Homework problems are taken from the 7th edition of Serway and Jewett text. 

The due dates/day for the homework, along with the problem numbers, are listed below.  Problems in bold red are graded in detail out of 5 points each.  Problems in green italics are not due. You are encouraged to attempt these problems before the solutions are put up. The rest of the problems are graded out of 2 points.  An almost correct solution or a valiant attempt gets 2 points while a decent attempt with the right method and equations and some progress, gets 1 point. The starred problems have extra questions associated with them. Look below the table for the extra questions associated with the starred problems.

Please make sure that all answers have the proper units. And refer to the homework instructions for guidelines to be used while solving these problems


Hw #





Week 1


July 14






July 15





July 16


CQ1, HW#1 16.2, 16.7, 16.13, 16.17 16.24, 16.31, 16.38, 16.37, 16.28, 16.46



July 17


CQ2, HW#2 17.12, 17.11, 17.17, 17.23, 17.22, 17.30, 17.39, 17.33, 17.35, 17.53



July 18


CQ3, HW#3 18.5, 18.7, 18.13, 18.15, 18.19, 18.22, 18.35, 18.31, 18.37, 18.44, 18.45






Week 2


July 21


CQ4, HW#4 23.2, 23.3, 23.7, 23.8, 23.10, 23.12, 23.14, 23.17, 23.21, 23.26, 23.27



July 22


23.37, 23.33, 23.45, 23.59, 23.55, 23.53, 23.40, 24.4, 24.9



July 23


CQ5, HW#6 24.6,24.13, 24.17,24.20, 24.21, 24.23, 24.47,



July 24


HW#7 24.50, 24.34


July 25


CQ25, HW#8

24.27, 24.39, 25.4, 25.5, 25.8, 25.11, 25.14, 25.19, 25.35, 25.36

Week 3

  July 28


  Turn in # 7 and 8


July 29 Tuesday CQ6, HW#9 25.32, 25.10, 25.15, 25.31, 25.39, 25.40, 25.44, 25.59, 25.60, 25.63
July 30 Wednesday HW#10 26.5, 26.6, 26.7, ,26.9, 26.17, 26.19, 26.23,
July 31 Thursday HW#11 26.51, 26.50, 26.56, 26.41, 26.37, 26.25, 26.62, 26.22


Aug 01 Friday CQ8, HW#12 27.6, 27.7, 27.10, 27.20, 27.21, 27.26,27.31, 27.23, 27.30,27.52,27.57,
Week 4 13 Aug 04
CQ9,HW#13 28.5, 28.7,28.9, 28.9, 28.13, 28.14, 28.23, 28.25, 28.17,28.16
Aug 05
HW#14 28.28, 28.29, 28.31, 28.32, 28.19, 28.50, 28.55, 28.56,28.33,28.35
Aug 06
HW#15 29.1, 29.4, 29.5, 29.7, 29.11, 29.15, 29.22, 29.20, 29.21
    Aug 07
Exam No homework
Aug 08
HW#16, CQ29 29.25, 29.27, 29.29, 29.31, 29.32, 29.35, 29.38, 29.39, 29.41, 29.42, 29.53, 29.56
Week 5
Aug 11

CQ30, HW#17

29.58, 30.2, 30.4, 30.5, 30.7, 30.8, 30.10, 30.23, 30.25, 30.21, 30.11, 30.15, 30.63
Aug 12
CQ34, HW#18 34.1, 34.5, 34.9, 34.12, 34.18, 34.23, 34.30, 34.33, 34.31
  19 Aug 13
CQ31, HW#19 31.2, 31.5, 31.8, 31.9, 31.15, 31.30, 31.28, 31.13, 31.48,
Aug 14
HW#20 31.49, 31.36, 31.37, 31.61, 31.62, 32.12, 32.16, 32.13
Aug 15
Week 6!
CQ35, HW#21 35.5, 35.9, 35.16, 35.21, 35.19, 35.36, 35.37, 35.49, 35.53, 35.51
CQ36, HW#22 36.7, 36.9, 36.13, 36.15, 36.17, 36.22, 36.24, 36.29, 36.33, 36.35, 35.54, 35.64
CQ37, HW#23 37.10, 37.9, 37.18, 37.21, 37.31, 37.33, 37.28, 37.25, 37.58,
CQ38, HW#24 38.3, 38.7, 38.19, 38.23, 38.40, 38.36, 38.41, 38.21
      REVIEW   16.7, 16.17, 16.31, 17.35, 17.12, 17.22, 18.13,18.35, 18.31,18.37, 18.45, 23.8, 23.12, 23.45, 23.59, 24.50, 24.39, 25.35, 25.36, 25.19, 25.59, 26.22, 26.41, 26.50, 28.11, 28.7, 28.13, 29.25, 29.58, 29.29,30.5, 30.4, 30.7, 30.10, 30.63, 31.8, 31.28, 31.30, 31.49, 31.62, 31.37




