4. With mp = 1.67 10 27 kg, we obtain
( 6.63 10 J.s ) h2 E1 = n2 = 2 8mL2 8m p (100 1012 m )
-34 2
(1)
2
= 3.29 10-21 J = 0.0206 eV.
Alternatively, we can use the mc2 value for a proton from Ta
Yale University Department of Physics Physics 181 University Physics, Spring 2011 Assignment #1 Solutions by Camille Avestruz
Problem 1: Isolated conducting spheres. HRW 21.2.
Figure 1: Problem 1 In (
Rework for Practice Introduction to Two-Source Interference Learning Goal: To gain an understanding of constructive and destructive interference. Consider two sinusoidal waves (1 and 2) of identical w
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then zero. If Q is the maximum charge on the capacitor, then the total energy is
2.90 10-
11. (a) It should be emphasized that the result, given in terms of sin(2 ft), could as easily be given in terms of cos(2 ft) or even cos(2 ft + ) where is a phase constant as discussed in Chapter 15.
12. We use Eq. 28-12 to solve for V: V=
( 23A )( 0.65 T ) iB = = 7.4 10-6 V. 28 3 -19 nle ( 8.47 10 m ) (150 m ) (1.6 10 C )
37. (a) The magnetic force must push horizontally on the rod to overcome
45. During charging, the charge on the positive plate of the capacitor is given by
q = C 1 - e - t ,
c
h
where C is the capacitance, is applied emf, and = RC is the capacitive time constant. The
6. (a) We use Eq. 25-17:
C = 4 0 40.0 mm 38.0 mm ab = = 84.5 pF. 2 b-a 8.99 109 NCm 40.0 mm - 38.0 mm 2
b
gb
g
d
ib
g
(b) Let the area required be A. Then C = 0A/(b a), or
A= C (b - a ) =
20. We imagine a cylindrical Gaussian surface A of radius r and unit length concentric q E dA = 2rE = enc . with the metal tube. Then by symmetry
A
0
(a) For r < R, qenc = 0, so E = 0. (b) For r >
3. The following diagram is an edge view of the disk and shows the field lines above it. Near the disk, the lines are perpendicular to the surface and since the disk is uniformly charged, the lines ar
1. From the time dilation equation t = t0 (where t0 is the proper time interval,
= 1 / 1 - 2 , and = v/c), we obtain
= 1-
FG t IJ . H t K
2 0
The proper time interval is measured by a clock at
1. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the
4. When S is barely able to see B the light rays from B must reflect to S off the edge of the mirror. The angle of reflection in this case is 45, since a line drawn from S to the mirror's edge makes a
2. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm, (b) and the larger wavelength to be approximately 610 nm. (c) From Fig. 33-2 the wavelength at which the eye is mos
2. (a) The frequency of yellow sodium light is c 2.998 108 m s f = = = 5.09 1014 Hz. -9 589 10 m (b) When traveling through the glass, its wavelength is
n = 589 nm = = 388 nm. n 152 .
(c) The li
3. Let R be the rate of photon emission (number of photons emitted per unit time) of the Sun and let E be the energy of a single photon. Then the power output of the Sun is given by P = RE. Now E = hf
Physics 123/5
z
z
You can find this lecture note at
http:/www.phys.uconn.edu/~kjoo/p123-5-Lecture1.pdf
If you have any question about this lecture, contact me at
[email protected]
Physics 123/5, Pg