos θ

r2

qd cos θ p cos θ=Φ ≈

4πε0r2 4πε0r2

∂ Φ 1 ∂ Φ 1 ∂ ΦE

= −∇Φ = −∂r

êr −r ∂θ

êθ −r sin θ ∂φ

êφ

∂ Φ p cos θ ∂ Φ p sin θ ∂ Φ= ,

∂r= −

2πε0r3 ,

∂θ−

4πε0r2 ∂φ= 0

p cos θ 1 p sin θE = êr + êθ

2πε0r3 r 4πε0r2

pE = [2 cosθ êr + sin θ êθ]

4πε0r3

D

1

r

dr

dθ1

=E rE θ

=2cos θ

sin θ= 2cot θ

1dr = 2cot θ dθ = dr = 2cot θ dθ

r⇒

r

ln r = 2 ln(sinθ) + k = r = r0

sin2 θ (when θ = π/2, r = r0)⇒r

= sin2 θr0

5

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Figure 5: The potential at any point P due to the electric dipole is equal to the sum of potentials of eachcharge alone. The equi-potential (dashed) and field lines (solid) for a point electric dipole calibrated for4πε0/p = 100.

In[1]:=

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Out[4]=

-2 -1 1 2

0.75

0.5

0.25

-0.25

-0.5

-0.75

r0 = 1r0 = .5

r0 = 0.25

r0 = 2

θ

E Field Lines

Figure 6: Mathematica Plot 1 – Electric field lines (Image by MIT OpenCourseWare.)

In[5]:= rp[phi_,theta_]:= Sqrt[Abs[Cos[theta]/(100*Phi)]]

In[6]:= pplot = PolarPlot[{rp[0.0025, theta2], rp[.01, theta2],

rp[.04,

theta2],

rp[.16,

theta2],

rp[.64,

theta2],

rp[2.56,

theta2],

rp[10.24,

theta2],

rp[40.96,

theta2]},

{theta,

-Pi,

Pi},

PlotRange

->

All]

Out[6]=

-1 1

Φ = .01

Φ = 0.04

Φ = 0

Φ = .0025

Φ = .16

-0.5 0.5

1

2

-2

-1

Equipotential

Lines

Figure 7: Mathematica Plot 2 – Equipotential lines (Image by MIT OpenCourseWare.)

In[7]:= tplot = Show[eplot, pplot]

7

http:///reader/full/rp[2.56http:///reader/full/rp[10.24http:///reader/full/rp[40.96http:///reader/full/rp[40.96http:///reader/full/rp[10.24http:///reader/full/rp[2.56

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C


Out[7]=

Figure 8: Mathematica Plot 3 – Electric field and equipotential Lines (Image by MIT OpenCourseWare.)

Problem 3.3

A

The bird acquires the same potential as the line, hence has charges induced on it and conserves charge when

it flies away.

B

The fields are those of a charge Q at y = h, x = Ut and an image at y = −h and x = Ut.

The potential is the sum of that due to Q and its image −Q.

Q 1 1Φ =

4πε0 (x − Ut)2 + (y − h)2 + z2 − (x − Ut)2 + (y + h)2 + z2

D

From this potential

∂ Φ=

Q

y − h y + h .E y = −

∂y 4πε0 [(x − Ut)2 + (y − h)2 + z2]3/2 − [(x − Ut)2 + (y − h)2 + z2]3/2

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Thus, the surface charge density is

Qε0

−h h

σ0

= ε0E y y=0

=z2]3/2 z2]3/2

|4πε0

[(x

−Ut)2 + h2 +

−[(x

−Ut)2 + h2 +

−Qh=

2π[(x − Ut)2 + h2 + z2]3/2

E

The net charge q on the electrode at any given instant is w l −Qh dxdzq = .

2π[(x − Ut)2 + h2 + z2]3/2

If w � h,z=0 x=0

l −Qhw dx

q = .

2π[(x − Ut)2

+ h2

]3/2

For the remaining integration, x� = (x − Ut), dx� = dx, andx=0

l−Ut −Qhw dx�q = .

2π[x�2 + h2]3/2 −Ut

Thus,

Qw l − Ut Utq = −

2πh (l − Ut)2 + h2 + (Ut)2 + h2 .

The dashed curves (1) and (2) in the figure 9(a) below are the first and second terms in the above equation.They sum to give (3).

l Ut

q

(1) (2)

(3)l /U t

v

(a) (b)

Figure 9: Curves for Problem 3.3 Part E. The net charge (a) and voltage (b) as a function of time on the

electrode in the y = 0 plane. (Image by MIT OpenCourseWare.)

F

The current follows from the expression for q as

dq Qw −Uh2 Uh2

i =dt

= −2πh [(l − Ut)2 + h2]3/2 + [(Ut)2 + h2]3/2

and so the voltage is then V = −iR = −R dq/dt. A sketch is shown in figure 9(b) above.

9

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C


Problem 3.4

Figure 10: Diagram for Problem 3.4. The image current from a line current I ̂ez a distance d above a perfect

conductor. (Image by MIT OpenCourseWare.)

A

By the method of images, the image current is located at (0, −d) with the current I in the opposite directionof the source current.For a single line current I at the origin, the magnetic field is

I IH

=2πr

êφ =2π(x2 + y2)

(−y êx + x êy).

