Question

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• The dimension of reciprocal of product of permeability and permittivity of free space is equal to
che dimension of​

Answer 1

Explanation:

square of the speed of light in vacuum.

Answer 2

The reciprocal of product of permeability and permittivity of free space is equal to the dimension of square of speed  [M⁰L²T⁻²].

Explanation:

• The permeability is the constant used while calculating the magnetic field strength of an object. It's value is μ$_{o} =4\pi$ × $10^{-7}$.

• The permittivity of the free space is the constant derived for an electric charge that produces a force or electric field around it. Its value is ε$_{o}=8.85$ × $10^{-12}$.

According to the question, the reciprocal of the product of the permeability will be, lets say $k$.

$k=\frac{1}{u_{o}*E_{o} }$

Substituting the values, we get

$k=\frac{1}{(4\pi *10^{-7})* (8.85*10^{-12}) }$

$k=\frac{1}{111.212*10^{-7-12} }$

$k=\frac{1}{111.212*10^{-19} }$

$k=0.009$ × $10^{19}$

$k=9$ × $10^{16}$

We can write this value as

$k=(3*10^{8} )^{2}$

We know the speed of light in vacuum $3$ × $10^{8}$ $m/s$ and is denoted by $c$.

Hence, we get

$(3*10^{8} )^{2} =\frac{1}{(4\pi *10^{-7})* (8.85*10^{-12}) }$

$c^{2} = \frac{1}{u_{o} E_{o} }$

Hence, the reciprocal of the product of permeability and permittivity in free space is the square of speed of light in vacuum.

We know, the SI unit of speed is meter per second.

The dimensional formula of speed will be $c=[M^{0} L^{1} T^{-1} ]$

The dimension of the reciprocal of the product of permeability and permittivity of free space will be $c^{2} =[M^{0} L^{2} T^{-2} ]$

Final answer:

Hence, the reciprocal of product of permeability and permittivity of free space is equal to the square of dimension of speed of light. $[M^{0} L^{2} T^{-2} ]$.