Step 0:

Draw an accurate diagram showing the positions and magnitudes of all the charges.

Step 1:

Find the magnitude and direction of each individual force vector acting on the charge of interest.

Step 2:

Resolve each individual force vector into its components along the x- and y-directions. (Be careful to distinguish between attractive and repulsive force vectors, and positive and negative components of vectors relative to your x-y coordinate system.)

Step 3:

Add the x-components and y-components separately to find the x- and y-components of the total force vector.

Step 4:

Use the Pythagorean Theorem and trigonometry to find the length and direction of the total force vector .

22-6P
Part (a):

Step 1: Draw a figure showing the location of q1, q2, and +Q. Note that +Q lies between q1 and q2.

Step 2: Make an assumption about the signs of q1 and q2 (both positive, both negative, or one of each).

Step 3: At +Q draw a vector that shows the direction of the force on +Q due to q1. Draw another vector that shows the direction of the force on +Q due to q2. Is there any chance that the two force vectors can be made to cancel each other? If not, change your assumption about the signs of q1 and q2, and redraw those force vectors.

Step 4: You now need to make the two force components equal in length, so that they cancel out. Use Coulomb's Law to find the length of each component (in symbols - not numbers!). Set the two lengths equal! You now have an algebraic equation that you can solve to find the ratio q1/q2.

22-9P (a):

Since the the q and 4q charges are free to move, they want to fly apart. You need a force component on each that will balance the repulsive force. This suggests that a negative charge is needed somewhere between the two positive charges. But the net force on this negative charge also must be zero! Use the latter condition to find the location at which the negative charge should be placed (requires solving a quadratic equation, and choosing the solution that places the negative charge between the two positive charges.) Next use the condition that the net force on one of the positive charges must equal zero to find the magnitude of the negative charge (in terms of q). Show that the net force on the other positive charge also equals zero with these choices for the location and magnitude of the negative charge.

22-9 (b):

To show that the equilibrium is unstable, displace one of the charges slightly and show that net force is in the direction of the displacement....thus the system of three charges tends to fly apart!

23-15P:

This is a fairly tricky problem. Here are a few hints. First, the tension in one of the threads is a force that is directed along the thread and away from the charged ball at the end of the thread. Second, this tension force must just balance the combined effect of the electrostatic and gravitational forces on the ball in order to have equilibrium. Third, the electrostatic force (which is determined from Coulomb's Law) is entirely in the horizontal direction. Fourth, the gravitational force (mg) is directed vertically downward. The vector sum of the electrostatic force component and the the gravitational force component must be along the same line as the tension in the thread if all three force components are going to add up to zero (equilibrium).

From this information deduce that tan(theta), which is approximately equal to sin(theta) = (x/2)/L, also is equal to the ratio of the electrostatic force to the gravitational force. This leads to the answer given.

Chapter 23 Hints:

Click here to view the strategy to use to find electric fields from continuous charge distributions.

23-12P:

Notice that the field vectors from the two +q charges at the point P lie along the hypotenuse of the triangle, but they are oppositely directed so they cancel each other out. Thus, you need only calculate the field from the +2.0q charge.

23-22P:

The field at any point z on the axis of a uniformly charged ring of radius R has been worked out in the text (Eq. 23-16). To find the point(s) where the field is a maximum, differentiate this expression and set the result equal to 0.

23-27P:

The field at any point z on the axis of a uniformly charged disk has been worked out in the text (Eq. 23-26). The quantity in parenthesis in that equation must be equal to 1/2 at the point where the field strength drops to half its value at the center of the disk.

Chapter 24 Hints:

Click here to view part 1 of the strategy for solving Gauss's Law problems.

Click here to view part 2 of the strategy for solving Gauss's Law problems.

24-19P (a):

Use as a Gaussian surface a cylinder concentric with the rod and the shell, and with a radius larger than that of the shell. Note carefully what the total charge per unit length is inside the Gaussian surface.

24-19P (b):

Note that the field within the metal of the cylindrical shell must be zero. Thus all field lines from the charge on the rod must terminate on the inside surface of the shell.

24-19P (c):

Now p lace the cylindrical Gaussian surface between the rod and the shell (and concentric with both).  Note that the only charge inside the Gaussian surface is the charge on the rod.

