How to Relate Formulas for Calculating Force Between Charges

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SUMMARY

The discussion focuses on calculating the force between charges when both the distance and the magnitudes of the charges are altered. Specifically, when the distance (r) between two charges is doubled, the force (F) is reduced to 0.25 of its original value. Simultaneously, if one charge is doubled and the other is tripled, the force increases by a factor of 6. The relationship can be mathematically expressed using the formulas Fe = k(q1q2)/r² and the proportionality of forces based on charge magnitudes and distance.

PREREQUISITES
  • Understanding of Coulomb's Law (Fe = k(q1q2)/r²)
  • Knowledge of proportional relationships in physics
  • Basic algebra for manipulating equations
  • Familiarity with the concept of electric charge
NEXT STEPS
  • Explore the implications of Coulomb's Law in different scenarios
  • Learn about the constant k in Coulomb's Law and its significance
  • Investigate the effects of varying distances on electric force
  • Study the concept of electric field strength and its relation to charge
USEFUL FOR

Students of physics, educators teaching electrostatics, and anyone interested in understanding the mathematical relationships in electric forces between charges.

chemboy
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if the following takes place simultaneously

distance between 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 between 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 don't 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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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?
 
Hi chemboy! :smile:
chemboy said:
I understand that when distance doubles (r) then the effect on net charge is .25 of original F
also
I understand that the force between 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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