Units & Dimensions

JEE Main level, single correct, medium

54 of 149

Question 54 of 149

If ϵ0\epsilon_0 is the permittivity of free space and E\mathrm{E} is the electric field, then ϵ0E2\epsilon_0 \mathrm{E}^2 has the dimensions :

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Answer: (D) [ML−1 T−2]\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]

E=KQR2E=Q4πε0R2ε0=Q4πR2E Now, ε0E2=Q4πR2E⋅E2=Q4πR2⋅E[ε0E2]=[QER2]=[Q][E][R2]=[Q][R2][W][Q][R]=[W][R3]=ML2 T−2 L3=ML−1 T−2\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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