Honors Homework Help

I wnet through a couple of dozen homeworks from the last week or so.

General comments:

Chapter 16 #41 was done in class

Many of you don't recognize the nature of a problem yet. For example, Chapt 16 problems 44,45, wer Gauss Law problems and should be approached accordingly.

Some of you still think a cm is a meter. If I lend you a penny will you pay me back a dollar? Well cm vs m are the same thing pennies vs dollars a factor of 1/100.

Details:

Honors HW help 11/2 through 11/16
Miscellaneous errors some of you did QT + QS instead of QTxQS in the equation for force between charges

many of you still do not distinguish between cm and meters 1 m squared = 10,000 cm squared so this mistake makes your answers come out very wrong!


Giancoli Chapt 16

# 44 flux = field passing through area = 580 x pi x.18^2 when field is perpendicular to area and flux = 0 when field is parallel to area surface , no field passes through

#45 Use Gauss Law flux = 4 pi Qtotal enclosed
so flux for A1 = 1.13x10^11 x( -1x10^-6 –2x10^-6)

flux =0 for A2

Giancoli Ch 17 [D = delta here ]


Question 5: a point exactly in the middle between two equal positive charges will have zero field since the field from each charge will cancel the other here and only here. The potential is nowhere zero since work would have to be done to bring a positive charge from far away to anywhere near these charges. The zero field point is surrounded by outward pointing fields that you would have to work against to bring a positive charge into this point.

problem #21
DKE = -DPE
when they are very far away from each other PE = 0 KE i = 0 here so KEf = PEi
Mass = 1 mg = 10^-6kg
PEi = kxQsxQT/r = 9x10^9 x 9.5x10^-6 x 9.5x10^-6/ .035

so KEf = 9x10^9 x 9.5x10^-6 x 9.5x10^-6/ .035 = 2.32X10^1 J
v = sqrt( 2KE/Mass) = sqrt (23/10^-6) = 4800 m/s

#70

Ignore capacitance
DV = 35x10^6 Dx = 1500m, so

E = DV/Dx = 23,700 V/m from Gauss: E Area = 4 pi k Qs = 1.13 x10 ^11 xQs

so Qs = E Area /1.13 x10 ^11

Area = 110sq km x10^6 sq m per sq km = 1.11x10^8 m^2

so Qs = 23,700 x1.11x10^8/1.13x10^11 = about 23 Coulombs

Gauss’ Law #2
Flux = E x area = 4 pi kc xQ enclosed by flux surface so Qenclosed = flux/(4 pi x9x10^9) Flux = surface area of box x (-)450 V/meter = [ 2 faces x .2x.2 + 4 faces .4 x.2] = .4 m2 x (-)450 = 180 Q = 180/1.13x10^-11 = 1.59 x10^-9 Coulombs

Giancoli Chapt 7

#7 total p before = total p after

12600 kg car moving at 18 m/s has p (in the horizontal direction) of 12600x18 =226800 pods gravel has no horizontal momentum so p total = 226800 and after the collision the loaded car has a mass of 17950 kg and still a momentum of 226800 so v’ = 226800/mass = 226800/17950= 12.6 m/s