# Dimensional analysis doubt!!!??

## A length-scale (l) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. ifnd the dimensions of length.

Aug 12, 2018

Let, $l \propto {\epsilon}^{a} {\left(k B\right)}^{b} {\left(T\right)}^{c} {n}^{d} {q}^{e}$

Now, length = $\left[{M}^{0} L {T}^{0}\right]$

permittivity = $\left[{M}^{-} 1 {L}^{-} 3 {T}^{4} {A}^{2}\right]$

Boltzmann constant = $\left[M {L}^{2} {T}^{-} 2 {\theta}^{-} 1\right]$

Absolute temperature =$\left[{M}^{0} {L}^{0} {T}^{0} \theta\right]$

Number per unit volume = $\left[{M}^{0} {L}^{-} 3 {T}^{-} 0 N\right]$

Charge =$\left[{M}^{0} {L}^{0} T A\right]$

N.B = Current is taken to be the fundamental quantity and its dimension is $\left[A\right]$ so, dimension of charge is $\left[A T\right]$ (as, $Q = I t$)

Now,we can put the values in the equation as,

$\left[{M}^{0} L {T}^{0}\right] = k {\left[{M}^{-} 1 {L}^{-} 3 {T}^{4} {A}^{2}\right]}^{a} {\left[M {L}^{2} {T}^{-} 2 {\theta}^{-} 1\right]}^{b} {\left[{M}^{0} {L}^{0} {T}^{0} \theta\right]}^{c} {\left[{M}^{0} {L}^{-} 3 {T}^{-} 0 N\right]}^{d} {\left[{M}^{0} {L}^{0} T A\right]}^{e}$ ($k$ is a constant)

Now,comparing the dimensional power of each fundamenta quantity we get,

$0 = - a + b \ldots \ldots .1$ (for mass)

$1 = - 3 a + 2 b - 3 d \ldots \ldots \ldots . .2$ (for length)

$0 = 4 a - 2 b + e \ldots \ldots \ldots \ldots \ldots \ldots . .3$ (for time)

$0 = - b + c \ldots \ldots \ldots \ldots \ldots \ldots \ldots 4$ (for temperature)

$0 = 2 a + e \ldots \ldots \ldots \ldots \ldots \ldots 5$ (for current)

$0 = d \ldots \ldots \ldots \ldots \ldots .6$ (for mole)

Solving them,we get,

a=-1, b=-1, c=-1, d=0, e=2

So,dimension of length comes out to be ${\epsilon}^{-} 1 {\left(k B\right)}^{-} 1 {\left(T\right)}^{-} 1 {n}^{0} {q}^{2}$