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 Table 37-3 (938 106 eV) and the hc = 1240 eV nm va
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 wavelength , period , and maximum amplitude . A sna
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 10-6 C Q2 U= = = 117 10-6 J. . -6 2C 2 3.60 10 F
c
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. The angular position of the rotating coil is mea
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 the force of friction, but it can be oriented so
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 equilibrium charge is qeq = C. We require q = 0.9
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 ) =
0
( 84.5 pF )( 40.0 mm - 38.0 mm ) = 191cm 2 .
(
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 > R, qenc = , so E (r ) = / 2 r 0 . With = 2.00 10
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 are uniformly distributed over the surface. Far away
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 rest relative to the muon. Specifically, t0 = 2.2
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 sides must be negative and exactly cancel the tota
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 45 angle relative to the wall. By the law of refl
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 most sensitive is about 555 nm. (d) Using the result
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 light speed when traveling through the glass is
v =
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 = hc/, where h is the Planck constant, f is the f
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 (a), both spheres have charge q, Q1 = Q2 = q (1 pts) (1)