1
FINALS May 10, 2010 Physics 201 Shankar 180 MINS
1. At t = 0 when the capacitor is uncharged the switch S in Fig.1 is closed. (i) What are
I (t = 0+ ) and I (t = ), the currents owing out of the battery just after the switch
is closed and at t = ? (ii)
Solutions to PS 4 Physics 201
1. (a) Below is a labelled diagram of the situation. We place the center of the sphere
which we are averaging over at the origin.
q
r'
Q
r
R
The potential due to the point charge q at a point on the surface of the sphere is
g
Solutions to PS 5 Physics 201
1. Force acting small part dl of the wire carrying current is given by
F = I dl B.
By integrating this over the entire loop, the total force is obtained:
Ftot = I dl B = I ( dl) B = 0.
(1)
(2)
Suppose two wires which draw two
Solutions to PS 6 Physics 401a
1. Let B point along the z axis. Then by circuar symmetry, we have
B(r, t) = B (r, t)ez
Using Faradays Law, we can nd the electric eld at radius r to be
d
E(r, t) dl = B(r, t) dA
dt
d 2 r
2 rE (r, t) =
B (r, t)rdrd
dt 0
0
(
Solutions to PS 7 Physics 201
1. The impedance of the circuit is given by
1
+ iL
iC
1
= R + i( L
).
C
(1)
Z ( ) = R +
(2)
Noting the relation between the amplitudes, |I | = |V |/|Z |, we have
|I ( )|
R
|I ( )|
=
=
|Imax |
|I (0 )|
R 2 + ( L
12
)
C
.
(3)
Solutions to PS 8 Physics 201
1. (a)
= kc = 6 109
rad
s
(1)
(b)
= 9.55 108 Hz
(2)
2
(c) Since the argument of the sin function is of the form ky + t, we know that
f=
k = 20j
(3)
and therefore the wave is propagating in the y direction
(d)
1
B = j k1000 s
Solutions to PS 9 Physics 201
1. Look at the gure. The answer is h/2.
h/2
h
FIG. 1:
2. By applying the Mirror Formula for concave mirrors, we have 1/u + 1/v = 1/f . To
have u = v , we need u = v = 2f = 60 cm.
3. (i) The position of the ball is given by zb
Solutions to PS 10 Physics 201
1. For a screen very far away from the aperture, the rays from each of the N slits emerge
at approximately the same angle. Thus, from the diagram below, we see that the path
length dierence between two adjacent slits is give
Solutions to PS 11 Physics 201
1. (i) The total probabily of the particle being found in the region of L/2 x L/2
should be 1. That is,
P (L/2 x L/2) =
L/2
5
| (x)|2 dx = A2 a + (A/2)2 a = A2 a = 1.
4
L/2
(1)
Therefore, we have
2
A= .
5a
(2)
(ii) Using the
Solutions to PS 12 Physics 201
1.
1 ipx
eh
L
1 ipx
p (x) = e h
L
p (x) =
(1)
(2)
= p (x)
(3)
p (x) (x)dx
(4)
p (x) (x)dx
=
p (x) (x)dx
(5)
Thus, we have, since (x) is real
Ap =
=
= A p
(6)
(7)
and hence
P (p) = Ap A
p
(8)
= A p Ap
(9)
= P (p)
(10)
2. For
Solutions to PS 3 Physics 201
1.
(x2 y )
y
=
x3
()
x 3
= x2 . That is,
Fx
y
=
Fy
.
x
Therefore, F can be written in the form
of F= U (x, y ) with some function U (x, y ), which means that F is conservative.
3
From U = x2 y , U = x2 y dx = 1 x3 y + C (y
Solutions to PS 2 Physics 201
1.
ke dq
r2
L
ke 0 xdx
=i
2
L L(x0 x)
ke 0 L (x x0 )dx
x0 dx
=i
+
2
L L (x x0 )
(x x0 )2
x0 L
x0
x0
k e 0
ln
+
=i
L
x0 + L
x0 L x0 + L
k e 0
x0 L
2x0 L
=i
ln
+2
L
x0 + L
x 0 L2
E=i
(1)
(2)
(3)
(4)
(5)
To nd the eld for x0 , w
Physics 201
Final Exam
Solutions
I.
i) The condition at equilibrium is
X
Qq 1
F=
mg = 0 = y0 =
40 y 2
r
Qq
.
40 mg
ii) To nd the spring constant due to the electric force, we write y = y0 + (y y0 ) = y0 + y where y
is the displacement from equilibrium an
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Midterm Physics 201 March 1, 2010 R.Shankar 75 mins
Questions 1-5 BOOK I, 6-9 BOOK II
1. From a sphere of radius R and charge density , I scoop out a sphere of radius R/2 as
shown in Fig.1. Find the electric eld at the center of the big sphere. 5
R
2R
F
Solutions to Midterm Physics 201
1. We can consider this situation as a superposition of a uniformly charged sphere of
charge density and radius R, and a second uniformly charged sphere of charge
density and radius
R
2
at the position of the cavity. At th
1
Faraday and Lenz: A loop in two frames R.Shankar February 17,
2010 .
I would like to clarify in some detail how it all works out for the loop in two frames.
The equation we need is
E=
(E + v B) dl =
d
B
v B dl
=
dA +
dt
S t
S
(1)
In the Lab frame, (pa
1
Geometrical Optics R.Shankar
Here are some notes on optics that you may not nd anywhere else. Please refer to Fig.
1 that shows a parabolic mirror with
y 2 = 4xf.
(1)
We know it can focus all rays from innity to the focal point F no matter at what y the
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Notes on Images and Potential Problems Physics 201b Spring
2010 Shankar
Here are some notes on solving for V using images. This supplementary material is are for
those who want more details. The homework problems and exams will be at the a much
simpler
PHYSICS 201b Quantum notes R. Shankar 2010
These notes, possibly containing some bugs were for students of Physics 201b.
They may not be reproduced for commercial purposes.
I.
THE DOUBLE SLIT EXPERIMENT
I am going to begin with the double slit experiment
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Quantum Cook Book April 14.
The basic ideas of quantum mechanics are very simple and sometimes get hidden in the
use of dierential equations and complex functions that arise when we do the real thing, as
in class. Here is a bare bone version that illust
1
Solutions to Problem Set 1 Physics 201b January 13, 2010.
1
1. (i) To produce a sphere with 8 C you need to rst let the two spheres touch each
other so that the charges redistribute equally on the two spheres. Now each sphere has
a charge of 1 C and gro
Solutions to PS 13 Physics 201
1. By plugging in the assumed form to the equation, we get
d2 (x)
1 d2 F (t)
F (t) = 2
(x).
dx2
v dt2
(1)
1 1 d2 F (t)
1 d2 (x)
=2
.
(x) dx2
v F (t) dt2
(2)
Dividing by F (t) (x),
The left hand side of this equation is a f