JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement

Let \([ {\varepsilon _0} ]\) denote the dimensional formula of the permittivity of vacuum. If \(M =\) mass, \(L=\) length, \(T =\) time and \(A=\) electric current, then:

A \(\;{\varepsilon _0}=M^{-1}L^{-3}T^2A\)
B \(\;{\varepsilon _0}=M^{-1}L^{-3}T^4A^2\)
C \(\;{\varepsilon _0}=M^{-1}L^2T^{-1}A^{-2}\)
D \({\varepsilon _0}=M^{-1}L^2T^{-1}A\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\;{\varepsilon _0}=M^{-1}L^{-3}T^4A^2\)

Step-by-step Solution

Detailed explanation

As we know \(F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{R^2}}} \Rightarrow {\varepsilon _0} = \frac{{{q_1}{q_2}}}{{4\pi F{R^2}}}\) Hence…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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