Use the superposition for a current I in the +z direction at y = d so that y is replaced by y − d and for thecurrent

−I in the

−z direction at y =

−d so that y is replaced by y + d. Then

I IHtotal =

2π(x2 + (y − d)2)(−(y − d) êx + x êy) − 2π(x2 + (y + d)2)(−(y + d) êx + x êy)

B

The surface current at the y = 0 surface is

−IdK z = −H x|y=0+ =⇒ K = π(x2 + d2) êz

The total current flowing on the y = 0 surface is

+∞ +∞ +∞−Id êz 1 −Id êz 1dx =

x−1Itotal = êz K z dx = ez.−I ˆtan

π d d=

(x2 + d2)π−∞ −∞ −∞

D

The force per unit length on the current I at y = d comes from the image current at y = −dµ0I 2

F = (I êz) × (µ0H(x = 0, y = d)) = êy.4πd

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______________________________________________________________________________





Reading Assignment: Sections 4.1, 7.1-7.4

Problem 4.1

, ,ε σ μ

Perfectly conducting coaxial cylindrical electrodes of length l , inner radius a, and outer radius b

are shown above. The material between the electrodes has dielectric permittivity ε , Ohmic

conductivity σ , and magnetic permeability μ . Note that there is no electric field in the region

r

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Problem 4.2

A pair of parallel plate electrodes with spacing s and voltage difference V 0 enclose an Ohmic

material whose conductivity varies with position asσ σ 0e x / s .= The permittivity ε of the

material is a constant and the system is in the DC steady state.

( ) /0

, x s

x eε σ σ =

a. Find the electric field and the resistance. b. What are the volume and surface charge distributions?

c. What is the total charge in the system, i.e., what is the sum of the total surface charge on

the electrodes and the total volume charge in the material?

Problem 4.3

xarea A

0

s,ε σ

0( 0, ) f t x s

ρ ρ

=

=

( , ) x E x ti(t )

Short circuited parallel plate electrodes of area A enclose a lossy dielectric of thickness s with

dielectric permittivity ε and Ohmic conductivity σ . The lossy dielectric at time t=0 has a free

volume charge density ρ f (t = 0, x) = ρ 0 x / s . Neglect fringing field effects.

a. What is the volume charge distribution for 0

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Problem 4.4

Consider an electric scalar potential in Cartesian coordinates that only depends on coordinates x

and y and that can be expressed as a product solution

Φ( , ) = ( ) ( ) (1) y X x Y y

In the region of interest, the volume charge density is zero, so the potential Φ( x, y) satisfies

Laplace’s equation2 2

2 ( , ) = ∂ Φ

2 + ∂ Φ

2 = 0

(2)∇ Φ x y ∂ x ∂ y

a. Using (1) in (2) show that (2) reduces to2 21 d X 1 d Y

+ = 0 (3) X dx2 Y dy2

Since each term in (3) is a function of x only or a function of y only, argue that each term can at

most be a constant ±k 2 where k 2 is called the separation constant2 21 d X 2 1 d Y 2

X dx2

= +k ,Y dy

2= −k (4)

b. Find solutions for X and Y when k 2 = 0 . These are called zero separation constantsolutions. Write down the general solution for Φ( , ) x y = X x Y y for the zero( ) ( )

separation constant solutions.

c. A hyperbolically shaped electrode whose surface shape obeys the equation xy = ab is at potential V 0 and is placed above a grounded right-angle corner as in the figure below.

The equipotential and field lines for a hyperbolically shaped

electrode ( xy ab)= at potential V 0 above a right-angle conducting

corner are orthogonal hyperbolas.

4

Figure 4.1 in Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.

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The boundary conditions are

( 0) = 0, Φ( y = 0) = 0, Φ( xy = ab) 0Φ = = V (5)

Using the zero separation constant solutions of part (b) find the electric scalar potential Φ( , y)

that satisfies the boundary conditions.

d. Find the electric field, E = −∇Φ and the surface charge distribution along the x=0 and

y=0 planes.

Find the equation y(x) of the electric field line that passes through the point ( , x y ) .e. 0 0

dy E Hint: = y

dx E x

f. Now find the non-zero separation constant (k 2 ≠ 0) solutions to eq. (4) and write down

the non-zero general solution for Φ( x, y) with spatially periodic solutions in the y

direction, and exponential solutions in the x direction.g.

2ε 1ε

x

y

0( 0, ) sin x y V ayΦ = =

A potential sheet with electric scalar potential

( 0, y) = 0

Φ =

V sin ay

is placed at x=0 separating dielectric media with permittivity ε 1 for x>0 and ε 2 for x

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6.013/ESD.013J — Electromagnetics and Applications Fall2005

Problem

Set

4

- Solutions

Prof.MarkusZahn MITOpenCourseWare

Problem 4.1

A

ByGauss’law,∇ D =∇ (εE) =ρ= 0 fora

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ProblemSet4 6.013,Fall2005

E

Use Ampere’s law:

A(t) I (t) I (t)H

ds

=I = 2πr =I (t) = A(t) = = H(r, t) = êφ· ⇒ r ⇒ 2π ⇒ 2πr

F

InductanceL= Φ/I

�

b I (t) µI (t)l b µl bΦ = B dS =l µ dr= ln = L= ln·

a 2πr 2πr a⇒

2π aS

G �b

ln 2πlε εRC= a = , thesameRCasparallelplates.

2πlσ ln �b

σ

a

H

LC=µl

b

2πlε=µεl2 , thesameLCasparallelplatesofdepthl.ln

2π a ln�b

a

Thespeedoflightinthematerialiscm = 1/√ µε,soLC=l2/c2 .m

Problem 4.2

A

∂ρ∇ J + = 0 = ∇ J = 0inDCsteadystate.·

∂t⇒ ·

∂J x∇ J = = J x =J 0 constant.·∂x

⇒J 0

σ(x)E (x) =J 0

= E (x) =σ0ex/s

⇒s s s

E (x) dx=σ

J 0

0

e−x/s dx=− J σ

0

0

se−x/s

=V 0 0 0

0

J 0s V 0σ0

σ0 (1− e−1) =V 0

=⇒ J 0

=s(1− e−1)

