Physics 105
Spring 2011
HW #1 Solutions
Q6) The net charge on a conductor refers to the difference between the number of
protons and the number of electrons. Neutral: difference =0; Positive: more protons than
electrons; Negative: more electrons than prot
Physics 105
Spring 2011
HW #5 Solutions
Ch. 19
Q5) The outlets in a double outlet are connected in parallel. Both outlets deliver the
same potential difference of 120 VAC and they operate independently. A lamp plugged
into one outlet can be turned on and
Physics 105
S2011
HW 4 Solutions
Ch 17
P32) Q = CV, so V = Q/C=16.510-8 C/950010-12 F = 17 V.
P39)
E = V/d. Know d, but need V. Use Q = CV so V = Q/C = 7210-6 C/0.8010-6 F = 90 V.
Thus E = 90 V/0.002 m = 45,000 V /m.
P46)
PE = 1/2CV2 = 0.5 220010-12 F (65
Physics 105
Spring 2011
HW #3 Solutions
Ch 17 Questions
Q6) Yes. A particle with a negative charge moving from lower potential
to higher potential will have its potential energy decrease. For example
suppose the a charge of -3 C moves from a potential of
Physics 105
Spring 2011
HW #2 Solutions
Ch 16
Q14) A negative charge can be used to determine the electric field. The
"
magnitude of E is still
"
F
, but the direction of E is opposite to the direction
q
of the force on the negative charge.
!
!
Q18) The e
Physics 105
Spring 2011
HW #6 Solutions
Ch. 19
Q19) No, the total energy supplied by the battery is not stored in the capacitor. Some of the
energy is dissipated as heat, as the charge flows from the battery to the capacitor through the
resistor. Recall f
Physics 105
Spring 2011
HW #7 Solutions
Ch. 20
Q8)
The kinetic energy will stay the same. The force on the charged particle due to the
magnetic field is perpendicular to the direction the particle is traveling and therefore no work is
done (W = Fdcos; = 9
Physics 105
Spring 2011
HW 12 Solutions
1) Ch23 P31
The angle between the light beam and the surface of the pool is given by tan = 1.3/2.7.
This gives = 25.7. The angle of incidence is 90 - 25.7 = 64.3. Using Snells law, the
$1
'
angle of refraction is gi
Physics 105
Spring 2011
HW 11 Solutions
Part 1
Ch. 23 P72
a) Image 1: The closest image is the one you see looking in any single mirror. Since
you are 1.5 m from the mirror, the image appears 1.5 m behind the mirror. Thus it is 3.0
m from you.
Image 2: Th
Physics 105
Spring 2011
HW 10 Solutions
1) a) C =
"0 A
. A = r2 = 0.0038 m2. D = 0.025 m. C = 1.36 10-12 F.
d
b) Q = CV = 3.4 10-10 coulombs.
!c) E = V/d = 250/0.025 = 10,000 V/m.
d) Flux through left side = EA =10,000 V/m (0.01m)2 = 3.1 Nm2/C. E is into
Physics 105
Spring 2011
HW #8 Solutions
Ch. 21
Q6)
a)
The current moves through RB from right to left. Using the right hand rule for
solenoids, the left side of coil B is a north pole. Over by coil A, the field due to coil B is pointing
from right to left
Physics 105
Spring 2011
HW #9 Solutions
1) a) FT = mg = .00444 kg 9.80 = 0.0435 N
b) Since the tension decreases, the force on the sphere due to the electric field must be up.
Since the sphere has a negative charge, the field must be down. Thus the top pl
Waves and Fields (PHYS 195): Week 6
Spring 2014 v1.5
Intro:
Well be studying the wave properties of sound. Most of the wave behavior is the same as on waves on
a string but there are some dierences. One is that sound waves normally travel in three dimensi
Waves and Fields (PHYS 195): Week 5
Spring 2014 v1.5
Intro:
This week start studying waves, beginning with transverse waves on a string and basic properties of
waves. We will work toward nding the equation of motion for waves on a string the wave equation
Waves and Fields (PHYS 195): Week 3
Spring 2014 v2.0
Intro:
This week we begin to study how the dynamics of oscillation changes due to damping. Then we will
add driving forces and encounter the phenomenon of resonance: In driven, lightly damped systems
th
Waves and Fields (PHYS 195): Week 2
Spring 2014 v2.1
Intro:
We will see on Wednesday (January 29) how all conservative systems exhibit SHM around equilibrium.
