Version A - Page 1 of 8

Version A - Page 2 of 8
Problem 1 (20)
Problem 2 (20)
Problem 3 (20)
Problem 4 (20)
Problem 5 (10)
Problem 6 (10)
TOTAL (100)

Version A - Page 3 of 8
Problem 1 (20 points): Mickey Mouse? No. It is a water molecule which has a dipole moment.
The water molecule (H
2
O) as a whole is electrically neutral, but the chemical bonds within the molecule
( q p
) of the water. Consider the water molecule to be a pair of charges (+ q
and –
) located at the origin (point
O
) whose dipole moment p
points in the + y direction. This is shown in the bottom figure. A chlorine ion (Cl
−
), of charge
−
1
.
60
×
10
−
19
C , is located at ( x
, y
) = (0,
R
). Assume that
R
is much larger than the separation d
between the charges in the dipole. Use the binominal expansion
(
1
+ δ ) n ≅
1
+ n δ + n
( n −
1 )
δ
2 /
2
+ L for
δ <
1 and n = ±
1
, ±
2
, L if necessary.
−
19
.
(a) (15 pts) Express the electric field at ( x
, y
) = (0,
R
) due to the water molecule’s dipole moment in terms of p
,
R and
ε
0
.
(b) (5 pts) Calculate the magnitude of the force on the chlorine ion exerted by the water molecule dipole if R = 3 .
00
×
10
−
9 m and
=
6 .
17
×
10
−
30
C
⋅ m .
Is this force attractive or repulsive?
21.62:
1)
2)
3)
Can the student identify the net E field from two point charges?
Can the student simplify the E field and relate to the dipole moment?
Can the student identify the electric force?
(a) The electric field at y = R due to two point charges (separated by a distance d ) is:
= E y
( y = R
) jˆ =
4
1
πε
0
⎡
⎢
(
R −
+ d q
/
2 )
2
+
(
R +
− d q
/
2 )
2
⎤
⎥ jˆ =
4
πε q
0
R 2
⎢
⎢
⎡
⎣
⎜
⎛ −
⎝
1
2 d
R
−
2
−
1
2 d
R
−
2
⎤
⎥
⎦ jˆ
Consider a binominal expansion and keep only the first two terms,
−
2
−
2
1 d
2
R
≅
1
+ d
R and
⎝
1 d
2
R
Hence E y
is given by
E y
( y = R
)
≅
4
πε q
0
R 2
⎡
⎢
⎛ +
⎝
1 d
R
− ⎜
1 d
R
⎤
⎥
=
4
πε q
0
R 2
⎡
⎢⎣
≅
1
−
2 d
R
⎤
⎥⎦
= d
R
2 qd
πε
0
R 3
=
2
πε
0
R 3
(b) E y
( 3 .
00
×
10
−
9 m)
=
∴ ≅
2
πε
0
R 3
6.17
2
πε
×
10
−
30
0
( 3
.
0
C
×
10
−
9
⋅ m m)
3
= qE y
=
(
−
1 .
60
=
×
10
−
19 jˆ (positive y -direction).
4 .
11
×
10
6
N
C)(4.11
×
10
6
C .
(positive y -direction).
N C )
= −
6.58
×
10
−
13
N . It is toward the The electric force F y water molecule (attractive OR negative y -direction).

Version A - Page 4 of 8
Problem 2 (20 points): I is the source of B. Wrong grammar? No. It is an important physics topic.
You find the magnetic field
B
produced by a straight currentcarrying conductor of length 2 a
(see the figure) using Biot-Savart’s law. The magnitude of
B
B =
is expresses as:
µ
4
0
π
I
a
− a
( x 2 + xdy y 2
3
)
/ 2
Here
( x 2 + xdy y 2
)
3 / 2
=
1 x x 2 y
+ y 2
. Now you apply for the above equation to find the magnetic field
B current loop at any point of the x
y
plane. Find
B direction) at point (2
L
, 0).
(magnitude and
1)
2)
Can the student see the net B field at (2L, 0) is a vector sum of B fields due to 4 current-carrying wires?
Can the student use the formula properly?

