Calculating F2 for Double and Triple Changes in Coulombs Law"

[wilson_chem90]

Homework Statement

Look back at the Example in the Using Coulombs Law-Comparative Analysis section. Calculate the magnitude of F2 if both changes occurred simultaneously-that is, if the distance between the charges doubled and the size of once tripled, while the size of the other doubled.

This is the original example:

If the electrostatic force between two charges is 6.0 x 10^-8 N, what effect would each of the following changes have on the magnitude of the force?
1) the distance between the charges is doubled
2)One charge is doubled and the other is tripled

Solution:
1)
F2 = (6.0x10^-8 N)xr1 / (2r1)^2
= (6.0x10^-8 N) / 4
= 1.5 x 10^-8 N

2) q1=2q1
q2=3q1

F1/F2 = q1q2/q1q2

(6.0x10^-8 N)/F2 = q1q2 / (2q1)(3q2)

F2 = (6.0x10^-8N)(6)
= 3.6 x 10^-7 N

This is the solution to the example!

Homework Equations

The Attempt at a Solution

I can do the actually solution to this problem, but my only question is what are the values of q1 and q2 and r? Since they're literally exactly the same in both questions, does it mean that they all just double and triple from the original? Meaning r = 4r1, q1 = 6q1, q2 = 6q2

Does anyone have any suggestions?

Thank You

 

[cepheid]

I don't understand where you get the factors of 4 and 6 from.

Double = multiply by a factor of 2.

Triple = multiply by a factor of 3.

The only difference is that, instead of making each of these changes separately, ALL three changes are now to be made at the same time so that (primed symbols represent the "new" quantities after the changes):

r' = 2r

q₁' = 3q₁

q₂' = 2q₂

 

[wilson_chem90]

ohh okay, sorry i thought it mean't something else.

Do i ended up setting F1/F2 = (q1q2/r1^2) / ((3q1)(2q1)/2r1^2)

I rearranged it to be F2 = 6F1/4r1 (after cancellations and rearranging)

Im not sure if i rearranged it correctly...

 

[wilson_chem90]

sorry i forgot to cancel r1, its actually F2 = 6F1/4