JEE Advance - Physics (2011 - Paper 1 Offline - No. 1)
Taking the electronic charge as 'e' and the permittivity as $$'{\varepsilon _0}'$$. Use dimensional analysis to determine the
correct expression for $${\omega _p}$$.
$$\sqrt {{{Ne} \over {m{\varepsilon _0}}}} $$
$$\sqrt {{{m{\varepsilon _0}} \over {Ne}}} $$
$$\sqrt {{{N{e^2}} \over {m{\varepsilon _0}}}} $$
$$\sqrt {{{m{\varepsilon _0}} \over {N{e^2}}}} $$
Explanation
We have $$[\omega ] = {T^{ - 1}}$$ and
$$[{\varepsilon _0}] = {{{{[QT]}^2}} \over {M{L^3}}}$$
Therefore,
$$\left[ {{1 \over {m{\varepsilon _0}}}} \right] = {{{L^3}} \over {{{[QT]}^2}}}$$
Now, $$\left[ {{{{e^2}} \over {m{\varepsilon _0}}}} \right] = {{{L^3}} \over {{{[T]}^2}}}$$
Also, the number N is defined as number of particle in unit volume, that is, N = n/V.
$$[N] = {1 \over {[V]}} = {1 \over {{L^3}}}$$
Therefore,
$$\left[ {{{N{e^2}} \over {m{\varepsilon _0}}}} \right] = {1 \over {{{[T]}^2}}}$$
Therefore, the quantity $$\sqrt {{{N{e^2}} \over {m{\varepsilon _0}}}} $$ has the dimension of $$\omega$$.
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