V 0E (x) =

(1− e−1)sex/sV 0

V 0

s(1− e−1)R= = =

i J 0lD lDσ0

B

dE −εV 0e−x/sρf =ε =

dx (1− e−1)s2 εV 0

σf (x= 0)=εE (x= 0)=(1− e−1)s

σf (x=s) =−εE (x=s) =(e

−−εV

1)0

s

2

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C s −εV 0e−x/s −εV 0 s−x/s dx

−ldεV 0Qv =ld dx=ld e =

(1

−e−1)s2 (1

−e−1)s2 s0 0

εV 0

Qs(x= 0)=(1− e−1)sld

Qs(x=s) =−εV 0

ld(e− 1)s

Qtotal =Qv +Qs(x= 0 )+ Qs(x=s) =−ldεV 0

+εV 0

ld +−εV 0

ld= 0s (1− e−1)s (e− 1)s

Problem 4.3

A

x ε−t/τ

, τ=s σρf (t) =ρ0

e

B

∇ E = = = E x =ρ0 ∂E x ρf (t) x

2

e−t/τ +C (t)·∂x ε

⇒2εss 2s s

−t/τE x dx=ρ0 e−t/τ +C (t)s= 0 = C (t) =−ρ0 e

6ε⇒

6ε0

2

2

2

x

εse−t/τ +ρ0

6

s

εe−t/τ =ρ0

2

1

εs−t/τ 2 − s

3e xE x =ρ0

C

σf (x= 0)=εE (x= 0)=−ρ0 s −t/τe6

σf (x=s) =−εE (x=s) =−ρ0 3

s−t/τe

D

i(t) ∂E x σs−t/τ 1 s −t/τ

A=σE x(x=s) + ε

∂t(x=s) =ρ0

3εe −

τ ρ0

3e = 0

Problem 4.4

A

∂ 2Φ ∂ 2Φ∇2Φ(x, y) =∂x2

+∂y2

= 0, Φ =X (x)Y (y)

d2X (x) d2Y (y)→ Y (y)dx2

+X (x)dy2

= 0

1 d2X (x) 1 d2Y (y)Rearranging + = 0→

X (x) dx2 Y (y) dy2 function

of

x only

function

of

y only

3

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� �

C


Theonlywaythetwotermscanaddtozeroforevery xandyvalue is if

1 d2X (x) 1 d2Y (y)

X (x) dx2 =k2 ,

Y (y) dy2 =−k2

B

Fork2 = 0,we have

d2X (x) d2Y (y)= 0, and = 0

dx2 dy2

Therefore,

X (x) =ax+bY (y) =cy+d

=⇒ Φ =Axy +Bx +Cy +D,

Φ =X (x)Y (y) ,anda,b,c,d,A,B,C,Darearbitraryconstants.wherewehaveused

BoundaryConditions:0; x= 0 (1)

Φ(x, y) = 0; y=0 (2)V 0; xy=ab (3)

Φ(x, y) =Axy +Bx +Cy +D(we know Φ(x, y) isofthisform)

Boundary

condition (1)Φ(x= 0, y) = 0 = Cy+D= 0.Thishastoholdforevery valueofy.This⇒

means that C= 0 and D= 0.

Boundary

condition

(2)Φ(x,y= 0)= 0 = Bx +D= 0.WealreadyknowthatD= 0,soBx= 0.⇒This has to hold forevery valueofx,so B= 0.Boundary

condition (3) Φ(x, y) suchthatxy=ab=V 0.We know D= 0, C= 0, B= 0,so Φ(x, y) =Axy

on the boundaryxy=ab.

V 0Φ(x, y) =Aab=V 0

= A=⇒ab

→ V 0Φ(x, y) = xyab

D

0

ẑ =

∂ Φ

∂z

∂ Φ ∂ Φ V 0

V 0E1

= =−∇Φ −∂x

x̂−∂y

ŷ− �

yx̂− xŷ−ab ab

Weusetheboundaryconditionn̂ [E1

−E2] =σs on thex=0planeandthenormal n̂ =x̂.·0 ε0V 0

E2]x̂ [ε1E1

− ε2·

E2]

σs =⇒ σs = +ε1E 1,x =−= yab

b/c perfect conductor

On they= 0 plane,n̂ =ŷ and

0 ε0V 0 ŷ [ε1E1

− ε2· x=σs=−ab

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E

dy x 1 2 1 2 dx y

=⇒ y dy=x dx =⇒2y =

2= x +C

1 2 2 2 2 2 2C=2(y0

− x0) =⇒ y − x = (y0

− x0)

F

1 1= +k2 and =−k2

d2X(x) d2Y (y)X (x) dx2 Y (y) dy2

Thesolutionis

X (x) =Aekx +Be−kx

Y (y) =C sin(ky) + D cos(ky)

whereA,,B,C,andDarearbitraryconstants.

Φ(x, y) =X (x)Y (y) = [asin(ky) + b cos(ky)]ekx + [csin(ky) + d cos(ky)]e−kx

wherea, b,c,anddarearbitraryconstants.

G

Φ1(x, y) = [a1 sin(ky) + b1 cos(ky)]ekx + [c1 sin(ky) + d1 cos(ky)]e

−kx

Φ2(x, y) = [a2

sin(ky) + b2

cos(ky)]e−kx + [c2

sin(ky) + d2

cos(ky)]ekx

Region (1)is forx≥ 0and Region (2)is forx≤ 0.Boundary Conditions:

(1) Φ1(x, y) = 0 ;x→∞(2) Φ2(x, y) =0;x→ −∞(3) Φ1(x, y) x=0 = Φ2(x, y) x=0 =V 0 sin(ay)| |

Boundary

condition (1) =⇒ noekx terms for Φ1(x, y) becausetheyblowupasx→∞,so

Φ1(x, y) = [c1 sin(ky) + d1 cos(ky)]e−kx.

