If $\epsilon_0$ is the permittivity of free space and $\mathrm{E}$ is the electric field, then $\epsilon_0…
If $\epsilon_0$ is the permittivity of free space and $\mathrm{E}$ is the electric field, then $\epsilon_0 \mathrm{E}^2$ has the dimensions :
- $\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$
- $\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}\right]$
- $\left[\mathrm{M}^{\circ} \mathrm{L}^{-2} \mathrm{TA}\right]$
- $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$
Solution
$\begin{aligned} & \mathrm{E}=\frac{\mathrm{KQ}}{\mathrm{R}^2} \\ & \mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}^2} \\ & \varepsilon_0=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2 \mathrm{E}} \\ & \text { Now, } \varepsilon_0 \mathrm{E}^2=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2 \mathrm{E}} \cdot \mathrm{E}^2=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2} \cdot \mathrm{E} \\ & {\left[\varepsilon_0 \mathrm{E}^2\right]=\left[\frac{\mathrm{QE}}{\mathrm{R}^2}\right]=\frac{[\mathrm{Q}][\mathrm{E}]}{\left[\mathrm{R}^2\right]}=\frac{[\mathrm{Q}]}{\left[\mathrm{R}^2\right]} \frac{[\mathrm{W}]}{[\mathrm{Q}][\mathrm{R}]}} \\ & =\frac{[\mathrm{W}]}{\left[\mathrm{R}^3\right]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^3}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\end{aligned}$
Asked in: JEE Main 2024 (08 Apr Shift 2)
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