24-28E:
The circular hole in the plate (from the viewpoint of charge) can be considered to be the superposition of a negatively charged disk on the flat plate.  The electric field at point P then will be the vector sum of the field at that point from the positively charged infinite sheet and the negatively charged disk.  Eq. 23-26 can be used to find the field from the negatively charged disk.

24-31P:

Use the given surface charge density to find E , which is constant. The value of E then can be used to find the force F on the electron, which also will be constant. The work done by this force must be just equal to the electron's initial kinetic energy in order to stop it.

Chapter 25 Hints:

25-8P:

By definition V(r) is equal to the integral of -Edr from the point at which the zero of potential is chosen (in this case the center of the sphere) to a point at radius r. While choosing the zero of potential at the center of the sphere rather than at infinity is a bit unusual, differences in potential between any two points will not be affected by this choice.

25-19P:

Treat the earth as a sphere of radius r. If the electric field at the surface is the same at all points, the equivalent charge distribution creating the field must be spherically symmetric. Use Coulomb's Law to determine how much charge you need to have a field of 100 V/m at the surface of the earth. Once you have this value for the charge, the potential is just that of a point charge of this magnitude at the appropriate radius.

25-25E:

Note that every point on the ring of charge is equidistant from the point P on the axis of the ring. Thus each charge element dq on the ring contributes an amount equal to Kdq/r to the total potential. Since r does not change as you sum over the charge distribution, the integral is very easy to evaluate.

Chapter 26 Hints:

26-15P: Call the distance between two of the plates x. Now the distance between the other two plates will be (a-b-x). Use the formula for two capacitors in series, i.e. C_tot = C₁C₂/(C₁ + C₂) with C₁ and C₂ being parallel plate capacitors with gaps of x and (a-b-x), respectively.

26-17P:

Since the 100 pF capacitor is charged to 50 V initially, you know the total charge on that capacitor from Q=CV. When the two capacitors are connected in parallel, the total charge must remain the same. The charge on the 100 pF capacitor can be found from Q₁ = 100 pF × 35 V. Subtract this from the total charge to find the charge Q₂ on the unknown capacitor, and use that information to find C₂ .

26-27P (a):

The initial stored energy is equal to Q² /(2C). When you connect the second capacitor in parallel the charge distributes equally, so what can you conclude about the size of the second capacitor. Now recalculate the energy stored in each and add the two results.

26-40P:

Think about the orientation of the field lines to justify treating the device as two capacitors in parallel.

Chapter 27 Hints:

27-2P:

Note that in one second all the charge on an area of the belt equal to 0.5 m by 30 m (i.e.) 15 sq. m. is delivered to the terminal of the Van de Graaff, and that corresponds to a current of 100 microamps.

27-11P:

Calculate the drift velocity of the electrons as we did in the example presented in class, then divide the length of the wire by the drift velocity to obtain the transit time.

27-23P:

As long as the wire has constant cross-section V = El, where E is the electric field and l is the length of the wire.  Since J = E/rho, it is easy to find the resistivity.

27-42P:

From the information given for the dimensions of the wire and from the given resistivity, you can find the resistance of the wire.  Then from P = I²R, you can find the current.  From P = IV you can find the voltage drop across the wire.  The current density can be found from the current and the cross-sectional area of the wire.

Chapter 28 Hints:

28-14P:

Since the current is known (50 A), and the voltage across the 12 V battery is measured at 11.4 V you can calculate the internal resistance of the battery from Ohm's Law. Likewise, knowing the voltage drop across the cable (3.0 V) and the current (50 A), you can calculate the resistance of the cable from Ohm's Law. The values you calculate can be compared with the maximum allowable values for these two components. If neither maximum allowable value is exceeded, then the motor must be the culprit.

28-26P:

In (a) note that this is equivalent to two 10 Ohm resistors in parallel with a 5 Ohm resistor.  In (b) you have the equivalent of a 10 Ohm resistor in parallel with a 5 Ohm resistor.  This combination, in turn, is in series with a 5 Ohm resistor; and, the whole combination is in series with a 5 Ohm resistor.

28-49P:

Use i = (EMF/R)e^-t/RC to find the current at t = 1s.  This is the rate at which the charge of the capacitor is increasing.  Energy storage in the capacitor is given by U = q²/2C, so dU/dt = q(dq/dt)/C.  q = CV and dq/dt is just the current at that instant of time.

Chapter 30 Hints:

30-6P:

Note that Fm=qVXB. Write out the cross product in component form noting that the x-component of B is zero. This will allow you to solve for the y- and z-components of B.