Answers to Even numbered problems

HW 1  Chapter 16
16.2  No numerical answer. Just graph      16.24 T= 631 N          16.28 q= 41.4o
16.38 P= 15.1 W, E= 3.02 J
16.46 a)  a_max= 375 m/s^2  b) F_max= 0.045 N , Tension= 46.9 N

HW 2 Chapter 17
17.12  a)  change in Pmax=1.27Pa         b) f=170 Hz   c)   wavelength= 2.00m        c)v=340m/s
17.22 No Numerical answer( just derive it)
17.30    P=0.123 w

HW4 Chapter 23
23.2  2.38 electrons
23.8 0.634d 
23.10 a T= Pi/2 Sqrt(md^3/kqQ)
23.12 d= 1.82m to the left of -2.50 charge
23.26 derivation to show E is like k*Q/r^2

23.40 (a)  36.9 deg,  53.1 deg
         (b)  167 nanosec, 221 nanose

24.6 (a) –55.7 nC
(b) The NEG charge has a spherically symmetric distribution, concentric with the shell.
24.20 (a)16.2 * 10^6 Newton/Coulomb, directed towards the filament
        (b)8.09 * 10^6 Newton/Coulomb, directed towards the filament
         (c)1.62 * 10^6 Newton/Coulomb, directed towards the filament

Chapter 25    Homework 9
25.10a) F=0N  b) E=0 N/C  c) V=45.0 kV
25.32 a )   EA>EB   b) EB=200N/C down   c) to the right
25.40 1.56 E 12 electrons
25.44 Derive it.

26.6 C =((2N-1)( Pi-Theta)*R^2)/d

Chapter 26   Homework 11
26.22 C= C0*( Sqrt(3)-1)/2
26.50b) 40 micro F
         c)         For the 50 micro F,  delta_V = 6.00 V  and Q =  30 micro C.
      For 30 micro F , delta_V =4.00 V.
      For 20 micro F  and 40 micro F , delta_V = 2.00 V.
      For 20 micro F,  Q =  40.0 micro C.
                      For 40 micro F,  Q = 80.0 micro C.
26.56 a) C= (epsilon)( L^2-L*x(k-1)/(d)
          b) U=Q^2 d/(2 epsilon* (L^2 -L*x(k-1))
          c) F=Q^2*d(k-1)/( 2 epsilon L^3 *k^2) in x   direction
          d) F=250 micro Newton in x direction
26.62 Cab=3.00 micro farad

Chapter 27( HW 12)
27.6    0.265C
27.10   0.130 mm/s
27.20  She can meet the design goal by choosing l1 = 0.898 m  and l2 = 26.2 m
27.26 (a) 17.3 A   (b) 22.4 MJ   (c)  $0.995
27.30 (a) 5.97 V/m    (b) 74.6 W    (c) 66.1 W

#14)     a)         V4>V3>V1>V2
            b)         V1=E/3,   V2=2E/9,   V3=4E/9,   V4=2E/3
             c)         I1>I4>I2=I3
d)         I1=I         I2=I3=I/3       I4 = 2I/3
             e) I4­­ increases while all the others decrease
f)         I1 = 3I/4              I2=I3=0         I4 = 3I/4

#16)     I1 = 0.714 Amperes,  I2 = 1.29 Amperes, E = 12.6 Volts

HW #14 
P28.28)   587 KiloOhms
P28.32)   a) 1.5 sec     b)  1.0 sec     c)  200 microA + 100 microA (e^(-t/1 sec)) where the 200 microAmps comes from the battery and the 100 comes from capacitor
P28.50)   a)  (28.9 Ohm –0.542R)microC/(2 Ohm +R)

  1. 1.96 microCoulomb
  2. YES;  R =53.3 Ohm
  3. 14.5 microCoulomb
  4. YES, it would be like cutting the wire that has R in it.  In that case Q = 0.542 microC.     