Our study of oscillation continues with a detailed look at more realistic models of oscillation,
Waves and Fields (PHYS 195): Week 1
Spring 2014 v2.0
Intro:
Our course starts with an in-depth study of oscillations: motion that returns to the same place at
the same momentum. The tools are familiar from last semester, e.g. Newtonian mechanics, although
Waves and Fields (PHYS 195): Week 4
Spring 2014 v1.0
Intro:
This week we nish our study of damped driven oscillators - and resonance. Next week well start on
our study of waves, beginning with transverse waves on a string. We will have one math interlude
Waves and Fields (PHYS 195): Week 8
Spring 2014 v1.5
Intro:
We will develop of the theory of static electric elds. Before break we will discuss Coulombs law,
the electric eld, and the close analogy with the gravitational eld. After break we continue study
Waves and Fields (PHYS 195): Week 7
Spring 2014 v1.0
Intro:
We have a bevy of aspects of waves to discuss, beats, interference in space, intensity and Doppler Shift.
Then it is onto elds! We will study gravitational, electric and magnetic (static) elds, a
Waves and Fields (PHYS 195): Week 9
Spring 2014 v1.5
Intro:
This week we nish our discussion of electric potential and electrostatics before moving on to a bit of
magnetostatics.
Reminder: We have Mid-term II in lab starting April 8. The topics include, p
Waves and Fields (PHYS 195): Solutions 2
Spring 2014 v1.0
Solutions:
(1) A suitably winterized right whale:
(a) F = ma gives kx = md2 x/dt2 which, expressed in standard form is
k
d2 x
d2 x
+ x = 0 or 2 + 2 x = 0
dt2
m
dt
(b) The period of the oscillation
Waves and Fields (PHYS 195): Solutions 7
Spring 2014 v1.0
Solutions:
(1) Fourier series by guess work:
(a) There are a number of dierent games so you might have dierent sketches. Heres the
one I worked with
(b) In all cases your result should be in terms
Waves and Fields (PHYS 195): Solutions 3
Spring 2014 v1.7
Solutions:
(1) Sine approximation
(a) Since all the derivatives are either sine or cosine and since we evaluate at 0, only the cosine
terms survive. Thus
sin(0) 2 cos(0) 3
sin = sin(0) + cos(0)
+
Waves and Fields (PHYS 195): Solutions 8
Spring 2014 v1.0
Solutions: There are a total of 3 possible Bonus points
(1) The intensity falls o as the surface area of the spherical wavefront. So for part (a)
P
2
2
5.97 105 W/m
I=
6 105 W/m
4r2
This is 78 dB.
Waves and Fields (PHYS 195): Solutions 6
Spring 2014 v1.0
Solutions:
2
(1) Practice with partials for u = ex + 2xy 2 .
(a)
2
u
= 2xex + 2y 2
x
(b)
u
= 4xy
y
(c)
2u
2u
=
= 4y
xy
yx
So they are equal, which is an example of the theorem mixed partials are eq
Waves and Fields (PHYS 195): Solutions 1
Spring 2014 v1.0
Solutions:
(1) (Taylor 2.3)
(a) Too many sig gs in the uncertainty. It should be h = 5.03 0.04 m.
(b) A silly number of sig gs in the best estimate. This is more clearly written as t = 1.5 1
s or t
Waves and Fields (PHYS 195): Solutions 4
Spring 2014 v1.8
Solutions:
(1) Phet spring sim: Using the 250 g mass on the third spring (set at 7 on the softness scale) I
observed an extension of 16.8 cm. The equilibrium condition kx = mg then gives
mg
k=
14.6