mes
_p(2L)_ pL_
R A/2 4K 412.,
That is, four times the original resistance.
@ a) RfR,~ = Ema2}) : Rf = 100 (2100 9(0.0004°C-1)(11.5°C) = 99.54 a.
b) R, R. = 346(1) 71-) : Rf = 0.0160 9 + 0.0160 0(0.0005°C-1)(25.8°C) =
0.0158 9. ,
(26-24? a) The current is

28-11: a) B = [( 73/2) and we can calculate the ux through each surface. Note
that there is no ux through any surfaces parallel to the yaxis. Thus, the total ux
through the closed surface is:
<I>B(abe) = 3-7? = ([(0.300T 0)] + [0.300T (2.00 T/m2)(0.300 mm

23-20: a)
Given E = (5.00 N / C m) xi + (3.00 N / C m) zk , edge length L = 0.300 m,
and nS1 = 1 = E n S1 A = 0. n S2 = + k 2 = E n S2 A = (3.00 N/C
j
m)(0.300 m)2z = (0.27 N/C m)z = (0.27 N/C m)(0.300 m) = 0.081 N/C m2.
nS3 = + 3 = E nS3 A = 0.
j
nS 4 =

24-16:
a)
b) V = 2
1 q
.
4 0 a
c) Looking at the diagram in (a): V(x) = 2
1 q
1
=2
4 0 r
4 0
9
a2 + x2
.
d)
e) When x> a, V =
24-34:
1 2q
, just like a point charge of charge +2q.
4 0 x
a)
1 1
kq kq
= kq .
r
ra rb
a rb
(i)
r < ra : V =
(ii)
ra < r < rb

Q2 $622 7
_ = [Al/0 7
20 280A. dd: 2 .
b) Increase the separation by dm => U = Elie0,119 : U0(1 + The change 1s
then 5% dx.
7 25-21;? a) U0:
/
c) The work done in increasing the separation is given by:
de2 Q2
280A =Fdx=>F= 260A.
(1) The force is not

2913;" The magnetic eld is itb the page at the origin; and the magnitude is
II
B=B+Bl=(+)
471' r2 r
z} B _ EB (4.0 x 10-6 C)(2.0 x 105 m/s) + (1.5 x 10-6 C)(8.0 x 105 m/s)
47: (0.300 m)2 (0.400 m)2
=> B = 1.64 x 10'6 T, into the page.
. F [1.012 1 1 #

33-5: a) +3] direction.
27rc 27rc 27r(3.00 x 108 m/s)
bw=27r -4
) f A ca (2.65 x 1012 rad/s) 7'11 X 10 m'
c) Since the electric eld is in the z-direction, and the wave is propagating in the
+ydirection, then the magnetic eld is in the -a:direction oc T5"

32-2: a) I = «21m = (2._10 A) = 2.97 A.
2 2
b) 1,3, = ;I = ;r
c) The root-meansquare voltage is always greater than the rectied average, because
squaring the current before averaging, then squarerooting to get the root-meansquare
value will always give a

31-7: a) [5] = L(di1/dt) = (0.260 H)(0.018O A/s) = 4.68 x 10-3 v.
b) Terminal a is at a higher potential since the coil pushes current through from b to
a and if replaced by a battery it would have the + terminal at a.
,uoNI _ uo(600)(2.50 A)
2m " 21r(0.0

22-14:
F = F1 + F2 and F = F1 F2 since they are acting in opposite directions at
x = 0 so,
4.00 x10 9 C 5.00 x109 C
1
9
6
6.00 x10 C
F=
(0.200 m) 2 (0.300 m)2 = 2.4 x10 N to the
4 0
right.
22-32:
A positive and negative charge, of equal magnitude q, a

30-4: From Ex. (303),
_ NBA _ (90)(2.05 T)(2.20 x 10-4 m2)
_ ._ - = 2.16 10"3 C.
Q R 6.80 n + 12.0 o X
dDB d 27ft _ 27TNABO , 27ft
30-7. 3.) E - - ma COS ' T Sln( T ) for
0 < t < T; zero otherWISe.
T
b)e=0att=.
2
27rNABo T _ 3T
T occursattZandt4.
d) Fro

22-40:
The earths electric field is 150 N/C, directly downward. So,
E 150
22-48:
300 0 2.66 x 10 9 C / m 2 , and is negative.
2 0
a)d = p/q= (8.9 x 10-30 C m)/(1.6 x 10-19 C) = 5.56 x 10-11 m.
b)max = pE = (8.9 x 10-30 C m)(6.0 x 105 N/C) = 5.34 x 10-24