Boundary

condition (2) = noe−kx terms for Φ2(x, y) becausetheyblowup⇒ asx→ −∞,so

Φ2(x, y) = [c2

sin(ky) + d2

cos(ky)]ekx.

Boundary

condition (3) =⇒

c1

sin(ky) + d1

cos(ky) =c2

sin(ky) + d2

cos(ky) =V 0

sin(ay).

Clearlyc1 =c2 andd1 =d2 becausesineandcosineare independent (youcan’tmakeasineequalacosineforally).Thatsaid,

c1 =c2 =V 0, d1 =d2 = 0, k=a

Φ1 =V 0 sin(ay)e−ax; x≥ 0

Φ2 =V 0 sin(ay)eax; x≤ 0

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H

0∂ Φ ∂ Φ ∂ Φ

E1

=

−∇Φ =

− ∂xx̂+

− ∂y −

∂x

ŷ+ ẑ

E1 =aV 0 sin(ay)e−ax x̂− aV 0 cos(ay)e−ax ŷ

E2 =−aV 0 sin(ay)eax x̂− aV 0 cos(ay)eax ŷ

Tofindthesurfacechargeweneedtousetheconditionn̂ [ε1E1 − ε2E2] =σs atx= 0,wheren̂ =x̂ =· ⇒ε1E 1,x|x=0 − ε2E 2,x|x=0 =σs.

E 1,x|x=0 =aV 0 sin(ay), E 2,x|x=0 =−aV 0 sin(ay)

σs =aV 0 sin(ay)(ε1 +ε2)

Forx >0,

E1 =aV 0e−ax[sin(ay) x̂ − cos(ay) ŷ]

dy cos(ay) sin(ay)

dx=−

sin(ay)=− cot(ay) =⇒ dx=−

cos(ay)dy

Letu= cos(ay) sothatdu=−a sin(ay) dy:1 du 1 1

dx= + = x= + ln(u) + C= + ln(cos(ay)) + Ca u

⇒a a

1C=x0 − ln(cos(ay0))

a

1

cos(ay)

(x− x0) = + lna cos(ay0)

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Fall Term 2005 Due: 10/12/05

Reading Assignment: Sections 1.3.2, 1.4, 1.6, 5.1, 5.3Quiz 1 on Thursday, October 20 at 10-11 a.m. Will cover material through Problem Set #5.

The Final Exam will be on Dec. 21, 1:30-4:30 p.m.

Problem 5.1

For the following electric fields in a linear medium of constant dielectric permittivity ε and

magnetic permeability μ , find the free charge density f ρ , magnetic field H , and current density

J .

a) 0 xE=E (x i + yy i ) sinω t

b) 0 xE=E (y i yx i ) cosω t−

Problem 5.2

An electric field is of the form:

6 2 j(4 10 t 4 10 z)xE=10Re e i

π π −×

−

×⎡ ⎤⎣ ⎦

volts/meter

a) What is the frequency f , wavelength λ , and speed of light in the medium?

b) If the medium has magnetic permeability μ0 =4π ×10−7 henry/meter, what are the

relative permittivity εr , wave impedance η

, and the magnetic field H ?

c) What is the Poynting vector, S=E×H?

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Problem 5.3

x

y

0

2

p

plasma

ε,μ

Jω εE

t

∂=

∂

•

0 0ε ,μ

j tx x0 0K= K cos K Re[e ]i t i

ω ω =

z

A sheet of surface current with the surface current density:

ω K=i K cosω t = i K Re[e j t ] amperes/meterx 0 x 0

is located at z 0= with free space extending from −∞ < z < 0 and a plasma extending for z > 0 .

The plasma region has dielectric permittivity ε and magnetic permeability μ0 of free space withcurrent constitutive law:

∂J 2= ω εE ∂t

p

a) If all fields are of the form:

ω ] J(z,t)=Re[J(z)e j t

l ω

]

E(z,t)=Re[E(z)e

j t

l ω ] H(z,t)=Re[H(z)e j t

what is the plasma complex conductivity,σ ( )ω , in the plasma region defined as:

lJ(z)= ( ) E(z) ?σ ω

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b) Ampere’s law in the plasma region becomes:

l l lH J(z) + jωεE jωε(ω)E .∇ × = =

What is the frequency dependent permittivity ε(ω) ?

c) Assume thatlE(z) is of the form:

l ⎪ˆ

p

-jk pz

x⎧E e i z > 0

E(z)=⎨⎪⎩

ˆ0

jk z0xE e i z < 0

What are the wavenumbers k and k ? 0

d) What is the general form of

l

E and E ?H(z) in each region in terms ofˆ

0

ˆ p

e) What are the boundary conditions at z=0 on the electric and magnetic fields?

f) Solve for l l pˆ ˆE0 , E , E(z), and H(z).

< >1 ⎡ ˆ ˆ *⎤ , for z 0 and z >g) Find the time average Poynting vector, S =

2

Re⎣⎢E × H

⎦⎥

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Problem 5.4

(1)

S0

1n

2n

45˚

(2)

Se

A glass prism in the shape of an isosceles right triangle has an index of refraction n . The1

surrounding environment has index of refraction n2 . Neglect multiple internal reflections within

the prism.

a) What is the minimum value of n1 for no time average power to be transmitted across the

prism hypotenuse when the prism is in free space (n2 = 1) or in water (n 2 =1.33) ?

b) If the incident light at point (1) has time average power per unit area S , what is the0

exiting time average power Se at point (2) in terms of the refractive index ratio n=n1/n2

assuming that n1 is above the minimum value for no time average power to be transmittedacross the prism hypotenuse. Evaluate S0/Se for the two minimum values of n1 found in

part (a) for free space or water surrounding the prism.