P28.56)   a)  222 microCoulombs

               b) 444 microCoulombs

Chapter29 HW16
32)     Fab = 0,  Fbc = 40 mN in neg. x direction, Fcd = 40 mn in neg. z direction, Fda = 40 mn in (x+z) direction, which is diagonal along x-z plane.   The forces on the four segments must sum up to zero, so one only needs to solve for three of the segments.
42)  (a) 37.7 mT    (b) 4.29*10^25 charge carriers per cubic meter, which is a lot.
56)   [lambda*g*Tan(theta)] over I

31.2 0.008mA
31.8 (a) (mu-oIL/2pi) ln(1 + w/h)    (b) – 4.80 microV; current is counterclockwise

31.28(a) to the right     (b) out of the plane of the paper  (c) to the right  (d) into the paper

31.30 Negative
31.48.3 mA down through 6.00 W, 860 mA down through 5.00 W, 923 mA up through 3.00 W



HW 1 Chapter 16 
16.7 Formula application
16.13 Simple extraction of parameters by comparison
16.17 This is the transverse speed you have to find, not wave velocity v. You know the displacement expression y(x,t). Can you find speed  and acceleration?
16.24 Simple formula application
16.31   We know the expression for μ = mL. Can you express m in terms of density ρ,  and volume V. Volume can be connected to length L and diameter of the wire? (wire is a cylinder).
16.38 Simple parameter extraction and formula application
16.28 Draw the freebody diagram of the point in between . Downward force is mg.  Solve for the tension in each rope. Will they be equal? You know the velocity. Can you relate the two?
16.46 similar to 16.17.

HW 2 Chapter 17
17.11 simple extraction of parameters by comparison,substitution and determine velocity from the displacement
17.12 simple extraction of parameters by comparison
17.17 Formula application
17.23 Formula application to find intensities and use total intensities to get the sound level.( don't simply add the sound level)
17.22 first determine how the intensity depends on distance and use simple algebra to derive it.
17.39 first determine the angle and use simple trigonometry to determine the horizontal distance .
17.33 make sure to get the correct sign of velocities for the observer and source for each case.
as discussed in class.
17.35 same as problem 17.33 and solve for velocity of source( velocity of observer is zero).
17.53  use the velocities of the observer and source relative to the water and can you determine the sign of the
of the velocities when the observer is moving in front and same direction? If object A has as velocity V_A with respect to the shore, and object B (here water) has velocity of V_B with respect to the shore, then the velocity of A with respect to B is V_A-V_B. The sign of each V_A and V_B is important (i.e., if they are + or -)

HW3 Chapter 18
18.5 can you determine  the form of the resultant wave? Extract the amplitude from general comparison
18.7 Relate the difference of the distance traveled  by sound from two speakers with the minimum  ( note minimum occurs for the odd multiple of half wavelength)
18.13 The speakers produce standing wave between them. Can you locate the nodes and anti-nodes?
18.15 First determine resultant wave equation. See that you get consistent answer for part b ,c and d
18.22 speed is the same for concert G and A. From this find the difference of effective length for G and       A.    make the first derivative of tension with respect to length to get the percentage change.
18.31 Simple formula application and solve for L
18.35 The resonant frequencies are equally spaced. Can you find list of frequencies including    fundamental frequency from this?  See how they are related to each other.
18.37 For the tube open at one end , the successive frequencies are multiple of odd number.
18.45 use the Doppler effect to find expression for the echo frequency. What is the sign of the velocities of student and reflected sound when the student moves away from wall?

