# I Why Are The Units of Coulombs Law What They Are?

1. Jul 17, 2016

### PurelyPhysical

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?

2. Jul 17, 2016

### David Lewis

Q1 times Q2 is cancelled out by Coulomb's constant (which contains a coulomb-2 term).

3. Jul 18, 2016

### PurelyPhysical

Thank you. That makes a lot of sense.

4. Jul 18, 2016

### Delta²

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.

5. Jul 18, 2016

### Bobby2

Delta[2]

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

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

6. Jul 18, 2016