Why Are The Units of Coulombs Law What They Are?

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Discussion Overview

The discussion revolves around the units used in Coulomb's Law, specifically the reasoning behind the units of force expressed as (N-M^2)/(C^2), the units of the permittivity constant, and the role of charge units in the final expression. The scope includes conceptual clarification and technical explanation of unit relationships in physics.

Discussion Character

  • Conceptual clarification
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why the units of force are used in Coulomb's Law, specifically asking about the units of the permittivity constant and the absence of charge units in the final expression.
  • Another participant suggests that the product of the charges (Q1 times Q2) is canceled out by Coulomb's constant, which contains a coulomb-2 term.
  • A further contribution explains that the relationship between the units is determined after establishing the physical law, indicating that the equality in Coulomb's Law necessitates a constant K to relate the units appropriately.
  • There is a reference to the inverse-square law and its historical context, mentioning figures like Newton and La Place.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the units in Coulomb's Law, with some agreeing on the cancellation of charge units while others explore the implications of unit relationships without reaching a consensus.

Contextual Notes

Participants do not fully resolve the implications of the units or the role of the constant K, leaving some assumptions and dependencies on definitions unaddressed.

Who May Find This Useful

This discussion may be useful for students and enthusiasts of physics seeking to understand the relationships between units in electrostatics and the conceptual underpinnings of Coulomb's Law.

PurelyPhysical
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Why are the units of force used when applying coulombs law (N-M^2)/(C^2)? This is actually a three part question.

1. Why are the units of the permitivity constant (C^2)/(N-M^2)?
2. Why do Q1 and Q2 not contribute to the final units? Each charge is measured in coulombs, but those units don't reflect in the final units.
3. Units of force are Newtons. Why then can we say that coulombs law equals force? What's going on with the other units that makes it so that we can still refer to it as a force?
 
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Q1 times Q2 is canceled out by Coulomb's constant (which contains a coulomb-2 term).
 
David Lewis said:
Q1 times Q2 is canceled out by Coulomb's constant (which contains a coulomb-2 term).
Thank you. That makes a lot of sense.
 
PurelyPhysical said:
Why are the units of force used when applying coulombs law (N-M^2)/(C^2)? This is actually a three part question.
3. Units of force are Newtons. Why then can we say that coulombs law equals force? What's going on with the other units that makes it so that we can still refer to it as a force?

There is nothing mysterious with the units that makes the final result to be expressed in units of force.
In physics the first thing are the physical laws. The relationship between the units (and the units of possible constants that are used, like the coulomb constant K) are worked out after we discover the physical laws.

In our case the physical law that Coulomb discover is that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance. In order to complete the equality we need a constant K so we can write down
##F=K\frac{Q_1Q_2}{r^2}## (1)
Now that we know this law holds, we can figure out the units of the constant K and the relationship between the units. So it will be because (1) holds

Newton=(units of constant K)*Coulomb*Coulomb/Meter^2. (2)

so the units of constant K have to be Newton*Meter^2/Coulomb^2 cause only if it is so then (2) holds.
 
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Likes   Reactions: Dale
Delta[2]

Do you mean F = [m1m2]/r^2?

Oh! What a popular equation! Newton, coulomb, La Place is there more?
 

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