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Units and Measurement question

2010 · Q188
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Units and Measurement question

2010 · Q188

NEETPhysicsUnits and MeasurementMCQ+4 / −1
The dimension of 12ε0E2{1 \over 2}{\varepsilon _0}{E^2}21​ε0​E2, where ε0{\varepsilon _0}ε0​ is permittivity of free space and E is electric field, is
  1. A
    ML2T−-−2
  2. B
    ML−-−1T−-−2
  3. C
    ML2T−-−1
  4. D
    MLT−-−1
View written solutionFree

Correct answer: B

12ε0E2{1 \over 2}{\varepsilon _0}{E^2}21​ε0​E2 = Energy density of an electric field E

Also Energy density = EnergyVolume{{Energy} \over {Volume}}VolumeEnergy​ = [ML2T−2][L3]{{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {\left[ {{L^3}} \right]}}[L3][ML2T−2]​ = [ML−1T−2]\left[ {M{L^{ - 1}}{T^{ - 2}}} \right][ML−1T−2]

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