Question

Find the dimension of $\dfrac{G}{4\pi {{\in }_{0}}}$ , where G is universal gravitational constant and ${{\in }_{0}}$ is permittivity of free space.
$\begin{align}
  & \text{A}\text{. }\left[ {{M}^{-2}}{{A}^{2}}{{T}^{2}} \right] \\
 & \text{B}\text{. }\left[ {{M}^{-2}}{{L}^{4}}{{T}^{4}}{{A}^{-2}} \right] \\
 & \text{C}\text{. }\left[ {{M}^{0}}{{L}^{6}}{{T}^{-6}}{{A}^{-2}} \right] \\
 & \text{D}\text{. }\left[ {{M}^{-2}}{{T}^{2}}{{A}^{2}} \right] \\
\end{align}$

Answer 1

Hint: G is a universal constant which comes in Newton's law of gravitation. To find the dimension of G, find the relation of G with the force between two massive bodies at a distance apart. To find the dimension of ${{\in }_{0}}$, find the relation with the force between two charges at a distance apart. Then we can find the required answer.

Complete step by step answer:
The dimension of G can be found from the law of gravitational force of attraction between two bodies.
The force of attraction between two bodies of mass m and M at distance R apart can be given by the mathematical expression,
$F=\dfrac{GmM}{{{R}^{2}}}$
Where, G is the gravitational constant.
$G=\dfrac{F{{R}^{{}}}}{mM}$
Again, the force of attraction between two charged particle of charge q each at a distance R apart is given by the mathematical expression,

$F=\dfrac{1}{4\pi {{\in }_{0}}}\dfrac{{{q}^{2}}}{{{R}^{2}}}$
$\dfrac{1}{4\pi {{\in }_{0}}}=\dfrac{F{{R}^{2}}}{{{q}^{2}}}$
So, we can write,
$\dfrac{G}{4\pi{{\in}_{0}}}=\dfrac{F{{R}^{2}}}{mM}\times\dfrac{F{{R}^{2}}}{{{q}^{2}}}=\dfrac{{{F}^{2}}{{R}^{4}}}{mM{{q}^{2}}}$
So, to find the dimension of $\dfrac{G}{4\pi {{\in }_{0}}}$, we need to find the dimension of the force, distance mass and charge.

The dimension of mass is $\left[ {{M}^{1}} \right]$
The dimension of distance s $\left[ {{L}^{1}} \right]$
The dimension of force can be given from the definition of force as the product of mass and the acceleration.

Dimension of acceleration is given as, $\left[ {{L}^{1}}{{T}^{-2}} \right]$
So, the dimension of force will be, $\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right]$
Again, charge can be mathematically expressed in terms of the current and time as,
$q=IT$
So, the dimension of charge is, $\left[ {{T}^{1}}{{A}^{1}} \right]$
Now, the dimension of $\dfrac{G}{4\pi {{\in }_{0}}}$can be given as,

$\begin{align}
  & \dfrac{G}{4\pi {{\in }_{0}}}=\dfrac{{{\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right]}^{2}}{{\left[ {{L}^{1}} \right]}^{4}}}{\left[ {{M}^{1}} \right]\left[ {{M}^{1}} \right]{{\left[ {{T}^{1}}{{A}^{1}} \right]}^{2}}} \\
 & \dfrac{G}{4\pi {{\in }_{0}}}=\dfrac{\left[ {{M}^{2}}{{L}^{6}}{{T}^{-4}} \right]}{\left[ {{M}^{2}}{{T}^{2}}{{A}^{2}} \right]} \\
 & \dfrac{G}{4\pi {{\in }_{0}}}=\left[ {{M}^{0}}{{L}^{6}}{{T}^{-6}}{{A}^{-2}} \right] \\
\end{align}$

So, the dimension of $\dfrac{G}{4\pi {{\in }_{0}}}$ is $\left[ {{M}^{0}}{{L}^{6}}{{T}^{-6}}{{A}^{-2}} \right]$
Hence, the correct answer is option C.

Note:
G is a universal constant with its value, $G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$ which is constant in all conditions. The permittivity or the dielectric constant of a material gives the opposition offered by the material to the formation of electric fields through the material.