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�

�

�

�

�

� ��

�

�

� � � ��

�

�

�

�

�

�

6.013/ESD.013J — Electromagnetics and Applications Fall2005

Problem

Set

5

- Solutions

Prof.MarkusZahn MITOpenCourseWare

Problem5.1

A

∂ ∂ ∂ρ=∇ D=∇ εE= x̂· ·

∂x+ŷ

∂y ∂z· εE 0(x̂x+ŷy)sinωt= 2εE 0sinωt+ ẑ

�x̂ ŷ ẑ

� �x̂ ŷ ẑ

�

∂

�

∂ ∂ ∂

� �

∂ ∂

� ∂y ∂x−

∂tµH=∇× E= ∂x ∂y ∂z�= ∂x ∂y 0�E 0sinωt=̂z ∂x− ∂y E 0sinωt= 0�

E x E y E z� �

x y 0�

H=C(r) =0⇔

Time-independentmagneticfieldthatcouldbesettozero,sinceitisnotgeneratedbythetimedependentelectricfield.

∂ ∂J=∇× H−

∂tεE=0−

∂tεE 0(x̂x+ŷy)sinωt=−ωεE 0(x̂x+ŷy)cosωt

B

∂ ∂ ∂ ∂y ∂xρ=∇ εE= x̂·

∂x+ŷ

∂y+ ẑ

∂z· εE 0(x̂y − ŷx)cosωt=

∂x−

∂yεE 0

cosωt= 0

�x̂ ŷ ẑ� �x̂ ŷ ẑ�∂ µH

=∇

∂ ∂ ∂ ∂ ∂−∂t

× E

=�

∂x ∂y ∂z� �

∂x ∂y0

��E 0cosωt=̂z(−2E 0cosωt)=�E x E y E z� �y 0�−x2E 0

� 0 2E 0

H=̂z sinωt + C( � r) =̂z sinωt⇔

µω � ωµ

∂J=∇× H−

∂tεE=0− [−ωεE 0(x̂y − ŷx)sinωt] =ωεE 0(x̂y − ŷx)sinωt,

wherethefirsttermis0becauseHdoesnotdependonposition.

Problem5.2

A

ω= 4π 106rad ω

= f= = 2 106 Hz= 2 MHz·sec

⇒2π

·

k= 4π 10−21 2π 1

= λ= = 102 m= 50m·m

⇒k 2

·

ω 4π 106 m mcn= =

·= 108

k 4π 10−2 sec sec·

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�

�

�

�

�

�

��

�

�

��

�

�

�

C


B

cn=c

n⇔ n= c

cn=

3 · 108 msec108 msec

=3

2n=√ εrµr=√ εr 1 εr=n = 9· ⇔

µ µrµ0 µr µ0 1 1η= = = = 120πΩ = 120πΩ= 40πΩ

ε εrε0 εr ε0 9·

3·

n= indexofrefractionNote:

η= impedance

�x̂ ŷ ẑ

�

∂ � � ∂E x ∂−∂t

× E=� ∂z��= +ŷ ∂z =ŷ ∂zE 0cos(ωt − kz)µ0H=∇ �0 0 ∂�E x 0 0�

∂ E 0k � 0=⇒ −∂t

µ0H=ŷ E 0k sin(ωt−

kz) =⇒

H=ŷ

ωµ0cos(ωt

−kz) + C( � r)

E 0 E 0=⇒ H=ŷ

cnµ0cos(ωt − kz) =ŷ

ηcos(ωt − kz)

1= H=ŷ cos(4π 106t− 4π 10−2z) Ampères/m⇒

4π· ·

� x̂ ŷ ẑ � �x̂ ŷ ẑ��H x H y H z� �0 H y 0�S=E× H=�E x E y E z�=�E x 0 0�=̂zE xH yE 0

1 W= S=̂zE 0⇒ η

cos2(ωt−

kz) =̂z10· 4π

cos2(4π·106t

−4π

·10−2z)

m2

2.5 W= S=̂z⇒

πcos2(4π· 106t− 4π· 10−2z)

m2

Problem5.3

A

∂ ∂J

=ω2εE = Re[Ĵejωt ] =ω2εRe[Êejωt ]∂t p

⇒∂t p

= Re[ jω Ĵejωt ] = Re[ω p2εÊejωt ] = jω Ĵ =ω p

2εÊ⇒ ⇒Ĵ ω p

2ε ω p2

= σ(ω) =⇒ Ê = jω =− jε ω

B

jωε(ω)Ê =Ĵ + jωεÊ =σ(ω)Ê + jωεÊ = jωε(ω) =σ(ω) + jωε⇒ω2 ω2

jωε(ω) =− jεω

p+ jωε =⇒ ε(ω) =ε 1−

ω

p

2

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=

�

�

C


k=ω ε(ω)µ0

=⇒k0

=ω√

ε0µ0

•

ω2

ω p2

ω

√ εµ0 1− ωω2

2

ω≥ ω p, pω2

• k p=ω ε 1−ω2

µ0 =ω√

εµ0 1−ω2 jω√ εµ0 ωp2 −

p

1, ω < ω p

D

� yE= ˆ H= ∂ ∇ ˆ − jωµ0H = ˆ 1 ��

�x̂

0

ŷ

0

ẑ�� 1 ∂E ˆx

=× ⇒ − jωµ0

��ˆ ∂z�� −ˆ jωµ0 ∂zE x 0 0• H0 =−ŷ

jωµ0E 0e =−ŷ

ωµ0ˆ jk0 ˆ jk0z k0 E ̂0e

jk0z, z 0

Note: The hatˆabovex,y,zdenotesaunitvector,butabovefieldcomponentsdenotesacomplexphasor.

