If varepsilon_0 is the permittivity of free s | vedclass

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Question

$\mathrm{E}=\frac{\mathrm{K}}{\mathrm{R}^2}$
$\mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}^2}$
$\varepsilon_0=\frac{\mathrm{Q}}{4 \pi

Vedclass · Accepted Answer

$\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

Vedclass · Answer

$\mathrm{E}=\frac{\mathrm{K}}{\mathrm{R}^2}$
$\mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}^2}$
$\varepsilon_0=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2 \mathrm{E}}$
$	ext { 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}$
${\left[\varepsilon_0 \mathrm{E}^2ight]=\left[\frac{\mathrm{QE}}{\mathrm{R}^2}ight]=\frac{[\mathrm{Q}][\mathrm{E}]}{\left[\mathrm{R}^2ight]}=\frac{[\mathrm{Q}]}{\left[\mathrm{R}{ }^2ight][\mathrm{Q}][\mathrm{R}]}}$
$=\frac{[\mathrm{W}]}{\left[\mathrm{R}^3ight]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^3}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}$

If $\varepsilon_0$ is the permittivity of free space and $E$ is the electric field, then $\varepsilon_0 E^2$ has the dimensions

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