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 bat
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 poin
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 =
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
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 ( )|
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 wav
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
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
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
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 hen
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 cons
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
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 ) =
1
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
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
dens
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 =
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 f
1
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 hom
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 EXPERIME
1
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 thin
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 t
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 b