24-16:
a)
b)V = 2
1 q
.
4 0 a
c)Looking at the diagram in (a): V(x) = 2
1 q
1
2
4 0 r
4 0
9
a2 x2
.
d)
e)When x> a, V =
24-34:
1
4 0
2q
, just like a point charge of charge +2q.
x
a)
1 1
kq kq
.
kq
r
ra
rb
a rb
(i)
r ra : V
(ii)
ra r rb : V
(iii)
r >

29-24:
Consider a coaxial cable where the currents run in OPPOSITE directions.
0 I
.
2r
a)
b)
29-44:
For a < r < b, Iencl = I
For r > c, the enclosed current is zero, so the magnetic field is also zero.
B dl B
d = 0I B2r = 0I B =
a)
I
i
b) B2 0 2
2r

27-14:
From the given currents in the diagram, the current through the middle branch
of the circuit must be 1.00 A (the difference between 2.00 A and 1.00 A). We
now use Kirchoffs Rules, passing counterclockwise around the top loop:
20.0 V (1.00 A)(6.00 1

24-61
a) V
1 Q1
V
x
E
(1
) ar
2
2
4 0 R1
x 2 0
x R2
V dv
a
0
b) E
1 rdr 2
2
2
4 0 x r
2 0
x 2 r 2 (
a
0)
( x2 r 2
2 0
x)
V
x
(1
)
x 2 0
x2 R2
24-77
1 Q1
1 Q1
V
2
4 0 R1
4 0 R1
a) E
b)
Conductor has
q2
same
electric
potential.
So: Q1 q1 q2
V
1 q1
1

23-20:
a)
Given E ( 5.00 N / C ) xi (3.00 N / C ) zk , edge length L = 0.300 m,
m
m
j
and n S1 1 E n S1 A 0. n S 2 k 2 E n S 2 A (3.00 N/C
m)(0.300 m)2z = (0.27 N/C m)z = (0.27 N/C m)(0.300 m) = 0.081 N/C m2.
n
S
n
S
n
S
3
4
5
S
6
b)
i
n
j
k
3
4
5
( 0

1
25-48:
a)
1
1
6
C eq
4.0 F 6.0 F 2.4 x 10 F
Q C eqV ( 2.4 x 10 6 F )( 660 V ) 1.58 x 10 3 C
and
V2 = Q/C2 = (1.58 x 10-3 C)/(4.0 F) = 395 V V3 = 660 V 395 V= 265 V.
b) Disconnecting them from the voltage source and reconnecting them to themselves we

PHY 207
Practice Test I
Name: _
Student ID: _
Answer all four problems. Partial credits are based on the clarity and the quality of the work you
show.
&
E=
&
kq
1
r , k=
=9x109 NM2C-2
2
r
4 o
*
F = qE
*
&
E dA =
Qin
o
kq
r
&
dV = E dl = ( E x dx + E y d

32-11
a) L
1
0
C
1
0
LC
b)
1
1
1
1
(0 L
) L
L
(0 )C
0 C
(0 )C 0C
0 (0 )C
When 0 X 0
0 0 0 so X 0
X (0 ) L
32-17
a) Z R 2 ( L
1 2
) 601
C
V
49.9mA
Z
1
L
o
c)
C 70.16
tan
R
VR IR 0.98v
b) I
d) VL IX L 4.99v
VC IX C 33.3v
VL
1
VL
1
1
e) VS
1 2

PHY 207
Practice Final
Prof. F. Zuo
Name: _
Student ID: _
On my honor, I have neither given nor received any aid on this examination
Signature: _
Answer six out the seven problems. Partial credits are based on the clarity and the quality of the
work you s

PHY 207
Practice
Final
Name: _
Student ID: _
On my honor, I have neither given nor received any aid on this examination
Signature: _
Answer ONLY five out of six problems. Partial credits are based on the clarity and the quality of
the work you show.
1._
2

I. Four charges of equal magnitude are arranged at the corners of a square of side I. as shown in
the Figure. {a} Find the magnitude and direction of the foree exerted on the charge in the lower
left corner by the other charges. {31} Show that the eleetri

PHY 207
Practice Test III
Name: _
Student ID: _
Answer the all four problems. Partial credits are based on the clarity and the quality of the work
you show.
&
&
*
kq
1
r , k=
=9x109 NM2C-2, F = qE ,
2
r
4 o
kdq
kq
V = ,V =
r
r
E=
dV =
dU
q
&
*
&
E dA =

PHY 207
Practice Test II
Name: _
Student ID: _
Answer the first four problems. Partial credits are based on the clarity and the quality of the work
you show.
&
E=
&
kq
1
r , k=
=9x109 NM2C-2
2
r
4 o
*
F = qE
&
*
E dA =
V =
Qin
o
kq
kdq
,V =
r
r
dV =
dU

Physics 207
QUIZ 1
Student Name_ ID_
Positive charge Q is distributed uniformly along the x-axis from x=0 to x=a. Calculate the
electric field at any point (x>a) on the x-axis.