Hints: HW 4 Chapter 23
23.2        Straightforward arithmetic
23.3        Drop the sig figs, just use exponents of 10
23.7        Break into vector components Fx  and F­y
23.8        Stable Equilibrium means it returns to equilibrium if it is slightly displaced.  You can explain this qualitatively.  Note:  “qualitative” does not mean “hand-waving”, you must use formulas to back up your explanation.
23.10     Restoring force => Simple Harmonic Motion.  Make a small-angle approximation to calculate the restoring force.
23.14 Similar to 23.7
23.17 Electric fields superimpose; for example, light rays pass right through each other without being altered, as evidenced by starlight.
Same integral as in example 23.6, only sideways
23.26Hint is given in problem
23.27 IIntegrate around the half circle, but you can use symmetry in the y-direction (not the x-direction).   Try  dq = lambda * r * d theta

Homework 5

23.33 Make a large drawing; don’t be afraid to use up half a sheet of paper for this sketch, it’s for credit
23.37 Read section 23.7
23.40 Same as a gravity/trajectory problem from Physics 141, only this time calculate the acceleration due to the E-field
23.45 Draw a free-body diagram with three different forces present: elec, mech, grav.
23.53 See the sol’n to problem 27 from the last hw.  The only difference is that here the charge density lambda varies along the length of the rod.
23.54 Again there are three forces to contend with:  electrical, mechanical (string tension) and gravitational.
23.59 Lots of algebra may be skipped by using lots of symmetry.

24.4 You may do it the long way, or just use Gauss’ law.
24.9 Imagine this picture in 3-D

24.6 Given E and r, find Q
24.13 The size L of the cube is totally irrelevant
24.17 a)Again, the size of the cube is irrelevant
         b)No calculation required, just think about the geometry
24.20 Apply Gauss’ Law
24.21 Use pillbox surface like the one in figure 24.13
24.22 use a couple of formulas, some Newtonian, some electrostatic
24.47 As you move from r = 0, to r = a, to r = b, c, etc.  the  E field changes continually.  Use Gauss’s Law at each stage to determine E.


25.15 calculate the electric fields from three charges. Which fields cancels out?
25.31 Simple formula application.
25.32 Electric field is gradient of potential difference. Is the electric field stronger when the distance
between the equipotential lines is closer? Note : the electric field is orthogonal to  equipotential lines.
25.39 first calculate the potential for the concentric spheres. Potential difference is scalar quantity. Take
gradient to get the electric field.
25.40 Determine the charge that produces the given potential. What is the charge of single electron?
25.44 Delta_V= -|E||d|cos(theta). Find an expression for field outside a cylindrical conductor (refer chapter 24, or its homework) and integrate it along radius to find potential difference using formula above. We have done it in class!
25.59 From the potential difference of the ring you can get the potential energy of the charge at point Q. use conservation of energy to get the speed of charge at infinity. The potential energy is zero at infinity.
25.63 take small ring of radius r and width dr.  what is the total charge of the ring?  Use this to calculate         potential of this thin ring. Integrate it to get  the potential of the disk. Note: charge is not constant. Follow example 23.8 and 23.7 but derive electric potential instead of electric field. Electric potential is simpler since it is not a vector. Once you get the expression, it is easy to set up the integral.

Chapter 26
26.5 Simple formula application
26.6 First determine the effective overlapping area. Use the formula for parallel plate capacitor to get  capacitance . Once you do that the you can get the equivalent capacitance of 2N-1 , which are in parallel capacitor
27.7 Simple formula application. Charge density is ratio of total charge to total area
26.9 Refer Example 26.1 to get the capacitor for Coaxial cylindrical cable. Delta_V= Q/C
26.17 Identify parallel and series connection and then find effective capacitance. The total charge is same for series connection. From this get the potential drop for each capacitor. For parallel connection, the potential drop is same.
26.19 First determine the total charge on C1 when S1 is closed( Q= V C1).  The total charge remains same after S2 is closed. The potential drop is same for C1 and C2, from this calculate Q1 and Q2
26.23 First identify series capacitor and determine effective capacitance. And then determine the effective capacitance of reduced parallel connection. At the end you get two capacitor connected in series, you can easily reduce this to get the over all effective capacitance b/n point a and b

26.22 The equivalent Capacitance b/n X and Y is same as A and B. Redraw the circuit replacing the infinite connection by equivalent capacitance C. And then solve for C
26.37 simple formula application
26.41 which capacitor break down first? Reduce the circuit to series circuit. The charge ll be same for series connection.
26.50 Simple formula application. Solve for C1
26.51 simple formula application( take the ratio of Emax and charge Q..find expression for area A)
26.56First determine the area filled with dielectric and also unfilled one. Calculate the equivalent capacitance. Are they connected in parallel? Why? Use this to get the total energy. Force is negative first  derivative of total energy.
26.62 From symmetry the potential difference across 8 micro farad is zero.