E

ẑ× (Ê p− Ê0) =0 = ŷ(E ̂p−E ˆ0) =0 = E ̂p=E ˆ0≡ E ˆ• ⇒ ⇒ẑ× ( ˆ H0) =K = H 0) =K 0x̂ = H 0 =• H p− ˆ ˆ ⇒ −x̂(H ̂p− ˆ ⇒ H ̂p− ˆ −K 0

F

ˆ ˆ D,E k0 ˆ k p ˆ ˆ ωµ0H 0

−H p=K 0 =

⇒ −ωµ0E

− ωµ0E=K 0 =

⇒E=

−k0+k pK 0

Therefore:

ωµ0 −jkpz−x̂k0+kp

K 0e , z >0Ê(z) =

x ωµ0 K 0ejk0z, z 0pH(z) = k0 jk0zŷ

k0+kpK 0e , z

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1 ωµ0 −jkpz

k p∗

+jkp

∗z• �S p =̂z2Re

k0

+k pK 0e ·

k0

+k p∗K 0e

1Re

ωµ0k p∗

K 2 2Im{kp}z =̂zωµ0

K 02Re

{k p}e2Im{kp}z=̂z 0e

2 |k0+k p|2

2|k0+k p|2

ωµ0 ,= = ẑ2|k0+kp|2K 0

2ω√

εµ0 1− ωω22

ω≥ ω p⇒ �Ŝ p0,

p

ω < ω p (because Re{k p} = 0)When the wavevector is purely imaginary inside a medium, thefields decayexponentially (theyare called“evanescent”) andnopoweriscarriedbythem.

Problem5.4

A

Figure 1: Diagram forProblem 5.4 Part A. (Image byMIT OpenCourseWare.)

For no power to be transmitted across the prism hypotenuse, the angle of incidence must be above the criticalangle.Ingeneral,fromSnell’sLaw:

n1

sin θ1

=n2

sin θ2

andfortotalinternalreflection:

n1 θ1=45◦sin θ2

≥ 1 = sin θ1

≥ 1 = n1

≥ n2√

2⇒n2

⇒

So for freespace (n2 = 1): n1,min =√ 2≈ 1.414and forwater (n2

= 1.33): n1,min

= 1.33√

2≈ 1.88

B

Thereflectioncoefficientat the inputsurface (1)is

n1

− n2

r(1) = =rn1+n2

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�

��

�

��


andat theoutputsurface (2)is

r(2) =n2

− n1

=−r.n2+n1

Thereforethereflectivityis

2 ��n1− n2�2 �nn21 − 1��2R= r = =� �

.� �

n1 + 1�| | n1

+n2

�

n2

Forn1

≥ n2√

2nopowerislostatthehypotenuse,sothepowertransmittedovertheinputis:

2 2S (2) n

n

2

1 − 1

S (2) = (1−R)(1−R)S (1) =⇒S (1)

= (1−R)2 =1− n1 + 1n2

2 2 2 2

+ 2n1 + 1 n1 − 2n1 + 1

S (2)

n1 −

n2 n2 n2 n2 n2 4n1= =

=

⇒S (1)

n1

2 n1

2

+ 1 + 1n2 n2

ForthevaluescalculatedinpartAn1 =n2√

2forbothcases,therefore:

S (2)

4√

22

S (1)=

(√

2+1)2≈ 0.943

forbothcases.

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Fall Term 2005 Due: 10/26/05

Reading Assignment:

Quiz 1: Thursday, Oct. 20, 2005 in lecture, 10-11AM. Covers material up to and

including P.S. #5.

Problem 6.1

An electric field is present within a plasma of dielectric permittivityε with conduction constituent

relation

2∂J f= ω 2 pε E , where ω p

2 =q n

∂t mε

with , and m being the charge, number density (number per unit volume) and mass of eachq n

charge carrier.

(a) Poynting’s theorem is

i

∂wEMf

∇ +S = −E Ji∂t

For the plasma medium, i

f E J , can be written as

E Jf

∂w k i = .

∂t What is w k ?

(b) What is the velocity v of the charge carriers in terms of the current density J f and

parameters , m defined above?q n and

(c) Write wk of part (a) in terms of v, q, n, and m. What kind of energy density is w ?k

(d) Assuming that all fields vary sinusoidally with time as:

E r ( ) , = Re⎣⎡ ˆ ( j t

⎦⎤t E r )e ω

write Maxwell’s equations in complex amplitude form with the plasma constitutive law.

(e) Reduce the complex Poynting theorem from the usual form

⎡1 ˆ ˆ * ⎤ 1 ˆ ˆ *⎢ r ⎥ j EM >= − E Jf ∇i E( )× H ( )r + 2 ω < w i⎣ 2 ⎦ 2

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to

⎡1 ˆ r ˆ * r⎤

ω w ) ∇i ⎢ E( )× H ( )⎥ + 2 j (< w EM > + < k > = 0 ⎣ 2 ⎦ What are < w EM > and < wk > ?

(f) Show that

2 1 2< w

EM

> + < wk

>=1

μ H − ( )ε ω E4 4

What is ε ( )ω and compare to the results from Problem 5.3b?

Problem 6.2

A TEM wave ( , E H ) propagates in a medium whose dielectric permittivity and magnetic x

y

permeability are functions of z , ( ) and μ z ε z ( ) .

(a) Write down Maxwell’s equations and obtain a single partial differential equation in H y .

(b) Consider the idealized case where ε ( ) z = ε ae+α z and μ ( ) z = μ ae

−α z . Show that the equation

of (a) for H y reduces to a linear partial differential equation with constant coefficients of

the form

∂2 H ∂ H ∂2 H y y y− β − γ = 0

∂ z 2 ∂ z ∂t 2

What are β and γ ?

