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Is the equation of $1/4πε_0$ in Coulombs law a scalar? Also, does this equation equate to Coulombs constant?

In addition, why does Jefimenko use $H=1/4π...$ in his equations? What is the significance of $1/4π$?

I'm looking at Jefimenko's equations and I can't tell if he's stating scalars are the 'cause' of vectors. Can someone help clear this up for me? I know that current and charge are scalars but I'm not sure if $∂B/∂t$ is suppose to translate to a weber somehow. He's using the terminology 'electric charge' and 'current' causing $E$ and $B$ fields yet he then has a part that says E is a consequence of $∂B/∂t$. So how does ∂B/∂t relate to charges and currents?

Is it true that $∂B/∂t$ is equal a scalar?

Sebastiano
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Zim
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2 Answers2

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This is the answer to the first part of the question (which, I think, should be split into two different questions.)

Factor $1/(4\pi\epsilon_0)$ is specific to the SI system of units. Many scientific works use other system of units, among which is cgs system, particularly popular in electrodynamics and condensed matter. In addition, there are systems of units specific to other fields, such as astronomy, particle physics, nuclear physics, etc.

The choice of a system of units is a matter of convenience, often dictated by some of the following reasons

  • having the scale of units compatible with the natural scales (i.e., to avoid writing too many factors $\times 10^{\pm x}$)
  • not having to write over and over physical constants (by setting $c=1, \hbar=1$, etc.)
  • having equations in a more convenient form.

The latter case is relevant here: in SI units the differential form of the Gauss' law and the resulting form of the Coulomb law are $$ \nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}\longrightarrow F = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}, $$ whereas in cgs units they take form: $$ \nabla\cdot\mathbf{E}=4\pi\rho\longrightarrow F = \frac{q_1q_2}{r^2}, $$ Factor $4\pi$ originates here from applying the Gauss law to a sphere with surface area $4\pi r^2$. Excluding this factor from the Maxwell equation, results in its reappearance in the expression for the Coulomb law and vice versa.

Roger V.
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"Is the equation of 1/4πε0 in Coulombs law a scalar?" an equation cannot be scalar.

"Also, does this equation equate to Coulombs constant?" a constant can equate an equation.

"In addition, why does Jefimenko use H=1/4π... in his equations? What is the significance of 1/4π?"

To understand 1/4π you should look at the whole equation i.e.

$E = \frac{q}{\epsilon \;4 \pi r^2 }$

You should be able to recall that $4 \pi r^2$ is area of a sphere. This is because the electric field which is created by the charge $q$ is spherically symmetric.

"I can't tell if he's stating scalars are the 'cause' of vectors." Some quantities are scalar such as mass, temperature, and all the constants in physics. However, the scalar and vector quantities can be related using differentiation. Gradient of a scalar field (not a scalar, but a scalar field) is a vector and divergence of a vector field (again not a vector, but a vector field) is a scalar.

"He's using the terminology 'electric charge' and 'current' causing E and B fields" there are different ways to create E and B field. electric charges can create E field but this is not the only way to create E field. B field can be created by current.

E field can also be created by ∂B/∂t i.e. changing the B field.

"So how does ∂B/∂t relate to charges and currents?" you can use current (which is moving charges) to create B. then if you change the current (the speed at which the charges are moving), you create ∂B/∂t and it gives you E.

"Is it true that ∂B/∂t is equal a scalar?" ∂B/∂t is a vector because B is a vector. when a vector is multiplied or divided by a scalar the result is still a vector.

Kian Maleki
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