Version A - Page 5 of 8
Problem 3 (20 points): Testing an effect of dielectric material.
Two square conducting plates with sides of length
L
are separated by a distance
D
. A dielectric slab with dielectric constant
K
with dimensions
L x L x D
is inserted a distance x into the space between the plates, as shown in the figure.
(a) (10 pts) Suppose that the capacitor is connected to a battery that maintains a constant potential difference
V
0
between the plates. Express the charge on the capacitor as a function of x
. Express the energy (
U
) stored in the capacitor in terms of
K
,
L
,
D
,
V
0
, x
, and
ε
0
.
(b) (5 pts) If the dielectric slab is inserted an additional distance
∆ x
into the space between the plates, express the change in stored energy (
∆ U
) in terms of
K
,
L
,
D
,
V
0
, x
,
∆ x
, and
ε
0
.
(c) (3 pts) Suppose that before the slab is moved by
∆ x
, the plates are disconnected from the battery, so that the charge on the plates remains constant. Then the dielectric slab is inserted an additional distance
∆ x
into the space between the plates. Express the change in stored energy (
∆ U
) in terms of
K
,
L
,
D
,
V
0
, x
,
∆ x
, and
ε
0
.
(d) (2 pts) There is an electric force on the slab by the charges on the plates when it is inserted an additional distance
∆ x .
∆ U should equal the work against this force to move the slab a distance
∆ x .
Applying this expression, show that the result of part (b) suggests that the electric force on the slab pushes it out the capacitor. Show that the result of part (c) suggests that the electric force on the slab pulls it into the capacitor.
24.77:
(a)
C
0
= C no dielectric
+ C dielectric
=
ε
0
D
(
L − x
)
L +
ε
0
D
KxL =
ε
0
L
D
∴ U =
1
2
C
0
V
0
2 =
1
2
ε
0
L
D
(
L +
(
K −
1 ) x
)
V
0
2
(b)
∆ U =
∴ ∆ U =
1
2
1
⎛
⎝
2
ε
(
∆ C
0
L
D
) V
0
2
∆ x
where
(
K −
1 )
∆ C
⎞
⎠
V
0
2 =
=
2 dC
0
V
0
2 dx
⎛
⎝
ε
∆ x
0
D
L
=
(
K
ε
0
D
L
( K
−
1 )
−
1 )
∆ x
⎞
⎠
∆ x .
(
L +
(
K −
1 ) x
)
.
1) Can the student identify two capacitors in parallel connection?
2) Can the student formulate the change in
U
as dU
/ dx x
∆ x
by maintaining
V in part (b)?
3) Can the student formulate the change in
U as dU / dx x
∆ x by maintaining Q in part (c)?
(c) If the charge is kept constant on the plates, then:
U =
1
2
CV 2 =
1
2
( CV )
2
1
C
=
1
2
Q
0
2
1
C
where Q
0
= C
0
V
0
∴ ∆ U
(d) “
=
Q
0
2
2 d
( 1
/ dx
C
)
∆ x =
0
∆ x =
Q
0
2
2
⎛
⎜⎜
−
C
1
0
2
⎞
⎠ ⎝ dC
0 dx
∆ x = −
Q
0
2
2
C
0
2
ε
0
D
L
( K −
1 )
⎞
⎠
∆ x = −
V
0
2
2
∆ U should equal the work against this force to move the slab a distance
∆ x .”
⇒
⎛
⎝
ε
0
D
L
( K −
1 )
⎞
⎠
∆ x
∆ U = − F ∆ x
For part (b), F = −
V
0
2
2
ε
0
L
D
( K −
1 ) . The force is in the opposite direction to the motion
∆ x , meaning that the slab feels a force pushing it out.
For part (c),
F = +
V
0
2
2 ⎛
⎝
ε
0
D
L
(
K −
1 )
⎞
⎠
. The force is in the same direction to the motion
∆ x , meaning that the slab feels a force pulling it in.

