Chapter 8 Electric & Magnetic Forces

Exercises 1,3,4,18,21,22,26

Problems 3

E1. The idenitcal materials get the same sign of static charge from friction, and like charges repel each other.

E3. Because of the Coulomb force. The opposite charges on the paint are attracted to charge on the metal surface.

E4. Because the like sign charges of all the paint particles repel each other and cause a nice even coat of paint to spread out on the surface. The repulsion of the like sign paint particles stops any clumping from occuring.

E18. No. You can not continue to divide the charges down smaller then the elementary unit charge of an electron.

E22. When a charged balloon (say negative charged) is put on a wall the negative charge of the balloon repels the electrons from the surface of the wall, leaving positive charge near the surface. The ballon sticks because the aatraction from the CLOSER positive charge is STRONGER then the repulsion from the FURTHER negative charge. If they were equal the net result would be no force.

E26. in order to make the volatge the same on the tip of the wire, the charge bunches near the tip and make a large electric field.

P3. Using the Coulomb force law (eq 8.1.1) F= k Q1 Q2
8.988x10^9 Nm^2/C^2 x 96,000C x 96,000C

--------- =
---------------------------------------------

R^2
1m^2

The attractive force would be 8.28x10^19 N or about 4 billion
billion pounds

Chapter 9 Electrodynamics

Exercises 1,2,6,12,13,14,22,23

Problems 1,2

E1. It should open (or break) the circuit connection stopping the current flow. Instead it staying closed allowing the current to flow through the circuit.

E2. The lamp will become dimmer. The file cabinet short circuits the current from the supply to the return before it goes through the lamp.

E6. Because the circuit is not complete, it is an open circuit. The current mast flow through a complete path from supply to return.

E12. Watts are a unit of power, which energy per unit time. Or, Joules/s. So, a kw-hr is the power, or energy per unit time x time, which tells you the total energy used.

E13. Because the power loss in any resistance is given by I^2 x R. even though the resistance R is small for a wire it is non-zero. So, if the current is large the power loss goes into heating the wire.

E14. Because the circuit is designed to carry only 30 amps before the I^2 R loss gets too high. if the fuse doesn't open the circuit at 30 amps the circuit can overheat from the loss and start a fire.

E22. the ratio of voltages from primary to secondary in a transformer is given by the ratio of turns. Since there are 1/3 the turns in the secondary the voltage will be 1/3 of the primary voltage or 1/3 x 120V = 40V

E23. The current will three times higher in the secondary or 9 amps. The total power V x I is the same in primary and secondary.

P1. Power used is given by voltage drop x current. or, 12 watts = 12 volts x ?? amps. Solving for current gives 1 amp.

P2. Power consumed is V x I = 10V x 5A = 50 watts