Question

Taking the electronic charge as $'c'$ and the permittivity as $'{ \in _0}'$, use dimensional analysis to determine the correct expression for ${w_P}$.
(A) $\sqrt {\dfrac{{Ne}}{{m{ \in _0}}}} $
(B) $\sqrt {\dfrac{{m{ \in _0}}}{{Ne}}} $
(C) $\sqrt {\dfrac{{N{e^2}}}{{m{ \in _0}}}} $
(D) $\sqrt {\dfrac{{m{ \in _0}}}{{N{e^2}}}} $

Answer 1

Hint:In order to solve this problem we use the dimensional formula of angular frequency ${w_P}$ which is
${w_P} = \dfrac{Q}{t} = [{T^{ - 1}}]$
After that we use the dimensional formulas of charge mass, permittivity ${ \in _0}$ and units of density N.

Complete Step by Step Answer:
We know that the angular frequency is given as
${w_P} = \dfrac{Q}{t}$
Where
Q $ = $ angular displacement
t $ = $ time
So, is the dimensional formula of ${w_P}$
${w_P} = [{T^{ - 1}}]$ …..(1)
We know that the dimensional formula of density N is
$N = \dfrac{1}{{{L^3}}} = [{L^{ - 3}}]$ …..(2)
The dimensional formulas of charge e, mass M and permittivity ${ \in _0}$ is given as
$ch\arg e(e) = current(I) \times time(t)$
$e = [{A^1}{T^1}]$ …..(3)
$Mass(M) = [{M^1}]$ …..(4)
$Permittivity({ \in _0}) = \dfrac{{{q_1}{q_2}}}{{4\pi F{r^2}}}$
The dimensional formula of force
$F = [{M^1}{L^1}{T^{ - 2}}]$
So, ${ \in _0} = \dfrac{{[{A^2}{T^2}]}}{{[{M^1}{L^1}{T^{ - 2}}][{L^2}]}}$
${ \in _0} = [{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}]$ …..(5)
To determine the relation between ${w_P}$ and e, ${ \in _0}$, M and N, we use dimensional analysis.
$[{T^{ - 1}}] = {[N]^a}{[e]^b}{[N]^c}{[{ \in _0}]^d}$ …..(6)
$[{T^{ - 1}}] = {[{L^{ - 3}}]^a}{[{A^1}{T^1}]^b}{[M]^c}{[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}]^d}$
$[{T^{ - 1}}] = [{M^{c - d}}][{L^{ - 3a - 3d}}][{T^{a + 4d}}][{A^{b + 2d}}]$
On comparing the powers of M, L, I & T
$c - d = 0$ $ - 3a - 3d = 0$
$c = d$ $a = - d$
$b + 2d = 0$ and $b + 4d = - 1$
$b = - 2d$ So, $ - 2d + 4d = - 1$
                    $2d = - 1$
$d = - \dfrac{1}{2}$ …..(7)
So, $b = - 2 \times \left( { - \dfrac{1}{2}} \right)$
$b = 1$ …..(8)
& $c = - \dfrac{1}{2}$ …..(9)
& $a = \dfrac{1}{2}$ …..(10)
From equation 6, 7, 8, 9 & 10
${w_P} = {N^{1/2}}{e^1}{M^{ - 1/2}} \in _0^{ - 1/2}$
${w_P} = \dfrac{{{N^{1/2}}e}}{{{M^{1/2}} \in _0^{1/2}}}$
${w_P} = \sqrt {\dfrac{{N{e^2}}}{{M{ \in _0}}}} $
Hence option C is correct answer $\sqrt {\dfrac{{N{e^2}}}{{m{ \in _0}}}} $

Note: In order to derive the relation between physical quantities we use the dimensional formula. Also we can check the correctness of physical expressions. You can make mistakes in applying the dimensional formula for some standard quantities and also in comparing it.