(c) Infinite magnetic permeability regions with zero magnetic field extend for zd. A

current sheet Re ⎡⎣ 0 j t

⎦⎤ is placed at z = 0. Take the magnetic field of the formi K eω

x

ˆ ω κ H = Re[i yHye( j t − z ) ]

and find values of κ that satisfy the governing equation in (b) for 0

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Problem 6.3

j t k yA sheet of surface charge with charge density σ

f = Re[σ ̂0e(ω −

y ) ] is placed in free space

( ,ε 0 μ 0 ) at z =0 .

j t k yσ f =Re[σ ̂0e

(ω −

y ) ]

ε 0 , μ 0

z

y

0 0,ε μ

The complex magnetic field in each region is of the form

⎧ ˆ ( y + z ) H e− j k y k z

i z >0 H ˆ =

⎪⎨

ˆ

1

( −

)

x

z

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Problem 6.4

A TM wave is incident onto a medium with a dielectric permittivity ε 2 from a medium with

dielectric permittivityε 1

at the Brewster’s angle of no reflection, θ B . Both media have the same

magnetic permeability μ = μ μ ≡ . The reflection coefficient for a TM wave is1 2

E ˆ η cosθ η −

cosθ r R 1 i 2 t = = E ˆ η cosθ η + cosθ i 1 i 2 t

(a) What is the transmitted angle θ t when θ i = θ B ? How are θ B and θ t related?

(b) What is the Brewster angle of no reflection?

1,ε μ 2 ,ε μ

i E

i H•

i Bθ θ =

t θ

(c) What is the critical angle of transmission θ when μ = μ μ ≡ ? For the critical angle toC 1 2exist, what must be the relationship between ε 1 and ε 2 ?

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(d) A Brewster prism will pass TM polarized light without any loss from reflections.

θ

i Bθ θ =

0 0,ε μ

0,ε μ

0

nε

ε =

For the light path through the prism shown above what is the apex angle θ ? Evaluate for

glass with n = 1.45.

(e) In the Brewster prism of part (d), determine the output power in terms of the incident

power for TE polarized light with n = 1.45 . The reflection coefficient for a TE wave is

E ˆ θ η η cos − cosθ r = = 2 i 1 t E ˆ

Rθ η η cos + cosθ

i 2 i 1 t

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C

6.013/ESD.013J — Electromagnetics and Applications Fall 2005

Problem Set 6 - Solutions

Prof. Markus Zahn MIT OpenCourseWare

Problem 6.1

A

∂ (Jf Jf )/∂t = 2Jf ∂ Jf /∂t · · ∂= (Jf Jf ) = 2Jf εω p

2E

∂ Jf /∂t = εω p2E

⇒∂t

· ·

Noting that E Jf

= ∂wk/∂t, we have that

·∂

(Jf Jf ) = 2εω p2 ∂wk ∂wk =

∂

Jf · Jf

= wk =Jf · Jf

=mJf · Jf

since ω p2 =

q 2n

∂t·

∂t⇐⇒

∂t ∂t 2εω p2

⇒2εω p

2 2q 2n mε

B

Jf Jf = ρv = nq v v =⇐⇒

nq

wk = m2Jq f2·nJf = m(2nq q 2|nv|)

2

=21mn|v|2

Therefore wk is the kinetic energy of the plasma charges.

D

∂ ω2

∂t(Jf · Jf ) = εω p2E =⇒ jω Ĵf = εω p2Ê ⇐⇒ Ĵf = − j ω

pεÊ. (1)

Maxwell’s equations in complex form are:

ˆ

E − jωµ ˆ − jωµ ˆ∇ ˆ = H (2)

∇ E =

H

×

ˆ ˆ ω2 ×

ˆ ω2

ˆ p∇ H

= Jf + jωεEˆ

(3)(1)

∇

ˆ p εˆ E

∇

H

= jωε 1 − ω2 E

∇ ×εÊ = 0 ⇐⇒ × H = − j ω E + jωεˆ ⇐⇒ ×εˆ ∇ E = 0

H µ ˆ∇

·· µ ˆ = 0 ∇ · H = 0.·

1

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E

Using (2) and (3) we have

1 1 1H∗) H∗ E

− H∗) H∗ H) −∇ 2

(Ê × ˆ =2( ˆ ·∇

× ˆ Ê ·∇

× ˆ =2[ ˆ · (− jωµ ˆ Ê · (Ĵf∗ − jωεÊ∗)]

1 1 1= −2 jω µ H ε E E f 4 |

ˆ |2 −4

| ˆ |2 −2

ˆ · Ĵ∗

⇐⇒ ∇ · Ŝ + 2 jω �wEM = −2

1Ê · Ĵ∗f

where

ˆ ˆŜ =1

2Ê × H∗, and �wEM = 1

4µ|H|2 − 1

4ε|Ê|2 (4)

But from (1):

1Ê Ĵf∗ = 1Ê j ω p2 Ê∗ = 2 jω 1εω p2 Ê 2 ,

2·

2·

ω 4 ω2| |

so

∇ Ŝ + 2 jω(�wEM + �wk ) = 0,·

where

�wk =4

1εω

ω p2

2

|Ê|2. (5)

F

From (4) and (5):

1 ˆ 2 1 ˆ 2 1ω p2

ˆ 2�wEM + �wk =4µ|H| −

4ε|E| +

4εω2

|E|1 ˆ 2 1 ˆ 2⇐⇒ �wEM + �wk =4µ|H| −

4ε(ω)|E| ,

where

ω2ε(ω) = ε 1 −

ω p

2

which is the result we found in Problem 5.3b.