30-18P:

As the strip moves through the B-field with velocity v, all the "free" electrons also move with this velocity so each one experiences a magnetic force equal to (-e)vB, since v and B are at right angles to each other. Thus electrons tend to pile up along the right edge of the metal strip. This creates an E-field that points from left to right across the strip, which in turn causes a force on the electrons to the the left. At equilibrium (-e)E = (-e)vB, so E = vB and the potential difference between the left and right side of the strip is V = Ed = vBd, where d is the width of the strip. Since V, B, and d are known, v can be solved for.

30-48P:

Use the formula i(lXB) to find the force on the moving wire. Since the current is constant, the force also is constant. Since the mass of the wire is given, you can calculate the acceleration. The velocity can then be found from the acceleration.

30-59P:

The force on a small segment of the wire is given by dF = i(dlXB), which points inward at an angle theta above the plane of the current loop. By symmetry the components of the force vectors in the plane of the loop cancel. Thus, the net upward force on the loop is obtained by integrating iBdl around the loop (easy) and multiplying by sin(theta).... Note that the current element and the B-field are at right angles to each other!

Chapter 31 Hints:

31-14P (a):

The B-field at point a can be considered to be the sum of three component, each of which points out of the paper. The first component is just that of a semicircle of radius R. You can use the Biot and Savart Law directly to find this field. The second and third components are the fields from the semi-infinite straight wire segments, which from symmetry must just each be half the field of a long straight wire.

31-14P (b):

The B-field at point b is just the sum of the fields from two long straight wires (to a very good approximation).

31-38P:

First note that the forces on the left and right legs of the loop cancel each other (use the right-hand rule to find the direction of the forces on small segments of the loop on each side). The force on the top segment of the loop points towards the wire carrying the 30 A current, while the force on the bottom segment points away from that wire. However, because the B-fields are not the same on the two segments the forces do not cancel.

31-50P (a):

Note that the current density in the wire is i/Pi(a^2 - b^2), where Pi stands for (3.14....). Now apply the hint given in the book.

Chapter 32 Hints:

32-10P (a):

First read sample problem 32-1 to see how to calculate the rate of change of the magnetic flux in the solenoid. Then note that the flux through the coil is the same as the flux in the solenoid, so it changes at the same rate. The emf generated in the coil is then just -N times the rate of change of the magnetic flux. The current that flows in the coil can then be found from Ohm's Law.

32-16P (a):

Since x is much greater than R you can use the value of the magnetic field on the axis of the big loop to find the magnetic flux through the small loop. Use Eq. 31-25 to find the field on the axis of the big loop, substituting x for z. The flux through the small loop is just its area multiplied by the on-axis field.

32-16P (b):

Differentiate the flux with respect to time to find the induced emf, noting that only x is changing with time, and dx/dt = v.

32-30P:

Note that the magnetic flux through the rotating loop is given by N times the dot product of the B-field and the area vector for the loop, so the flux is equal to NBAcos(theta), where theta is the angle the loop makes with the B-field. But theta is changing at a constant rate (w = 2Pi x 1000 x 60 radians per second), so theta = wt and the rate of change of the flux is wNBAsin(wt).

32-36P:

First note that the diagram for this problem is on the following page (page 896). Since B is given by B = 4.0yt^2, the magnetic flux must be obtained by integrating BdA = 4.0yt^2dxdy. This will be a double integral, but the integration over x is trivial, and the integration over y is easy. Note that you can regard t^2 as a fixed parameter when doing the integration over the area of the loop. When you have found the flux, which depends on t, simply differentiate it and use Faraday's Law to find the emf.

32-43P:

Use Faraday's Law and the symmetry of the fields (axial) to find E at each of the points a, b, and c.

Chapter 33 Hints:

33-56P (a):

Note that the same current flows through each inductor, and that the back emf is given by -L(di/dt).

33-5P (b):

Consider the effect of mutual inductance - if any.

33-5P (c):

This should be obvious from parts (a) and (b).

33-22P (a):

Note that the loop theorem requires that the voltage across the inductor at any time be (Vb)(exp(-t/tau)), where Vb is the battery voltage and tau is the time constant. Thus, and easy way to find Vb would be to make a semi-log plot of the data and extrapolate back to t=0.

33-22P (b):

Once you have found Vb, the expression for the voltage across L can be evaluated at a particular time (after t=0). Then take the log of both sides of that expression to solve for tau.

