1.Units, Dimensions and Measurement
medium

જો $\varepsilon_0$ મુક્ત અવકાશની પરાવૈધતાંક અને $\mathrm{E}$ વિધુત ક્ષેત્ર હોય તો $\varepsilon_0 \mathrm{E}^2$ નું પરિમાણ. . . . . . . . .છે.

A$\left[\mathrm{M}^0 \mathrm{~L}^{-2} \mathrm{TA}\right]$
 
B$\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$
C$\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$
D$\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}\right]$
(JEE MAIN-2024)

Solution

$\mathrm{E}=\frac{\mathrm{K}}{\mathrm{R}^2}$
$\mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}^2}$
$\varepsilon_0=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2 \mathrm{E}}$
$\text { Now, } \varepsilon_0 \mathrm{E}^2=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2\mathrm{E}} \cdot \mathrm{E}^2=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2} \cdot \mathrm{E}$
${\left[\varepsilon_0 \mathrm{E}^2\right]=\left[\frac{\mathrm{QE}}{\mathrm{R}^2}\right]=\frac{[\mathrm{Q}][\mathrm{E}]}{\left[\mathrm{R}^2\right]}=\frac{[\mathrm{Q}]}{\left[\mathrm{R}{ }^2\right][\mathrm{Q}][\mathrm{R}]}}$
$=\frac{[\mathrm{W}]}{\left[\mathrm{R}^3\right]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^3}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}$
Standard 11
Physics

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