JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Application +4 | -1

\varepsilon_0

is the permittivity of free space and

E

is the electric field, then

\varepsilon_0 E^2

has the dimensions:

View answer and explanation

Correct option: B

\left[{M} {L}^{-1}{T}^{-2}\right]

Option B ([M][L]^-1[T]^-2) is correct because it represents the dimensions of energy density, which is equivalent to \

\varepsilon_0 E^2

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