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Animation 13.3: Radiation From a Quarter-Wave Antenna

Animation 13.3: Radiation From a Quarter-Wave Antenna

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Question 10: Calculate the flux



G G

S

∫∫ ⋅ dA of the Poynting vector evaluated at r = a



through an imaginary cylindrical surface of radius a and height d, with area A = 2πab ,

i.e. over the sides of the capacitor. Your answer should involve Q, a, I, d, π , and ε o .

What are the units of this expression?

.



εo Area ε o π a 2

=

.

d

d

Rewrite your answer to Question 2 above using the capacitance C. Your answer should

involve only Q, I, and C.

Question 11: The capacitance of a parallel plate capacitor is C =



Question 12: The total electrostatic energy stored in the capacitor at time t is given by

1 Q(t ) 2

. Show that the rate at which this energy is increasing as the capacitor is charged

2 C

is equal to the rate at which energy is flowing into the capacitor through the sides, as

calculated in Question 3 above. That is, where this energy is coming from is from the

flow of energy through the sides of the capacitor.



Question 13: Suppose the capacitor is discharging instead of charging, i.e. Q(t ) > 0 but

dQ(t )

now

> 0 What changes in the picture above? Explain.

dt



Friday 4/22/2005



Solving9-6



MASSACHUSETTS INSTITUTE OF TECHNOLOGY

Department of Physics

Tear off this page and turn it in at the end of class !!!!



Note:



Writing in the name of a student who is not present is a Committee on Discipline



offense.





Problem Solving 9: Displacement Current and Poynting Vector

Group ___________________________________ (e.g. 6A Please Fill Out)

Names ____________________________________

____________________________________

____________________________________

Question 1: Use Gauss’ Law to find the electric field between the plates as a function of time t ,

in terms of Q(t), a, ε 0 , and π .

Answer:



G



Question 2: Using your expression for E above, calculate the electric flux through the flat disc

of radius r < a.

Answer:



Question 3: Calculate the Maxwell displacement current I d = ε 0



dΦ E

through the disc.

dt



Answer:



Question 4: What is the conduction current through the flat disc of radius r < a?

Answer:



Question 5: Calculate the line integral of the magnetic field around the circle,



v∫



G G

B ⋅ ds .



circle



Answer:



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Question 6: What is the magnitude of the magnetic field at a distance r < a from the axis. Your

answer should be in terms of r, I(t), µo , π , and a.

Answer:



Question 7: If you use your right thumb to point along the direction of the electric field, as the

plates charge up, does the magnetic field point in the direction your fingers curl on your right

hand or opposite the direction your fingers curl on your right hand?

Answer:



Question 8: Would the direction of the magnetic field change if the plates were discharging?

Why or why not?

Answer:



Question 9: What is the Poynting vector for r ≤ a ?

Answer:



Question 10: Calculate the flux



G



G



∫∫ S ⋅ dA of the Poynting vector evaluated at r = a through an



imaginary cylindrical surface of radius a and height d, with area A = 2πab , i.e. over the sides of

the capacitor. Your answer should involve Q, a, I, d, π , and ε o . What are the units of this

expression?

Answer:

Question 11: Rewrite your answer to Question 2 above using the capacitance C.

Answer:

Question 12: The total electrostatic energy stored in the capacitor at time t is given by

Q 2 (t ) / 2C . Show that the rate at which this energy is increasing as the capacitor is charged is

equal to the rate at which energy is flowing into the capacitor through the sides, as calculated in

Question 3 above. That is, where this energy is coming from is from the flow of energy through

the sides of the capacitor.

Answer:

Question 13: Suppose the capacitor is discharging instead of charging, i.e. Q(t ) > 0 but now

dQ(t ) / dt < 0 What changes in the picture above? Explain.



Answer:



Friday 4/22/2005



Solving9-8



MASSACHUSETTS INSTITUTE OF TECHNOLOGY

Department of Physics



Problem Solving 10: Double-Slit Interference

OBJECTIVES

1. To introduce the concept of interference.

2. To find the conditions for constructive and destructive interferences in a double-slit

interference experiment.

3. To compute the intensity of the interference pattern.

REFERENCE: Sections 14.1 – 14.3, 8.02 Course Notes.

Introduction

When ordinary light is emitted from two different sources and passes through two narrow slits,

the plane waves do not maintain a constant phase relation and the light shows no interference

pattern in the region beyond the openings. In order for an interference pattern to develop, the

incoming light must satisfy two conditions:





The light sources must be coherent. This means that the plane waves from the sources

must maintain a constant phase relation.







The light must be monochromatic. This means that the light has just one wavelength.



When a coherent monochromatic laser light falls on two slits separated by a distance d , the

emerging light will produce an interference pattern on a viewing screen a distance D away from

the center of the slits. The geometry of the double slit interference is shown in the figure below.



Figure 1 Double slit interference



Friday 5/6/2005



Solving10-1



Consider light that falls on the screen at a point P a distance y from the point O that lies on the

screen a perpendicular distance D from the double slit system. The light from the slit 2 will

travel an extra distance r2 − r1 = ∆r to the point P than the light from slit 1. This extra distance is

called the path length.



Question 1: Explain why constructive interference will appear at the point P when the path

length is equal to an integral number of wavelengths of the monochromatic light.

∆r = mλ, m = 0, ± 1, ± 2, ± 3, ... constructive interference



Answer:



We place the screen so that the distance to the screen is much greater than the distance between

the slits, D >> d . In addition we assume that the distance between the slits is much greater than

the wavelength of the monochromatic light, d >> λ . Then the angle θ is very small, so that



sin θ



tan θ = y D .