Homework # 12
27.6 Simple formula application.
27.7 Simple formula application. Note : total charge is same Avogadro's number times charge of an electron.
27.10 First determine the number of mobile charge per unit volume for aluminum(n). To get this compute mass of single aluminum atom. How is  n related to density and mass of single atom? Then just use formula to get  drift velocity.
27.20 Try to get two equations. First equation: how  do R1 and R2 depend on L1 and L2? For second equation : how do R1 and R2 depend on L1 and L2  for delta _temperature(20 C)?  Refer to Table 27.2 for resistivity and temperature coefficient. Carbon has negative temperature coefficient. Main idea is as temperature increases resistance of one material increases while the other decrease. Then just solve for L1 and L2
27.21Simple formula application
27.23 Simple formula application.
27.30 Simple formula application. How does the resistance depend on temperature?
27.31Simple formula application and conversion of units. Total potential energy stored in the battery is charge multiplied by potential difference.
27.57 Take wedge of infinitesimal length dx. What is the cross-sectional area of sliced wedge? Find how the length of cross sectional area (y) depends on length of the wedge. This is the same as finding equation of slanted  top side of wedge. It is just linear function. Just integrate it to get total resistance. Reason out why you integrate it.

HW #13 Hints

#5)       break the circuit up into smaller components then use V = I/R
#7)       Same as for #5 above
#9)       The entire current must pass through the first resistor
#13)     Lots of parallels
#14)     Follow the current flow
#16)     Kirchoff’s Laws (pronounced like in football, “Kick-Off”)
#17)     More fun with Kirchoff’s Laws        
#23)     Kirchoff again
#25)     Traverse the loops and calculate the voltage drops

HW #14 Hints
P28.19)   Use the results from p28.17
P28.28)   Use the formula for voltage of a capacitor as it charges up
P28.32)   time constant = RC.  For c) find the current through the battery and the 100 kOhm resistor, add them together
P28.50)   Once the capacitor is in equilibrium (fully charged) there is no current in that branch so you just have parallel circuits.  First find all the voltages, with R the only unknown. 
P28.55)   Power is 60 Watts; use the eqn relating power, voltage and resistance.
P28.56)   Where does the current flow when the switch is closed?  When it’s open? 

P29.25)   Same as Experiment 6 from Laboratory
P29.29)   The force on the rod is independent of its motion.  Work = Force x Distance.  Use the Work-Energy theorem, where you account for both translational and rotational, initial and final.  This is really a PHYS 141 problem.
P29.31)   Cross-products
P29.32)    First find the magnetic moment “mu” of the coil
P29.35)    Start by calculating just the lengths of the sides for each geometry
P29.42)   Straightforward formula application; refer to section 29.6
P29.53)   Don’t forget there are two springs in parallel; this is mostly a PHYS 141 problem
P29.56)    Draw a free body diagram including the tension in the wires; another PHYS 141 problem

31.2 find induced voltage form Faraday’s law.  I= V/R
31.5 simple formula applications
31.8 First find magnetic field of the long wire inside rectangle. From that find flux Note that the magnetic field is not uniform. Integrate:  B(r) is constant for strip of small rectangle of width dr.
31.9 First find rate of change of magnetic flux inside the ring. Then use Faraday’s Law
31.13simple Faraday’s law
31.15 first determine magnetic flux and use Faraday’s law. What is direction of emf?
31.28 for Lenz’s law induced magnetic field opposes the change in external magnetic filed.
31.30 use Lenz’s law.  Does field increases as magnet moves closer? How does induced field react to this?
31.48 find induced Voltage from the two infinite long solenoids using Faraday’s Law.  Then use loop rule and Kirchhoff’s junction rule to find current in each resistor

HW r