Vacuum permittivty with alternative Standard-Unit for meters

[GoodQuestion]

TL;DR

Value of Permettivity when 1m would be 1,10m

I was wondering: what would be the value of Vacuum Permettivity in the case 1 meter (say 1m") would be defined as the distance we nowadays see as 1,10 meters.

At first this looks easy: $\varepsilon0 = 8.8541878128 \cdot 10{^{12}}$ F / m with normal meters
so $\varepsilon0" = \varepsilon0 \cdot 1.1$
$\varepsilon0" = 9.739660659408 \cdot 10{^{12}}$ F / m" with converted meters.

However, when I look at Farads, this is defined as $$\frac{s^4 \cdot A^2}{m^2 \cdot kg}$$ so Farads should be converted too.

And Amperes are defined as:
"The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one metre apart in vacuum, would produce between these conductors a force equal to $2 \cdot 10^{−7}$ Newtons per metre of length."

which also involves meters, and Newtons, which are defined as $$\frac{kg \cdot m}{s^2}$$

this all makes it too tough for me.

So, in short, my question is: is the calculated version of E0" correct?

 

[Dale]

And Amperes are defined as:
"The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one metre apart in vacuum, would produce between these conductors a force equal to 2⋅10−7 Newtons per metre of length."

That is the old definition. The modern definition is: "The ampere, symbol A, is the SI unit of electric current. It is defined by taking the fixed numerical value of the elementary charge e to be $1.602176634×10^{−19}$ when expressed in the unit C, which is equal to A s, where the second is defined in terms of ##∆ν_{Cs}##."

see: https://www.bipm.org/utils/common/pdf/si-brochure/SI-Brochure-9-EN.pdf

I was wondering: what would be the value of Vacuum Permettivity in the case 1 meter (say 1m") would be defined as the distance we nowadays see as 1,10 meters.

Under the current definitions of the SI the meter is defined as: "The metre, symbol m, is the SI unit of length. It is defined by taking the fixed numerical value of the speed of light in vacuum c to be 299792458 when expressed in the unit m/s, where the second is defined in terms of $∆ν_{Cs}$."

So, if 1 METRE = 1.1 metre then c = 299792458 metre/second = 272538598 METRE/second

Then, since in the current SI we have $$\epsilon_0=\frac{e^2}{4 \pi \alpha \hbar c}$$ Thus, keeping all other SI defining constants the same and keeping physics the same, we simply have a reduction in the numerical value of c by a factor of 1.1 and thus an increase in the numerical value of $\epsilon_0$ by a factor of 1.1

So, in short, my question is: is the calculated version of E0" correct?

Yes, but you shouldn't write it in units of F/m since you have changed the units.

 

[GoodQuestion]

Shouldn't $\hbar$ change also, because there's a meter in there as well?

Assuming the value of KG en second stays the same, shouldn't this lead to $\varepsilon0$" becomes $\varepsilon0 \cdot 1,1{^3}$ ?

 

[Dale]

Shouldn't $\hbar$ change also, because there's a meter in there as well?

It is not necessary, but you certainly could do that if you wish.