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Dimensional formula of permittivity? Of space as well as vaccum.

Dimensional formula of permittivity?
Of space as well as vaccum.

Grade:12

2 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
4 years ago
Vikas TU
14149 Points
4 years ago
Dear student 
Permittivity of free space depends on the hand-inputted value of the vacuum permeability μ0 = 4π * 10^-7 Henrys (inductance) per meter (i.e., Newton/Ampere^2) with respect to the classical relationship c^2 = 1/(μ0ε0).
Since [Henrys per meter] times [Farads per meter] (with the latter constituting the dimensionality of ε0) makes [s^2/m^2], its inverse equates to the dimensions of lightspeed squared on the LHS of the equality [c^2 = 1/(μ0ε0)].
This also means that, based on μ0, the permittivity of free space ε0 is exactly 10^7/(4π|c|^2) Farads per meter — where |c|^2 is the modulus of lightspeed squared, or just the number 299,792,458^2 — which is not quite obvious at first glance.
M^-1 L^-3 T^4 A^2 is the dimensional formula for permittivity of vacuum. Here M is Mass, L is length, T is time and A is current

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