Question 2: Based on the geometry of the double slits, show that the condition for constructive

interference becomes

d sin θ = mλ , m = 0, ± 1, ± 2, ± 3, ... constructive interference.



Answer:

Question 3: Explain why destructive interference will appear at the point P when the path

length is equal to an odd integral number of half wavelengths

1⎞



d sin θ = ⎜ m + ⎟ λ, m = 0, ± 1, ± 2, ± 3, ... destructive interference.

2⎠





Answer:



Question 4: Let y be the distance between the point P and the point O on the screen. Find a

relation between the distance y , the wavelength λ , the distance between the slits d , and the

distance to the screen D such that a constructive interference pattern will occur at the point P .

Answer:



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Question 5: Find a similar relation such that destructive interference fringes will occur at the

point P .

Answer:



Intensity of Double-Slit Interference

Suppose that the waves are emerging from the slits are sinusoidal plane waves. The slits are

located at the plane x = −D . The light that emerges from slit 1 and slit 2 at time t are in phase.

Let the screen be placed at the plane x = 0 . Suppose the component of the electric field of the

wave from slit 1 at the point P is given by



E1 = E0 sin ( ωt ) .

Let’s assume that the plane wave from slit 2 has the same amplitude E0 as the wave from slit 1.

Since the plane wave from slit 2 has to travel an extra distance to the point P equal to the path

length, this wave will have a phase shift φ relative to the wave from slit 1,



E2 = E0 sin ( ωt + φ ) .

Question 6: Why are the phase shift φ , the wavelength λ , the distance between the slits, and the

angle related θ by

φ=





d sin θ .

λ



As a hint how are the ratio of the phase shift φ to 2π and the ratio of the path length

∆r = d sin θ to wavelength λ , related?

Answer:



Question 7: The total electric field at the point P is the superposition of these two

fields E = E1 + E2 . Using the trigonometric identity

⎛ A+ B ⎞

⎛ A− B ⎞

sin A + sin ( B ) = 2 sin ⎜

⎟ cos ⎜

⎟,

⎝ 2 ⎠

⎝ 2 ⎠



show that the total component of the electric field is



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φ ⎞ ⎛φ ⎞



E = E1 + E2 = 2 E0 sin ⎜ ωt + ⎟ cos ⎜ ⎟ .

2⎠



⎝2⎠

Answer:



The intensity of the light is equal to the time-averaged Poynting vector

I= S =



1

EìB .

à0



Since the amplitude of the magnetic field is related to the amplitude of the electric field by

B0 = E0 c . The intensity of the light is proportional to the square of the electric field,



φ⎞

⎛φ ⎞



⎛φ ⎞

I ∼ E 2 = 4 E0 2 cos 2 ⎜ ⎟ sin 2 ⎜ ωt + ⎟ = 2 E0 2 cos 2 ⎜ ⎟ .

2⎠

⎝2⎠



⎝2⎠

Let I max be the amplitude of the intensity. Then the intensity of the light at the point P is

⎛φ⎞

I = I max cos 2 ⎜ ⎟

⎝ 2⎠



Question 8: Show that the intensity is maximal when d sin θ = mλ, m = 0, ± 1, ± 2, ± 3, ... .

Answer:



Question 9: Graph the intensity pattern on the screen as a function of distance y from the point

O for the case that D >> d and d >> λ .

Answer:



Friday 5/6/2005



Solving10-4



MASSACHUSETTS INSTITUTE OF TECHNOLOGY

Department of Physics

Tear off this page and turn it in at the end of class !!!!

Note:

Writing in the name of a student who is not present is a Committee on Discipline offense.

Group



___________________________________ (e.g. 6A Please Fill Out)



Names ____________________________________

____________________________________

____________________________________



Problem Solving 10: Double-Slit Interference

Question 1: Explain why constructive interference will appear at the point P when the path length is

equal to an integral number of wavelengths of the monochromatic light.



∆r = mλ, m = 0, ± 1, ± 2, ± 3, ... constructive interference

Answer:



Question 2: Based on the geometry of the double slits, show that the condition for constructive

interference becomes



d sin θ = mλ , m = 0, ± 1, ± 2, ± 3, ... constructive interference.

Answer:



Question 3: Explain why destructive interference will appear at the point P when the path length is

equal to an odd integral number of half wavelengths



1⎞



d sin θ = ⎜ m + ⎟ λ, m = 0, ± 1, ± 2, ± 3, ... destructive interference.

2⎠



Answer:



Friday 5/6/2005



Solving10-5



Question 4: Let y be the distance between the point P and the point O on the screen. Find a relation

between the distance y , the wavelength λ , the distance between the slits d , and the distance to the

screen D such that a constructive interference pattern will occur at the point P .

Answer:



Question 5: Find a similar relation such that destructive interference fringes will occur at the point P .

Answer:



Question 6: Why are the phase shift φ , the wavelength λ , the distance between the slits, and the angle

related θ by φ =





d sin θ .

λ



Answer:



⎛ A+ B ⎞

⎛ A− B ⎞

⎟ cos ⎜

⎟ , show that

⎝ 2 ⎠

⎝ 2 ⎠



Question 7: Using the trigonometric identity sin A + sin ( B ) = 2sin ⎜

the total component of the electric field is



φ ⎞ ⎛φ ⎞



E = E1 + E2 = 2 E0 sin ⎜ ωt + ⎟ cos ⎜ ⎟ .

2⎠



⎝2⎠

Answer:



Question 8: Show that the intensity is maximal when d sin θ = mλ,



m = 0, ± 1, ± 2, ± 3, ... .



Answer:



Question 9: Graph the intensity pattern on the screen as a function of distance y from the point O for

the case that D >> d and d >> λ .

Answer:



Friday 5/6/2005



Solving10-6



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