Problem 6.2

A

∂ ∂ Since neither ε,µ nor the excitation depend on x,y, we can set ∂x = ∂y = 0 and then:

∂ x̂ ŷ ẑ ∂ ∂ ∂E x ∂H y∇ × E = −

∂tµH ⇐⇒

0 0 ∂z = − ∂t µ(ŷH y) ⇐⇒ ∂z = −µ ∂t (1)E x 0 0 2

8/20/2019 Problem Set 2005

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C


∂ x̂ ŷ ẑ ∂ ∂ ∂H y ∂E x∇ × H =∂t

εE

⇐⇒

0 0 ∂z

=∂t

ε(x̂E x) ⇐⇒∂z

= −ε∂t

(2)

0 H y 0

x∂ (1) = ∂ 2E =

∂ 2H y

∂t ⇒ ∂t∂z −µ(z) ∂t2 ∂ 1 ∂H y ∂ 2H y

∂ 2 =⇒ ε(z)∂z ε(z) ∂z − ε(z)µ(z) ∂t2 = 0 (3)∂ (2) = ∂ 1 ∂H y = E x ∂z ⇒ ∂z ε(z) ∂z − ∂z∂tB

Using now ε(z) = εae+αz and µ(z) = µae

−αz we get:

∂ 2H y d 1 ∂H y ∂H y(3) =⇒

∂z2+ ε(z)

dz ε(z) ∂z− ε(z)µ(z)

∂t= 0

∂ 2H y ε′(z) ∂H y ∂ 2H y

⇐⇒ ∂z2

− ε(z) ∂z −ε(z)µ(z)

∂t2

= 0

∂ 2H y εaαe+αz ∂H y +αz −αz ∂

2H y⇐⇒∂z2

−εae+αz ∂z

− εae µae∂t2

∂ 2H y ∂H y ∂ 2H y⇐⇒

∂z2− α

∂z− εaµa

∂t2= 0 (4)

Therefore β = α and γ = εaµa.

Trying now the solution H = Re H ˆyej(ωt+κz)ŷ we get

(4) = (+κ)2H ˆy

−α(+κ)H ˆy

−εaµa( jω)

2H ˆy = 0

⇒ ⇐⇒ (κ2 −

ακ + ω2εaµa)H ˆy = 0

α + α2 − 4ω2εaµa α − α2 − 4ω2εaµa= κ1 = , κ2 = (5)⇒

2 2

D

Boundary Conditions:

0× [ ˆ 0+) − ̂�

� � � ˆ H y K 0• @ z = 0 : ẑ H(z = H(z � = 0−)] = K ⇐⇒ − ˆ (z = 0+) = (6)

0• @ z = d : ẑ H(z � = d+) − H(z = d−)] = 0 ⇐⇒ H y(z = d−) = 0 (7)× [ ̂� � � � ˆ ˆ

E

The general solution is of the form

H ˆy = H ˆ1eκ1z + H ˆ2e

κ2z,

so applying the boundary conditions we get

(6) =⇒ −(H ˆ1 + H ˆ2) = K 0 ⇐⇒ H ˆ1 + H ˆ2 = −K 0

(7) = H ˆ1eκ1d + H ˆ2eκ2d = 0 H ˆ2 = −H ˆ1e(κ1−κ2)d⇒ ⇐⇒

3

8/20/2019 Problem Set 2005

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Together, these give

H ˆ1[1 − e(κ1−κ2)d] = H ˆ1 = −K 0−K 0 ⇐⇒1 − e(κ1−κ2)d

and

H ˆ2 = +K 0

e(κ1−κ2)d = +K 0

=−K 0

,1 − e(κ1−κ2)d e−(κ1−κ2 )d − 1 1 − e−(κ1−κ2)d

where κ1 − κ2 = α2 − 4ω2εaµa ≡ ∆ from (5).

F

Therefore

H ˆy = −1 −

eκ1z+

1 −eκ2z

K 0,e(κ1−κ2)d e−(κ1−κ2)d

and from (2)

∂Ĥ y = − jωε(z)E ˆx∂z

= E x = − −1 κ1eκ1z

+κ2eκ2z

K 0 = +K 0 κ1e∆z/2

+κ2e−∆z/2

eˆ −αz/2⇒ jωεae+αz 1 − e(κ1−κ2)d 1 − e−(κ1−κ2)d jωεa 1 − e∆d 1 − e−∆d

Problem 6.3

A

|k|2 = ω2ε0µ0 =⇒ ky2 + kz2 = ω2ε0µ0 ⇐⇒ kz = ± ω2ε0µ0 − ky2 (1)

B

∂H ˆx ∂H ˆxˆ ˆ ˆ∇ × H = jωε0E ⇐⇒ ŷ∂z

− ẑ

∂y= jωε0E

= Ê =H ˆ1

[ŷ(− jkz z(− jky)] e−j(ky y+kz z) = −H ˆ1(ŷkz − ẑky)e−j(ky y+kz z), z > 0⇒

jωε0) − ˆ

ωε0and

Ê =H ˆ2

[ŷ( jkz z(− jky)] e−j(ky y−kz z) = H ˆ2

(ˆ )e−j(ky y−kz z), z < 0. jωε0

) − ˆωε0

ykz + ẑky

Boundary conditions at z = 0 :

E ˆy(z = 0+) = E ˆy(z = 0

−)H ˆ2

kze−jky y =

H ˆ1kze

−jky y = H ˆ2 = H 1 (2)• ⇐⇒ωε0

−ωε0

⇒ − ˆ

ε0E ˆ(z = 0+) − ε0E ˆ(z = 0−) = σ̂0e−jky y ε0 H

ˆ2

ky − ε0 −H ˆ1(−ky) = σ̂0 = H ˆ2 − H ˆ1 = ωσ̂0 (3)• ⇐⇒

ωε0 ωε0⇒

ky

From (2) and (3), we have that

H ˆ2 = −H ˆ1 = ω2k

σ̂

y

0.

4

C

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D

For the fields to be evanescent in z, kz must be imaginary, so from (1)

ω2ε0µ0 − k2 < 0 =⇒ ω < √ εk

0

Problem Set 2005

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