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# A length scale $\left( l \right)$ depends on the permittivity $\left( \varepsilon \right)$ of a dielectric material, Boltzmann constant $\left( {{k_B}} \right)$ , the absolute temperature $\left( T \right)$ , the number per unit volume $\left( n \right)$ of certain charged particles and the charge $\left( q \right)$ carried by each of the particles. Which of the following expression(s) for $l$ is (are) dimensionally correct?This question has multiple correct options.A) $l = \sqrt {\left( {\dfrac{{n{q^2}}}{{\varepsilon {k_B}T}}} \right)}$B) $l = \sqrt {\left( {\dfrac{{\varepsilon {k_B}T}}{{n{q^2}}}} \right)}$C) $l = \sqrt{\dfrac{q^2}{\varepsilon n^{2/3} k_b T}}$D) $l = \sqrt{\dfrac{q^2}{\varepsilon n^{1/3} k_b T}}$

Last updated date: 19th Sep 2024
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Answer
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Hint: The dimension of $l$ is $\left[ L \right]$ . So, find the dimensions of RHS of all the options and match with that of the length. You have to find out the dimensions of $\dfrac{{{q^2}}}{\varepsilon },{k_B}T$ and $n$ . Option A and B are reciprocal of each other so they can’t be correct simultaneously. Similarly, in option C and D, the raised power of $n$ is different so they also can’t be correct simultaneously.

Complete step by step answer:
As this question has multiple correct options. So, the best way to solve it to eliminate the incorrect options or to check all the options.
We know that the dimension of $l$ is $\left[ L \right]$, so we have to find the dimensions of RHS of all the options and match with that of the length. As all the options contain some common terms that are $\dfrac{{{q^2}}}{\varepsilon },{k_B}T$ and $n$ , so we to find out the dimensions of these quantities.
We know that the potential energy of two equal charges $q$ separated by a distance $r$ from it is given by $P.E = \dfrac{{{q^2}}}{{4\pi {\varepsilon _0}r}}$
So, $\dfrac{{{q^2}}}{\varepsilon } = P.E \times r \times \left( {4\pi } \right)$
For calculating the dimension we can take ${\varepsilon _0}$ as $\varepsilon$ as both will have the same dimension. And $4\pi$ is a dimensionless quantity.
Dimension of $P.E = \left[ {M{L^2}{T^{ - 2}}} \right]$
Dimension of $r = \left[ L \right]$
Therefore, the dimension of $\dfrac{{{q^2}}}{\varepsilon } = \left[ {M{L^2}{T^{ - 2}}} \right]\left[ L \right] = \left[ {M{L^3}{T^{ - 2}}} \right]$
Now, we know that the kinetic energy $KE = \dfrac{3}{2}{k_B}T$ and dimension of $KE = \left[ {M{L^2}{T^{ - 2}}} \right]$ .
As $\dfrac{3}{2}$ is a dimensionless quantity then the dimension of ${k_B}T = \left[ {M{L^2}{T^{ - 2}}} \right]$
Now, $n$ is the number per unit volume and the dimension of volume is $\left[ {{L^3}} \right]$
Therefore the dimension of $n = \left[ {{L^{ - 3}}} \right]$ .
Now, we will find the dimension of all the options.
For option A, dimension will be $\sqrt {\dfrac{{\left[ {{L^{ - 3}}} \right]\left[ {M{L^3}{T^{ - 2}}} \right]}}{{\left[ {M{L^2}{T^{ - 2}}} \right]}}} = \left[ {{L^{ - 1}}} \right]$
For option B, dimension will be $\sqrt {\dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {{L^{ - 3}}} \right]\left[ {M{L^3}{T^{ - 2}}} \right]}}} = \left[ L \right]$
For option C, dimension will be $\sqrt{\dfrac{[ML^3 T^{-2}]}{[L^{-3}]^{2/3 }[ML^2 T^{-2}]}}= [L^{3/2}]$
For option D, dimension will be $\sqrt{\dfrac{[ML^3 T^{-2}]}{[L^{-3}]^{1/3 }[ML^2 T^{-2}]}}= [L]$
Therefore the dimensions of option B and D are matched with the dimension of length.

Hence, option B and D are correct.

Note: Dimensions of any physical quantity are those raised powers on base units to specify its unit. Dimensional formula is the expression which shows how and which of the fundamental quantities represent the dimensions of a physical quantity.