What Is the New Force Between Two Charges if the Distance Is Doubled?

[oceanflavored]

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

 

[konthelion]

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?

 

[oceanflavored]

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?

 

[alphysicist]

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.

 

[oceanflavored]

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?

 

[konthelion]

Yes, that's correct.

 

[tiny-tim]

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!

yessssss!