# Why is k = 9 * 10^9 and how is it related to Pi?

• -Castiel-
In summary, the use of k = 1/[4(pi)(epsilon)] and epsilon = 8.8 * 10^(-12) in Coulomb's law and subsequently, k = 9 * 10^9 is due to the desire to expose fundamental properties and relationships between physical models. The inclusion of pi comes from the symmetry of the situation described by the equation. While it is possible to use different values for k and epsilon, keeping them as defined constants allows for easier calculations and reveals the underlying symmetries of the laws being studied.
-Castiel-
Why use k = 1/[4(pi)(epislon)] and epsilon = 8.8 * 10^(-12) and subsequently, k = 9 * 10^9

It could simply be k = 9 * 10^9, and different k for different medium instead of different permittivity for different medium.

What I mean is why does Pi, (I can handle the 4) comes into the equation, one reason I can think of is that it comes if you use Gauss Law to derive Coulomb's law but I am looking for something more convincing.

The Permittivity of free space is a constant experimentally derived. Why would one derive that in the first place? If the force between two charges was directly proportional to some constant K. One would experimentally see that it is 9 * 10^9. Where was the idea of breaking it down to 1/[4(pi)(epislon)] entertained?

Many problems in this area of physics will have a 4pi in the numerator. So a 4pi in the denomenator will simplify things. Perhaps that is the reason.

grzz said:
Many problems in this area of physics will have a 4pi in the numerator. So a 4pi in the denomenator will simplify things. Perhaps that is the reason.

That is the reason I thought of seconds after I clicked the Submit button, it does seem a likely reason but I will wait and see if someone knows something better.

Your epsilon is usually $\epsilon_o$ - the permitivity of free space. We can modify the basic equation for different media by including the relative permitivity which is different for different materials.

It is possible to adopt a system of units where you don't have these constants - however, doing so hides some of the physics. For instance, there is an equivalent number for magnetism $\mu_o$ which is the permiability of free space. Together they make the speed of light: $c^2=1/\epsilon_o\mu_o$

The idea is to expose the interaction of fundamental properties and the relationships between physical models.

The pi is more interesting - it comes from the symmetry of the situation the equation describes: spherical.

Last edited:
Simon Bridge said:
the speed of light: $c^2=1/\epsilon_o\mu_o[/itex I cannot resist posting this comment: Is not [itex]c^2=1/\epsilon_o\mu_o$

a beautiful formula.

It shows the permitivity on which the electric field depends and the permeability on which the magnetic field depends. And that is what light is - an electromagnetic wave. (At least in classical Physics)

Simon Bridge said:
Your epsilon is usually $\epsilon_o$ - the permitivity of free space. We can modify the basic equation for different media by including the relative permitivity which is different for different materials.

You can use K here too. epsilon(naught)/epsilon = K/K(naught)

Simon Bridge said:
It is possible to adopt a system of units where you don't have these constants - however, doing so hides some of the physics. For instance, there is an equivalent number for magnetism $\mu_o$ which is the permiability of free space. Together they make the speed of light: $c^2=1/\epsilon_o\mu_o$

No it doesn't really hide it.

k = 1/[4(pi)(epislon)]

Let l = mu/4(Pi) (I know l doesn't exist but bear with me.)

(I really need to learn how to type those formulae.)

then

epsilon = 1/[4(pi)(k)]

mu = 4(Pi)l

$\epsilon_o\mu_o = kl$

You can now put this back in your formula.

Simon Bridge said:
The idea is to expose the interaction of fundamental properties and the relationships between physical models.

Like I said before. It is just a ratio. Whatever ratio epsilon is in, K is in inverse of that ratio. So relationships between models isn't affected.

Simon Bridge said:
The pi is more interesting - it comes from the symmetry of the situation the equation describes: spherical.

grzz said:
a beautiful formula

I completely agree. But my point is that permittivity and permeability are experimentally derived.

Why not use 9 * 10^9 and 10^(-7) directly. As I proved earlier they do not deform the equation in any major way.

bump.

-Castiel- said:
I completely agree. But my point is that permittivity and permeability are experimentally derived.

Why not use 9 * 10^9 and 10^(-7) directly. As I proved earlier they do not deform the equation in any major way.

Permittivity and permeability are not experimentally derived, they are defined constants. Permeability is defined as 4\pi e-7 and permittivity is defined using c and the permeability of freespace. The value for the permittivity and permeability is purely a choice of your unit system. As previously stated, we can use a Gaussian system where the permeability and permittivity of free space are unity. In addition, we can move the 4\pi from the factor k to Maxwell's equations (as we do in Gaussian units). Placing the 4\pi off of the Maxwell equations is called rationalized units.

We do not use a numerical value for k directly since k is dependent upon the permittivity of the background. It is not a constant across all possible problems.

Why not use 9 * 10^9 and 10^(-7) directly. As I proved earlier they do not deform the equation in any major way.
When using the equations to do calculations, this is what you do.

When you want to derive general relations, discovering new physical laws, the math works out easier if you keep the geometric constants separate from the physical ones. It's also more rigorous philosophically, and exposes the symmetries of the laws you discover.

Much the same sort of reasoning is behind using radiens instead of degrees for angles.

## 1. What is Coulomb's Law and Gauss Law?

Coulomb's Law and Gauss Law are two fundamental principles in the study of electromagnetism. Coulomb's Law states that the force of attraction or repulsion between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Gauss Law, on the other hand, states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.

## 2. What is the difference between Coulomb's Law and Gauss Law?

While both laws deal with the concept of electric charges, Coulomb's Law describes the force between two point charges, whereas Gauss Law describes the electric flux through a closed surface due to an enclosed charge. Coulomb's Law also follows an inverse square relationship, while Gauss Law is independent of distance.

## 3. What is the significance of Coulomb's Law and Gauss Law in physics?

Coulomb's Law and Gauss Law play a crucial role in understanding and predicting the behavior of electric charges and fields. They are used in various areas of physics, such as electrostatics, electromagnetism, and even in the design of electronic devices.

## 4. How are Coulomb's Law and Gauss Law related to each other?

Coulomb's Law can be derived from Gauss Law by applying the divergence theorem. In simpler terms, Coulomb's Law is a special case of Gauss Law where the electric field is constant and the distance between charges is large.

## 5. Can Coulomb's Law and Gauss Law be applied to both positive and negative charges?

Yes, both laws can be applied to both positive and negative charges. Coulomb's Law takes into account the sign of the charges in its calculation, while Gauss Law considers the net charge enclosed by a closed surface, regardless of its sign.

• Electromagnetism
Replies
5
Views
778
• Electromagnetism
Replies
2
Views
2K
• Electromagnetism
Replies
3
Views
826
• Introductory Physics Homework Help
Replies
6
Views
814
• Introductory Physics Homework Help
Replies
6
Views
1K
• Electromagnetism
Replies
8
Views
16K
• Electromagnetism
Replies
1
Views
2K
• Electromagnetism
Replies
5
Views
7K
• Classical Physics
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K