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Studying for my SAT II Physics

  • #1
i'm studying for my SAT II Physics on June 7, 2008. and taking practice tests from REA.
i had trouble understanding this question. please help :) it would be VERY appreciated.

Homework Statement


Two charges are separated by 2 m. The force of attraction between them is 4 N. If the distance between them is doubled, the new force between them is...
A) .5 N
B) 1 N
C) 2 N
D) 4 N
E) 8 N

Homework Equations


Force = [K(Q1)(Q2)] / (R^2)
the formula is on http://www.sparknotes.com/testprep/books/sat2/physics/chapter13section2.rhtml under Coulomb's Law

The Attempt at a Solution


i'm sorry, i really couldn't figure out. it's probably a simple question, and i just can't see it.

PLEASE HELP!
thank you SO much :biggrin:
 

Answers and Replies

  • #2
238
0
Hello,
You are given the distance = 2m. In Coulomb's Law the distance is R.
Now, if the distance doubles, then will the force increase or decrease(i.e. will it be greater or equal to 4N)? and by what factor?
 
  • #3
Ummm....the force would decrease because the force is inversely proportional to the square of distance. And it would decrease by a factor of 1/16, right?
 
  • #4
alphysicist
Homework Helper
2,238
1
Hi oceanflavored,

It would decrease, and the new denominator is 16, but that's not the factor that the force decreased by. You have to take into account what the force was originally.
 
  • #5
eureka!! (i think!)

okay, i think i get this:
if the distance increases by a factor of 2, the force would decrease by a factor of (1/2)^2 or (1/4)
so the new force would be 4 x (1/4) = 1 N
yessssss???
 
  • #6
238
0
Yes, that's correct.
 
  • #7
tiny-tim
Science Advisor
Homework Helper
25,832
249
okay, i think i get this:
if the distance increases by a factor of 2, the force would decrease by a factor of (1/2)^2 or (1/4)
so the new force would be 4 x (1/4) = 1 N
yessssss???
Hi oceanflavored! :smile:

yessssss!!! :biggrin:
 

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