Readings (PH213)

Just the reading questions for PH213. The practice problems are still at

Questions, comments, and suggestions, email me.

Works best in Opera: the fastest and most secure browser ever!

Home< PH213 | PH 212 | Equations | Readings

Equations and Formulas

By number (in numerical order): Show

By topic (in alphabetical order): Show

It was suggested that for future reference, I put the Reading Questions here with answers.

A. Apr 3rd: 3. Why can electric field lines never cross?

Answer: The definition of the electric field at a point is the net force a positive test charge placed at that point would experience divided by it’s charge (21-6, page 554), and the definition of the net force is all the forces combined. So if two field lines cross, that would mean a test charge placed at the intersection would experience forces in two different directions, in which case the sum of these two forces—the net force—would be in neither of these directions. But then that would mean the electric field wouldn’t be in either of these directions, and since field lines represent the electric field at every point along it’s length, if there’s a point where the field lines do not represent the direction of the electric field (ie the intersection, where we just established that the direction of the electric field would be tangent to neither of the field lines), those two crossing lines must not be field lines. Hence, now that we’ve proven that any two crossing lines must not be field lines, this implies that field lines never cross (same logic as "dead people can’t breath, so someone breathing can’t be dead").

Back to Top - Equations and Formulas - Email Me

B. Apr 8th: 1. What can you say about the flux through a closed surface that encloses an electric dipole?

Answer: Zero, or more correctly, there is no flux through such a closed surface. The flux, by Gauss’s Law, must be equal to Qencl/∈o(22-2, page 578). What is Qencl? Zero, since the definition of a dipole is a net neutral object that has a separation of charges (21-11, page 565). Does the placing of the charges have to do with anything? No, so we ignore it and observe that since the the dipole has a net charge of zero, Qencl/∈o=0 and hence the flux is also zero.

2. A conductor carries a net positive charge Q. There is a hollow cavity within the conductor, at whose center is a negative charge -q. What is the charge on (a) the outer surface of the conductor and (b) the inner surface of the conductor?

Answer: This is almost exactly the same as Conceptual Example 22-8, page 585. Remember why, according to Coulomb, an electric field cannot exist within a conductor? The mobile charge carriers would move until they neutralized the field (21-9, page 562). If there’s a +q charge inside a cavity, how would they neutralize the field? -q’s worth of charge carriers would go to the surface of the cavity, and by conservation of charge (21-1, page 546, though I can’t imagine anyone really needing to read up on that), +q’s worth of charge carriers would goto the surface, hence (a) the outer surface has Q+q and (b) the inner surface has +q.

Back to Top - Equations and Formulas - Email Me

C. Apr 10th: 1. A cube has sides of length L. On one face of the cube, the electric field is uniform with magnitude E and has a direction pointing directly into the cube. The total charge on and within the cube is zero. Which statements are necessarily true?
A. The electric field on the opposite face of the box has magnitude E and points directly out of the face.
B. The total flux through the remaining five faces is out of the box and equal to EL².
C. At least one other face of the cube must have an inward flux.
D. None of the previous statements must be true.

Answer: Only B is necessarily true, although this does not preclude A or C from being true. Firstly, why is B necessarily true? Gauss’s Law. Because the total enclosed charge as given in the problem is zero, the net flux through the cube must be zero as well. What is the flux through that one face of the cube? Φ=E⋅A=EA⊥=-EL². Since the net flux must be zero, the total flux out of the remaining five faces must be +EL².

Back to Top - Equations and Formulas - Email Me

D. Apr 15th: 1. Electric field lines are perpendicular to equipotential lines. Why?

Answer: Any motion of a test charge with a nonzero component parallel to an electric field line will either speed up or slow down due to the force from the electric field, and if the velocity changes, the kinetic energy changes, which implies that work is being done on or by the test charge, which in turn means the potential energy is decreasing or increasing, ultimately meaning that the potential isn't equal. So it follows that since the potential is never equal if a line is not perpendicular to the electric field line, hence equipotential lines are always perpendicular to field lines.

