Solutions to Homework #4 for Physics 104, Summer 2003

Chapter 8

Exercises 16, 17, 22, 26,32

16. When it is dark, the photoconductor behaves as an insulator, due to the lack of photons to excite the motion of electrons. Thus, the light will turn on when the photoconductor becomes an insulator

17. Red light corresponds to low energy in the visible spectrum. When photons of frequency corresponding to redl light hits the photoconductor, nothing happens because the energy is insufficient to excite the electrons, across the energy gap between the valence and the conduction band.

22. The attraction between a charged body and a neutral but polarized object is premised on the fact that the opposite charges induced on the neutral but polarized body will, over all, attract the charged body. The charges on the polarized body that are of similar sign as the charged body must reside on the farther side. The strength of this repulsion must be less than that of the attraction between the charged body and the oppositely charged particles on the closer surface. If electrostatic force weren't dependent on separation, there would be no overall attraction.

32. When the copper coin passes by a strong magnet, the changing magnetic field it sees causes induced currents to flow in the coin. The action of these currents is to be in such a direction that generates a field that opposes such a change in the field. This induced magnetism and the resulting magnetic force between the two magnets slows down the copper coin's roll.

Problem 5
Using F=k(q1)(q2)/r^2, where q1=q2=1C , k=9x10^9 N-m^2/C^2, F=1N, we get r=94,800 m.

Case 6

a. No light on the photoconductor keeps the electrons in the valence band confined there, w/out enough energy to jump to the conduction band. Thus, there is no electric current.

b. Time will come when the potential on the photoconductor surfaces will be at the same potential as the battery terminal connected to it. Then no charges will flow. Another way of thinking about this is the charges initially moved by the battery to the photoconductor surfaces generates their own field that repels additional charges.

c. Inside the photoconductor, light has arrived with enough energy to kick the charges from the valence to the conduction band.

d. The light helped close the ciruit, by exciting electrons to the conduction band, and allowing them to flow.

e. Red light doesn't have enough energy to kick charges from the valence to the conduction band.

Chapter 9

Exercises 4, 8, 17, 24, 36

4. Current flows from objects at higher voltage or potential to ones of lower potential if there is a conducting path available. Thus, the bird perched on a SINGLE power line is held at a high potential but current flow isn't possible since it is not in contact with another object of lower potential.

8. The end through which positive current flows into is the end with the higher voltage. The drop in voltage across the strips is due to the electrical resistance provided by the conducting strip.

17. The additional load, represented by the hairdryer, draws current from the main line. Thus, when it is on, the total amount of current available to the lighting system is reduced and causes the dimming.

24. The end of the paper clip that touches the magnet's north pole becomes a magnetic south.

Problem 1,5, 13
1. The power P (watts) dissipated by a conductor is the product of the voltage V(volts) across it and the current I(A) passing through it. Thus P = IV Since P = 12 W and V = 12 volts, then current I = 1 A.

5. THe power delivered by a battery of voltage V and current I is P = IV. In the first flashlight, the battery is delivering P = (2A)(1.5 v) = 3 Watts. In the second flashlight, the current passing thru each battery must be 2A (since it goes thru each battery connected in a chain) and the voltage across is the same. Thus. an identical power of 3 Watts is delivered by EACH battery.

13. P = IV = (600A)(400kV) = 240 megawatts

Case 1

a. The principle of a step-up transformer is based on the electromagnetic induction (Len'z law) which requires the presence of a CHANGING magnetic field. However, a DC or direct current voltage is NOT changing.

b. The current flowing in/out of the battery is DC (fixed or direct current) while that flowing in/out of the transformer is AC (alternating current).

c. Power between the primary and secondary coils of the transformer is transferred through the changing magnetic fields (i.e. Changing magnetic field carries power from one to the other).

d. The power dissipated in the primary (P=IV) must be identical to that in the secondary (P'=I'V'). Thus, IV=I'V' or (20A)(12v) = I' (20v) or I' = 12 A.

e. The second battery must be connected in parallel to the first. That is, the 2 positive terminals of the batteries need to be connected to each other; the same is done with the 2 negative terminals. This way, the 2 batteries generate a total of 12 volts still, but the amount of current that can be drawn from them is more than if you had only a single battery.