Question

Find the dimensional formula for ${\mu _ \circ }$ and ${\varepsilon _ \circ }$ :

Answer 1

Hint: First let us see what ${\mu _ \circ }$ and ${\varepsilon _ \circ }$ is:
${\varepsilon _ \circ }$ stands for permittivity of free space.
${\mu _ \circ }$ stands for vacuum permeability in free space.

Complete step by step answer:
The permittivity of free space ${\varepsilon _ \circ }$ , is a physical constant often used in electromagnetism. It reflects the power of vacuum to permit the use of electrical fields. It is also related to the energy contained in the electrical field and to the capacitance.
The permittivity of free space is given mathematically as:
${\varepsilon _ \circ } = \dfrac{1}{{{\mu _ \circ }{c^2}}} \approx 8.8542 \times {10^{ - 12}}F{m^{ - 1}}$

Where \[{\mu _ \circ }\] is the permeability of free space and $c$ is the speed of light.
An electrical field, $E$ , in an area of space, has field energy associated with it, which is energy density.
$\dfrac{{Energy}}{{volume}} = \dfrac{{{\varepsilon _ \circ }{E^2}}}{2}$

The energy stored in a capacitor is also given by:
$energy = \dfrac{{{\varepsilon _ \circ }A}}
{{2d}}{V^2}$

Where
$A$ is the area of the plates,
$d$ is the distance between the plates
$V$ is the voltage between the plates
The permittivity of free space can also be used to detect the Coulomb power. The constant shows how intense the force is between two charges separated by a distance:
$F = \dfrac{1}
{{4\pi {\varepsilon _ \circ }}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$

Where
$F$ is the Coulomb force
${q_1}$ and ${q_2}$ are two charges
$r$ is the separation between the charges.
$k = \dfrac{1}{{4\pi {\varepsilon _ \circ }}}$

Thus, the dimension of ${\varepsilon _ \circ }$ is:
$
  {\varepsilon _ \circ } = \left[ {{k^{ - 1}}} \right] \\
   = {\left[ {M{L^3}{T^{ - 4}}{A^{ - 2}}} \right]^{ - 1}} \\
   = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right] \\
$

The permeability of free space ${\mu _ \circ }$ is a physical constant often used in electromagnetism. It is defined to have an exact value of $4\pi \times {10^{ - 7}}\,N{A^{ - 2}}$ . It is linked to the energy contained in a magnetic field.
It is related to the speed of light as:
$c = \dfrac{1}{{\sqrt {{\mu _ \circ }{\varepsilon _ \circ }} }}$

Where
$c$ is the speed of light
${\varepsilon _ \circ }$ is the permittivity of free space
Magnetic field $B$ in the space area has field energy correlated with it. The density of energy is defined as:
$energy = \dfrac{{{B^2}}}{{2{\mu _ \circ }}}$

It also gives the usage of magnetic force. The constant shows how intense the force is between two electrical currents divided by a distance:
$F = \dfrac{{{I_1}{I_2}}}{{2\pi {\mu _ \circ }r}}$

Where
$F$ is the magnetic force
${I_1}$ and ${I_2}$ are two currents
$r$ is the separation between the wires
So, the dimension of ${\mu _ \circ }$ is given as:
$
  {\mu _ \circ } = {\left[ {{c^2}{\varepsilon _ \circ }} \right]^{ - 1}} \\ = \left[ {ML{T^{ - 2}}{A^2}} \right] \\
$

Note:
Although the permeability and permittivity may appear the same, they are different.
Permittivity measures the obstruction created by the material in the creation of an electrical field, while
The permeability is the ability of the material to allow magnetic lines to conduct through it.