Question

Taking the electronic charge as '$e$' and the permittivity as ${}^{\prime }{ϵ}_0^{\prime }$, use dimensional analysis to determine the correct expression for ${\omega }_p$.

• A. $\sqrt{\frac{Ne}{m{ϵ}_0}}$
• B. $\sqrt{\frac{m{ϵ}_0}{Ne}}$
• C. $\sqrt{\frac{N{e}^2}{m{ϵ}_0}}$
• D. $\sqrt{\frac{m{ϵ}_0}{N{e}^2}}$

Answer 1

Answer: (C)

$\sqrt{\frac{N{e}^2}{m{ϵ}_0}}$

We have ${\omega }_p$ as the angular frequency.
Thus its unit would be given as $\frac{\theta }{t}=\left[T^{-1}\right]$
We have the units of density $N$ as $\frac{1}{L^3}$, charge $e$ as $IT$, mass as $M$ and ${ϵ}_o$ is evaluated using the force equation $F=\frac{1}{4\pi {ϵ}_o}\frac{q_1{q}_2}{r^2}$ as $\frac{I^2{T}^4}{M{L}^3}$

By evaluating the units of each given option we get.
From dimension analysis: $[T{]}^{-1}=[N{]}^x\left[e{]}^y\right[M{]}^z[{ϵ}_0{]}^t$...(i)

Substituting the units:
$[T{]}^{-1}=[\frac{1}{L^3}{]}^x\left[IT{]}^y\right[M{]}^z[\frac{I^2{T}^4}{M{L}^3}{]}^t$

Comparing the powers of $M,L,I$ and $T$.
$z-t=0$ ...(ii)
$-3x-3t=0$ ...(iii)
$y+2t=0$ ...(iv)
$y+4t=-1$ ...(v)

Solving equations $y=1;t=-\frac{1}{2};z=-\frac{1}{2};x=\frac{1}{2}$

Substituting $x,y,z$ and $t$ in equation (i)
${\omega }_p=\frac{\sqrt{N}e}{\sqrt{m}\sqrt{{ϵ}_0}}=\sqrt{\frac{N{e}^2}{m{ϵ}_0}}$