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Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 8 April, Shift 2 — Question 31

If ϵ0\epsilon_{0} is the permittivity of free space and EE is the electric field, then ∈0E2\in_{0} E^{2} has the dimensions :

  1. Option A:

    [M0 L−2 T\left[\mathrm{M}^{0} \mathrm{~L}^{-2} \mathrm{~T}\right. A]

  2. Option B:

    [ML−1 T−2]\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]

    Correct
  3. Option C:

    [M−1 L−3 T4 A2]\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]

  4. Option D:

    [ML2 T−2]\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-2}\right]

Answer: B

Step-by-step solution

E=KQR2E=\frac{K Q}{\mathrm{R}^{2}}

E=Q4πε0R2\mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}

ε0=Q4πR2E\varepsilon_{0}=\frac{Q}{4 \pi R^{2} E}

Now, ε0E2=Q4πR2E.E2=Q4πR2.E\varepsilon_{0} \mathrm{E}^{2}=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^{2} \mathrm{E}} . \mathrm{E}^{2}=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^{2}} . \mathrm{E} [ε0E2]=[QER2]=[Q][E][R2]=[Q][R2][W][Q][R]\left[\varepsilon_{0} \mathrm{E}^{2}\right]=\left[\frac{\mathrm{QE}}{\mathrm{R}^{2}}\right]=\frac{[\mathrm{Q}][\mathrm{E}]}{\left[\mathrm{R}^{2}\right]}=\frac{[\mathrm{Q}]}{\left[\mathrm{R}^{2}\right]} \frac{[\mathrm{W}]}{[\mathrm{Q}][\mathrm{R}]}

=[W][R3]=ML2 T−2 L3=ML−1 T−2=\frac{[\mathrm{W}]}{\left[\mathrm{R}^{3}\right]}=\frac{\mathrm{ML}^{2} \mathrm{~T}^{-2}}{\mathrm{~L}^{3}}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 2
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis

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