Back to Top - Equations and Formulas - Email Me

E. Apr 17th: 1. What is the process of calculating the potential energy of three point charges brought together from infinity to three specific positions?

Answer: Integrating the potential energy of two of the point charges, plus integrating the potential energy of the third with each of the other two.

Back to Top - Equations and Formulas - Email Me

F. Apr 22th: 3. Does a light bulb, or other resistor, in a circuit use up current passing through it? (Give an explanation and not simply yes or no.)

Answer: No. It uses up the energy—the voltage, the electric potential, the potential energy in the electric potential, but doesn't use up any of the charge. If you connect too many light bulbs in series, they will combine to use up too many volts and none will light (or at least they will light very dimly). But if you connect too many light bulbs in parallel, they won't combine to use too much current (until the battery runs out). So long as the battery can keep pumping charge, charge can go through the light bulb and light it, forever.

Back to Top - Equations and Formulas - Email Me

G. Apr 24th: 1. Car batteries can be rated in ampere-hours (Ah). What aspect of the battery is being rated?

Answer: The charge that can be pumped through it over the lifetime of the battery. Look at the units: current=charge/time, so current⋅time=time⋅charge/time, the times cancel out each other and you get charge.

2. Can a copper wire and an aluminum wire of the same length have the same resistance? Explain.

Answer: Yes, depending on their thickness and temperature.

Back to Top - Equations and Formulas - Email Me

H. Apr 29th: 1. Why can birds sit on a power line safely, but it would be dangerous to touch a powerline with a metal pole?

Answer: There's no potential difference between their feet, so no current would pass through them. However, a short circuit would go through the wire into the a grounded metal pole, creating a temporary blackout and, if the wires melt, a permanent one.

2. Suppose you have two light bulbs with resistances R1 and R2, with R1>R2, connected in series to a battery. Which one will be brighter? If they were connectred in parallel, which would be brighter? Explain.

Answer: P=IV=V²/R=I²⋅R, so in series, they have equal I but R1 has a greater R, so R1 is brighter, and in series, the opposite is true, since they have equal V but R2 has less R and so V²/R is greater.

Back to Top - Equations and Formulas - Email Me

I. May 13th: 1. In what direction are the magnetic field lines beneath a straight wire carrying a current that is moving directly toward you?
1. toward you
2. right
3. left
4. up
5. down

Answer: Choice 2, to the right. Take your right hand, make it into a thumbs-up sign, point the thumb at yourself, and see where the fingers below your hand are pointing.

2. In the vector relation F = Il×B, which pairs of the vectors F, l, B are always at right angles?
1. F and l & F and B
2. F and B
3. l and B
4. none of these vectors need always be at right angles

Answer: Choice 1, F is always perpendicular to both l and B by the definition of the cross product (you may want to review Chapter 10, page 239.

Back to Top - Equations and Formulas - Email Me

J. May 15th: 1. Would it be possible, as in the case of the electric field, to define a magnetic field so that the direction of the magnetic field at a point is defined as the direction of the force on a positive charge at that point? Explain why or why not.

Answer: No, because the direction of the force on the test charge would depend on the velocity of the of the test charge. In other words, the force is not unique for every test charge at a point (it depends on the velocity of the test charge), so it would never work.

2. How could you tell whether a moving charge in a region of space is being deflected by an electric field or a magnetic field?

Answer: If you're given the choice between a uniform magnetic field or a uniform electric field, the answer would be that a moving charge in a uniform electric field, if deflected, will move parabolically, whereas in a u. m. f., if deflected, it will move in a helix. Otherwise, however, if it's some random field and you just know the mass, charge, and initial velocity of the particle, and are given it's motion, there's really no way you can tell for sure whether the field is magnetic or electric. Too many variables.

Back to Top - Equations and Formulas - Email Me

K. May 20th: 1. Suppose you have 1 kg of copper from which you can make copper wire for a solinoid. To get the strongest magnetic field, should you make the wire long and thin or short and fat, or something else? Explain.

Answer: Long and thin. The longer the wire, longer you can make the solenoid.