# Electrostatic Constant Written Funny

## Main Question or Discussion Point

So the electrostatic constant k = 9*10^9, but I've also seen it written in my Fundamentals of Phys book (Resnick) and Wikipedia as k = 1/(4*pi*e0). Why is this? Thanks.

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No one knows?

Doc Al
Mentor
So the electrostatic constant k = 9*10^9, but I've also seen it written in my Fundamentals of Phys book (Resnick) and Wikipedia as k = 1/(4*pi*e0). Why is this? Thanks.
The ε0 is a constant called the electric permittivity of free space. Read about it here: http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elefie.html#c3"

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The capacitance of a parallel plate air capacitor of area A and plate separation d is

C = e0 A/d

where the permittivity of free space e0 = 1/(u0 c2) = 8.85 x 10-12 Farads per meter,
and the permeability of free space is
u0 = 4 pi x 10-7 Henrys per meter

jtbell
Mentor
It's partly for historical reasons, and partly a matter of minimizing the number of equations that have factors of $4 \pi$.

Using $\epsilon_0$ Coulomb's Law is more complicated:

$$F_{elec} = \frac {1} {4 \pi \epsilon_0} \frac {q_1 q_2} {r^2}$$

but other equations like Gauss's Law and the parallel-plate capacitor equation are simple:

$$\vec \nabla \cdot \vec E = \frac {\rho} {\epsilon_0}$$

$$C = \frac {\epsilon_0 A}{d}$$

Whereas using k, Coulomb's Law is simpler:

$$F_{elec} = k \frac {q_1 q_2} {r^2}$$

but you have to insert factors of $4 \pi$ into other equations:

$$\vec \nabla \cdot \vec E = 4 \pi k \rho$$

$$C = \frac {A}{4 \pi k d}$$