Question 32: Units , Dimensions and Error Analysis

Question

The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equal to that of:
($ \mu_0 $ = Vacuum permeability and $ \epsilon_0 $ = Vacuum permittivity)

Accepted Answer

The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ can be understood as follows:
Vacuum permeability ($ \mu_0 $) and vacuum permittivity ($ \epsilon_0 $) relate to the properties of inductance and capacitance, respectively, in a vacuum. We start by considering the formulas for inductance ($ L $) and capacitance ($ C $):
Inductance ($ L $)
is given by:
$   L = \frac{\mu_0 \cdot 	ext{Number of turns} \cdot 	ext{Area}}{	ext{Length}}   $
Capacitance ($ C $)
is given by:
$   C = \frac{	ext{Area} \cdot \epsilon_0}{	ext{Distance}}   $
From these, the ratio $\frac{L}{C}$ can be expressed as $\frac{\mu_0}{\epsilon_0}$:
$ \frac{L}{C} \propto \frac{\mu_0}{\epsilon_0} $
Taking the square root of this ratio, we have:
$ \sqrt{\frac{\mu_0}{\epsilon_0}} \propto \sqrt{\frac{L}{C}} $
This simplifies further to considering the relationship between time constant ($	au$) and resistance ($R$):
$ \frac{L}{C} = \frac{	au R}{(	au / R)} = R^2 $
Thus, by taking the square root:
$ \sqrt{\frac{\mu_0}{\epsilon_0}} = R $
In conclusion, the dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equivalent to the dimension of resistance, $ R $.

Answer

Voltage

Answer

Inductance

Answer

Resistance

Answer

Capacitance

Units , Dimensions and Error Analysis

The dimension of \sqrt{\frac{\mu_0}{\epsilon_0}} is equal to that of: ( \mu_0 = Vacuum permeability and \epsilon_0 = Vacuum permittivity)

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