Version A - Page 6 of 8
Problem 4 (20 points): Enjoy at PHYS-208 lab!
Your group carries out two experiments on a circuit containing two inductors (
L
= 15 mH and 5.0 mH), a resister (
R
= 125
Ω
), two capacitors (
C
= 25.0
µ
F and 35.0
µ
F), and an ideal battery with
ε
= 75.0 V.
The capacitors are initially uncharged.
(a) (4 pts) [Preparation] Find the equivalent inductance for two inductors.
Find the equivalent capacitance.
(b) (6 pts) [Experiment 1] Switch S is in position 1 and you wait for a very long time. Let the current through the inductors be i
1
, the current through the capacitors i
2
, and q
2
? i
2
, and the charge on the capacitor q
2
. What are values of i
1
,
(c) (10 pts) [Experiment 2] The switch is now suddenly flipped to position 2 and you observe an energy oscillation between the inductors and capacitors.
Note that the charge on the equivalent capacitor is generally expressed as q
( t
)
= q max sin
( ω t + φ )
. (i) Find
ω
,
φ
, and q max
; (ii) Find the period of the
30.74: oscillation; (iii) What is the numerical value of the current in the circuit 1.2 ms after the switch is connected on position 2.
(a)
L eq
= 20 mH;
1
C eq
=
25
1
µ
F
+
1
35
µ
F
⇒ C eq
=
14
.
6
µ
F
(b) Using Kirchhoff’s Rules on the left branch when switch S is on position 1:
ε − i
1
∴ i
1
=
R
ε
R
−
1
(
L eq
−
Steady state di
1 dt
=
0
⇒ e
−
( R / L eq
) t
( t → ∞
)
:
) i
1
R i
1
=
ε
R
+
=
L eq di
1 dt
75 .
0
125
V
Ω
= ε .
=
0
.
600 A
Two capacitors are disconnected from the battery.
∴ i
2
=
0 and q
2
=
0 .
(c) (i)
ω =
1
/ LC =
1
/
( 20
×
10
−
3
H) ( 14
.
6
×
10
−
6
F)
=
1
.
851
×
10
3 s
−
1
φ
using the initial condition of q = 0 at t = 0.
∴ φ =
0
Since i max
=
ε i ( t )
R
=
=
0
.
dq dt
600
= ω q max cos
A .
∴ q max
( )
=
, i max
ω i max
=
= ω q max
. We determine
0 .
600 A
1.851
×
10
3 s
−
1
= q max
using the initial condition of
3
.
24
×
10
−
4
C
(We obtain the same answer using the energy conservation.)
(or 324
µ
C ).
(ii)
T
(iii)
= i ( t )
1
/ f
= ω
=
1
⎛
⎝ q max cos
ω
2
π
⎞
⎠
( )
, where
∴ T
=
= ( 1 .
851
×
10
3
3
.
39 s
−
1
×
10
−
3 s .
)( 3 .
24
×
10
−
4
C )cos[( 1 .
851
×
10
3 s
−
1
)( 1 .
2
×
10
−
3 s )] =
−
0 .363
A

Version A - Page 7 of 8
Problem 5 (10 points): EM Wave
A plane electromagnetic wave has a wave intensity of 500 dimensions 50 cm
×
80 cm
J/s
⋅ m
2
. A flat, rectangular surface of is placed perpendicular to the direction of the plane wave. The surface absorbs
20% of the energy and reflects 80%. Take the speed of light in air to be c =
3
.
0
×
10
8 m/s . Find the radiation pressure.
Based on Lecture Problem
The radiation pressure is p rad
=
F
A
=
1
A
∆ p
∆ t
.
Here
∆ p
∆ t
∴ p rad
=
( p incident w ave
)
−
∆ t
( p reflected wave
)
=
IA c
∆ t
−
⎛
⎜⎜
−
( 0 .
80 )
⎝
∆ t
IA ∆ t c
⎞
⎟⎟
=
1
.
80
IA c
=
1
.
80
I c
=
1
.
80
×
(500 J/s
⋅
3 .
0
×
10 8 m/s m
2
)
= 300
×
10
−
8
J/m
3
= 3
.
0
×
10
−
6
N/m
2

Version A - Page 8 of 8
Problem 6 (10 points): PHYS 208 can be useful when you examine a diamond …
You get a diamond whose index of refraction is n
= 2.419. You borrow a flashlight and shine the light into it from surface 1 shown in the figure. Determine the angle of the incident angle
θ
when the light strikes surface 2 at the critical angle.
=
.
1) Can the student identify the critical angle using Snell’s law?
2) Is the student able to relate the refracted angle on surface 1 and the critical angle?
3) Can the student find the incident angle using Snell’s law (again)?
Based on 33.46 and Lecture Problem