Force of Repulsion Between Charged Bodies: Calculating with Coulomb's Law

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SUMMARY

The discussion centers on calculating the force of repulsion between two similarly charged bodies using Coulomb's Law. When the distance between the charges is increased from 5 cm to 10 cm, the force of repulsion decreases to one-fourth of its original strength due to the inverse square relationship defined by the law. Specifically, as the distance (r) doubles, the force (F) is reduced to F/4, confirming the principles of Coulomb's Law. This calculation is crucial for understanding electrostatic interactions in physics.

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  • Coulomb's Law
  • Understanding of electrostatic forces
  • Basic algebra for manipulating equations
  • Concept of inverse square relationships
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Deebu R
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Homework Statement


two similar charged bodies are kept 5 cm apart in air. If the second body is shifted away from the first another 5 cm, there force of repulsion will be?

Homework Equations


Coulomb's law? F= k (q1q2)/r^2?
 
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What do you think? What are your thoughts on this problem? What does Coulomb's law govern?
 
force of attraction or repulsion is directly proportional to product of charges and inversely proportional to square of the distance between them. So...if the r is moved by another 5 cm r^2 also increases.
When r= 5 r^2 = 25
When r=10 then r^2 is 100
Since it is inversely proportional the force reduces by 1/4. True?
 
Deebu R said:
force of attraction or repulsion is directly proportional to product of charges and inversely proportional to square of the distance between them. So...if the r is moved by another 5 cm r^2 also increases.
When r= 5 r^2 = 25
When r=10 then r^2 is 100
Since it is inversely proportional the force reduces by 1/4. True?

Yep! When you double the distance, you reduce the force to 1/4 of its original strength. If you triple the distance, the force is reduced to 1/9 the strength.
 
I was not sure if my answer was correct or not. Now I know. Thank you for your time.
 

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