If _0 is the permittivity of free space and E is the electric field, then _0 E^2 has the dimensions :

Correct answer: D. [M L^ -1 ~T^ -2 ] aligned & E= KQ R^2 & E= Q 4 π _0 R^2 & _0= Q 4 π R^2 E & Now, _0 E^2= Q 4 π R^2 E · E^2= Q 4 π R^2 · E & [ _0 E^2]=[ QE R^2 ]= [Q][E] [R^2] = [Q] [R^2] [W] [Q][R] & = [W] [R^3] = ML^2 T^ -2 L^3 =ML^ -1 T^ -2 aligned

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Units and Dimensions

JEE Main 2024 · 8 April · Evening Shift · Physics · Units and Dimensions

If _0 is the permittivity of free space and E is the electric...

\epsilon_0

is the permittivity of free space and

\mathrm{E}

is the electric field, then

\epsilon_0 \mathrm{E}^2

has the dimensions :

Options

A $\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$
B $\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}\right]$
C $\left[\mathrm{M}^{\circ} \mathrm{L}^{-2} \mathrm{TA}\right]$
D $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$

Correct answer

D. $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$

Step-by-step 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}

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