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

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Homework Help Overview

The problem involves calculating the force of repulsion between two similarly charged bodies using Coulomb's Law, specifically considering the effect of increasing the distance between them from 5 cm to 10 cm.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of Coulomb's Law, questioning how changes in distance affect the force of repulsion. Some participants explore the mathematical relationship between distance and force, while others express uncertainty about their reasoning.

Discussion Status

The discussion includes various interpretations of how distance affects the force of repulsion, with some participants confirming the mathematical reasoning presented. There is acknowledgment of the relationship between distance and force reduction, but no explicit consensus on the final answer.

Contextual Notes

Participants express uncertainty about the correctness of their reasoning and calculations, indicating a need for further clarification on the application of Coulomb's Law in this context.

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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