General Physics Ques 53
Question 4
- Taking the electronic charge as $e$ and the permittivity as $\varepsilon_{0}$, use dimensional analysis to determine the correct expression for $\omega_{p}$. (a) $\sqrt{\frac{N e}{m \varepsilon_{0}}}$ (b) $\sqrt{\frac{m \varepsilon_{0}}{N e}}$ (c) $\sqrt{\frac{N e^{2}}{m \varepsilon_{0}}}$ (d) $\sqrt{\frac{m \varepsilon_{0}}{N e^{2}}}$
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Answer:
Correct Answer: 53.(c)
Solution:
- $N=$ Number of electrons per unit volume
$$ \begin{aligned} \therefore \quad[\mathrm{N}] & =\left[\mathrm{L}^{-3}\right],[\mathrm{e}]=[\mathrm{q}]=[\mathrm{It}]=[\mathrm{AT}] \ {\left[\varepsilon_{0}\right] } & =\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right] \end{aligned} $$
Substituting the dimensions, we can see that
$$ \omega_{p}=\sqrt{\frac{\mathrm{Ne}^{2}}{m \varepsilon_{0}}}=\left[\mathrm{T}^{-1}\right] $$
Angular frequency has also the dimension $\left[\mathrm{T}^{-1}\right]$.
BETA
