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Question: Taking the electronic charge as \('c'\) and the permittivity as \('{ \in _0}'\), use dimensional ana...

Taking the electronic charge as ′c′'c' and the permittivity as ′∈0′'{ \in _0}', use dimensional analysis to determine the correct expression for wP{w_P}.
(A) Nem∈0\sqrt {\dfrac{{Ne}}{{m{ \in _0}}}}
(B) m∈0Ne\sqrt {\dfrac{{m{ \in _0}}}{{Ne}}}
(C) Ne2m∈0\sqrt {\dfrac{{N{e^2}}}{{m{ \in _0}}}}
(D) m∈0Ne2\sqrt {\dfrac{{m{ \in _0}}}{{N{e^2}}}}

Explanation

Solution

In order to solve this problem we use the dimensional formula of angular frequency wP{w_P} which is
wP=Qt=[T−1]{w_P} = \dfrac{Q}{t} = [{T^{ - 1}}]
After that we use the dimensional formulas of charge mass, permittivity ∈0{ \in _0} and units of density N.

Complete Step by Step Answer:
We know that the angular frequency is given as
wP=Qt{w_P} = \dfrac{Q}{t}
Where
Q == angular displacement
t == time
So, is the dimensional formula of wP{w_P}
wP=[T−1]{w_P} = [{T^{ - 1}}] …..(1)
We know that the dimensional formula of density N is
N=1L3=[L−3]N = \dfrac{1}{{{L^3}}} = [{L^{ - 3}}] …..(2)
The dimensional formulas of charge e, mass M and permittivity ∈0{ \in _0} is given as
charg⁡e(e)=current(I)×time(t)ch\arg e(e) = current(I) \times time(t)
e=[A1T1]e = [{A^1}{T^1}] …..(3)
Mass(M)=[M1]Mass(M) = [{M^1}] …..(4)
Permittivity(∈0)=q1q24πFr2Permittivity({ \in _0}) = \dfrac{{{q_1}{q_2}}}{{4\pi F{r^2}}}
The dimensional formula of force
F=[M1L1T−2]F = [{M^1}{L^1}{T^{ - 2}}]
So, ∈0=[A2T2][M1L1T−2][L2]{ \in _0} = \dfrac{{[{A^2}{T^2}]}}{{[{M^1}{L^1}{T^{ - 2}}][{L^2}]}}
∈0=[M−1L−3T4A2]{ \in _0} = [{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}] …..(5)
To determine the relation between wP{w_P} and e, ∈0{ \in _0}, M and N, we use dimensional analysis.
[T−1]=[N]a[e]b[N]c[∈0]d[{T^{ - 1}}] = {[N]^a}{[e]^b}{[N]^c}{[{ \in _0}]^d} …..(6)
[T−1]=[L−3]a[A1T1]b[M]c[M−1L−3T4A2]d[{T^{ - 1}}] = {[{L^{ - 3}}]^a}{[{A^1}{T^1}]^b}{[M]^c}{[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}]^d}
[T−1]=[Mc−d][L−3a−3d][Ta+4d][Ab+2d][{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=0c - d = 0 −3a−3d=0 - 3a - 3d = 0
c=dc = d a=−da = - d
b+2d=0b + 2d = 0 and b+4d=−1b + 4d = - 1
b=−2db = - 2d So, −2d+4d=−1 - 2d + 4d = - 1
2d=−12d = - 1
d=−12d = - \dfrac{1}{2} …..(7)
So, b=−2×(−12)b = - 2 \times \left( { - \dfrac{1}{2}} \right)
b=1b = 1 …..(8)
& c=−12c = - \dfrac{1}{2} …..(9)
& a=12a = \dfrac{1}{2} …..(10)
From equation 6, 7, 8, 9 & 10
wP=N1/2e1M−1/2∈0−1/2{w_P} = {N^{1/2}}{e^1}{M^{ - 1/2}} \in _0^{ - 1/2}
wP=N1/2eM1/2∈01/2{w_P} = \dfrac{{{N^{1/2}}e}}{{{M^{1/2}} \in _0^{1/2}}}
wP=Ne2M∈0{w_P} = \sqrt {\dfrac{{N{e^2}}}{{M{ \in _0}}}}
Hence option C is correct answer Ne2m∈0\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.