
1)Considertwochargedparticlesacertaindistanceapart,e.g.twoelectrons.
a)Whatislikelytobelarger,thegravitationalortheelectricalforcebetweenthem?
b)Howwouldtheelectricforcebetweenthemchangeiftheyarebroughttohalftheiroriginaldistance
ofseparation?Whataboutthegravitationalforce?
a)Theelectricalforceisfarlargerthanthegravitationalforce,sincekqeqeismuchlargerthanGmeme,as
showninclass.
b)Theelectricforcevarieswithseparationastheinverse‐squarelaw,1/d2.Soifdwashalved,thend2is
quartered,and1/d2is4xasmuch,i.e.theelectricalforcebecomes4timesaslarge.Thegravitational
forceobeysthesamedistance‐dependence,soitwouldalsobe4timesaslarge.
2)Ifyouarecaughtoutdoorsinathunderstorm,whyshouldyounotstandunderatree?Canyouthink
ofareasonwhyyoushouldnotstandwithyourlegsfarapart?Orwhylyingdowncanbedangerous?
(Considerelectricpotentialdifference).
Thetreeislikelytobehitbecauseitprovidesapathoflessresistancebetweenthecloudoverheadand
theground.Thetreeandthegroundnearitarethenraisedtoahighpotentialrelativetotheground
fartheraway.Ifyoustandwithyourlegsfarapart,onelegonahigher‐potentialpartofthegroundthan
theother,orifyouliedownwithasignificantpotentialdifferencebetweenyourheadandyourfeet,
youmayfindyourselfaconductingpath.That,youwanttoavoid!
3)Willthecurrentinalightbulbconnectedtoa220‐Vsourcebegreaterorlessthanwhenthesame
bulbisconnectedtoa110‐Vsource?
Thecurrentwillbeless:Current=Voltage/Resistance,soiftheVoltageishalvedthen,therewillbehalf
asmuchcurrentthroughthesamebulb.
4)Ifelectronsflowveryslowlythroughacircuit,whydoesitnottakeanoticeablylongtimeforalamp
toglowwhenyouturnonadistantswitch?
Howquicklyalampglowsafteranelectricalswitchiscloseddoesnotdependonthedriftvelocityofthe
conductionelectrons,butdependsonthespeedatwhichtheelectricfieldpropagatesthroughthe
circuit—aboutthespeedoflight.
5)Considerapairofflashlightbulbsconnectedtoabattery.Willtheyglowbrighterconnectedinseries
orinparallel?Willthebatteryrundownfasteriftheyareconnectedinseriesorinparallel?Notethat
thebrightnessofabulbmeasuresthepower.
Thebulbswillglowbrighterwhenconnectedinparallel,forthevoltageofthebatteryisimpressed
acrosseachbulb:Power=VoltagexCurrent=(Voltage)2/Resistance.Whentwoidenticalbulbsare
connectedinseries,halfthevoltageofthebatteryisimpressedacrosseachbulb.Thebatterywillrun
downfasterwhenthebulbsareinparallelbecausethisdrawsmorepowerfromthebattery.