Q99 / 200 · NEET 2010 (Screening) · Physics

The dimension of $\frac{1}{2} \varepsilon_0 \mathrm{E}^2$, where $\varepsilon_0$ is permittivity of free space and $E$ is electric field, is

  • A
    $\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
  • Correct answer
    B
    $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$
  • C
    $\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$
  • D
    $\left[\mathrm{MLT}^{-1}\right]$

Solution

$$ \begin{aligned} & \text { Dimensions of } \varepsilon_0=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right] \\ & \text { Dimensions of } \mathrm{E}=\left[\mathrm{MLT}^{-3} \mathrm{~A}^{-1}\right] \\ & \therefore \text { Dimensions of } \frac{1}{2} \varepsilon_0 \mathrm{E}^2=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right] \\ & \quad=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-3}\right] \quad \times\left[\mathrm{M}^2 \mathrm{~L}^2 \mathrm{~T}^{-6} \mathrm{~A}^{-2}\right] \end{aligned} $$

Hence, the correct answer is option: 'B'.

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