33-36P (a):

Use the information given to find L for the solenoid, and note that the finite (10 ohm) resistance of the solenoid can be considered to be equivalent to a resistance in series with the L of the solenoid. The energy stored is then just (1/2)Li^2, where i is the current flowing at 5.00 ms.

33-36P (b):

The energy supplied by the battery also has gone into Joule heating in the 10 ohm resistance. The i^2R losses have to be integrated from t=0 to t=5.00 ms using the formula for i vs t for the LR circuit.

33-49P (a):

Note that in the orientation shown in the figure the windings of both coils go counter-clockwise looking in from the left. Thus, the fields generated by each coil are in the same direction, so the field lines from one coil link the other coil is the same way that the self-field of that coil does. Hence, the mutual induction adds to the self-induction.

33-51P:

From the way the coil is wound the flux through each of its turns is exactly the same as the flux through the solenoid. Follow the derivation on p. 901 to calculate that flux, then use the basic definition of mutual inductance in terms of number of turns, flux and current to find the mutual inductance of the solenoid and coil.

Chapter 35 Hints:

35-23P:

Since f = 1/2*PI*(LC)^(1/2), you can solve for C as a function of frequency. Then you can use the fact that the dial must cause a linear change in frequence from 2.0x10^5 Hz to 4.0x10^5Hz as the angle rotates from 0 deg to 180 deg (i.e. f = 2.0x10^5(1 + theta/180) Hz), to find what C must be as a function of theta.

35-33P:

Since the maximum energy stored in the capacitor in any one cycle is just q^2/2C, the exponential term in the equation for charge storage in the damped LC oscillator (eqn. 35-17) must be squared and set equal to 1/2. This then can be solved for the corresponding time.

Chapter 37 Hints:

37-4P:

From the cylindrical symmetry of the problem, you can deduced that the magnetic field lines must be circular. From Ampere's Law (with Maxwell's displacement current added), 2PiRB = mu0(id) where id is the "displacement current", which in turn is equal to epsilon0 x d/dt(electric flux). Use the value of the electric field for a parallel plate capacitor to find the flux...note that this must be multiplied by sin(wt) where w = 2 x Pi x f and f = 60/sec.

37-13P (a):

The "displacement current" must be numerically equal to the real current...in this case 2 A, in order for Ampere's law to hold.

37-13P (b):

Use the fact that id = epsilon0 x A x dE/dt to solve for dE/dt.

37-13P (c):

Since the electric field is uniform between the capacitor plates, the "displacement current" in the small rectangle will just be equal to the total id x A1/A where A1 is the area of the small rectangle.

37-13P (d):

From Ampere's law the value of the integral is just mu0 x id x A1/A, where id is the total "displacement current".

37-17P (a):

The proof follows from the fact that the line integral around path abcdefa is equal to the sum of the line integrals around paths abefa and bcdeb, and that the total flux through abcdefa is just the sum of the fluxes through abeda and bcdeb. The first result can be shown by noting that segment be is traversed in opposite directions on the two smaller paths. The proof for part (b) is similar.

Chapter 38 Hints:

38-14P (a):

This follows by direct substitution of E0 x sin(wt-kx) and B0 x sin(wt-kx) in the corresponding wave equations, provided w/k = c^2.

38-14P (b):

In this case you must apply the "chain rule" for derivatives symbolically since you don't know what the functional form of f is.

38-26P:

In class we showed that the instantaneous energy stored in the E-field is equal to the instantaneous energy stored in the B-field. Since the two fields are coupled, they must have the same time dependence. Thus the time-averaged energy densities also are equal.

38-39P (a):

lamda x f = c, so knowledge of the wavelength gives the frequency and vice-versa for an electromagnetic wave.

38-39P (b):

The B-field must be perpendicular to the E-field so must be in the z-direction. Its magnitude is (300 V/m)/c.

38-39P (c):

k = 2 x Pi/lamda, w = 2 x Pi x f (see part a).

38-39P (d):

The time-averaged value of Poynting's vector is given by (1/2)Em x Bm/mu0.

38-39 (e):

The rate at which momentum is delivered is just the force exerted on the surface (of area 2 m^2). The force is just the radiation pressure x the area, so F = IA/c, where I is the time averaged value of the Poynting's vector from part d (also called the intensity I), and A is 2 m^2. The radiation pressure is just F/A.