Help with question

  • Thread starter chemboy
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  • #1
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Main Question or Discussion Point

if the following takes place simultaneously

distance betwen two charges is doubled and one charge is doubled and the other is trippled how do I set this up mathmatically

I understand that when distance doubles (r) then the effect on net charge is .25 of original F
also
I understand that the force betwen charges when increased as mentioned above, there will the force will increase by 6X

Im thinking simply the answer is 6X the force X .25 but there must be a way to relate the formulas Fe = 1/r^2 and Fe = q1q2/r^2 to show this

If I knew what r was then I could use Fe = kq1q2/r^2

Im worried this is a mathmatical problem and not that I dont understand the theory behind the physics.

any help??

will it end up something like

F = (9.0 X 10^9N*m^2/C^2) X 6(6.0X10^-8 N) / 4
 
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Answers and Replies

  • #2
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Consider this:

if a=2b and c=4b then what mathematical operation does one have to do to know how many times c is larger than a?
 
  • #3
tiny-tim
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Hi chemboy! :smile:
I understand that when distance doubles (r) then the effect on net charge is .25 of original F
also
I understand that the force betwen charges when increased as mentioned above, there will the force will increase by 6X

Im thinking simply the answer is 6X the force X .25 but there must be a way to relate the formulas Fe = 1/r^2 and Fe = q1q2/r^2 to show this
Yes … that's completely correct! :smile:

You looked at the formula q1q2/r², and you correctly said "it's proportional to q1 and to q2, and inversely proportional to the square of r … so it's 2 x 3 / 4".

This "proportion" method works for any formula! :biggrin:

Why are you worried? :confused:
 

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