Question

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

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

Answer 1

The correct option is C $\sqrt{\frac{N{e}^2}{m{\epsilon }_0}}$

Dimension of $N=\left[L^{-3}\right]$

Dimension of $e=\left[AT\right]$

Dimension of ${ϵ}_o$, using $F=\frac{1}{4\pi {ϵ}_o}\frac{qQ}{r^2}$

So dimension of ${ϵ}_o=\frac{\left[AT\right]\left[AT\right]}{\left[ML{T}^{-2}\right]\left[L^2\right]}=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

Dimension of angular frequency, ${\omega }_p=\left[T^{-1}\right]$

Using dimension analysis, let's say,

${\omega }_p=N^w{ϵ}^x{e}^y{m}^z$

$\left[T^{-1}\right]=\left[L^{-3}{]}^w\right[M^{-1}{L}^{-3}{T}^4{A}^2{]}^x\left[AT{]}^y\right[M{]}^z$

$\left[T^{-1}\right]=\left[M{]}^{-x+z}\right[L{]}^{-3x-3z}\left[T{]}^{4x+y}\right[A{]}^{2x+y}$

Comparing the two sides,

$-x+z=0⟹z=x$

$4x+y=-1$

$2x+y=0$

$-3x-3w=0⟹w=-x$

Solving second and third equation, $x=-\frac{1}{2},y=1$

Hence, $z=-\frac{1}{2},w=\frac{1}{2}$

${\omega }_p=\sqrt{\frac{N{e}^2}{{ϵ}_{o}m}}$