Dimensions: Practice Problem & Solution
The dimension of \(\frac{1}{2}\epsilon_0 E^2\), where \(\epsilon_0\) is permittivity of free space and \(E\) is electric field, is: [2010 Pre]
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Dimensions. Understanding the underlying formula is key to arriving at the correct answer below:
The expression \(\frac{1}{2}\epsilon_0 E^2\) represents electric energy density, which is Energy per unit Volume. \([\text{Energy density}] = [\text{Energy}]/[\text{Volume}] = (ML^2T^{-2})/L^3 = ML^{-1}T^{-2}\).
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