The dimension of left- dfrac - 1 - 2 - right- - varepsilon - 0 - E - 2 - is- varepsilon - 0 - permittivity of space- Electric field- ML - -1 - T - -2 - ML - 2 - T - -1 - MLT - -1 - ML - 2 - T - -2 -

Question

Toppr Admin · Accepted Answer

As we know that-xA0-Energy Density-energyvolume-12-x3B5-0E2Hence -xA0-Dimesion of -12-x3B5-0E2 is-xA0- -xA0- -xA0-ML2T-x2212-2-L3-ML-x2212-1T-x2212-2

(a) Angular Momentum	(e) $[{M L}^{2} T^{- 3}]$
(b) Impulse	(f) $[{M L}^{2} T^{- 1}]$
(c) Pressure	(g) $[M L T^{- 1}]$
(d) Power	(h) $[{M L}^{- 1} T^{- 2}]$

The dimension of $(\frac{1}{2}) ε_{0} E^{2}$ is( $ε_{0}$ : permittivity of free space, Electric field)
${M L}^{- 1} T^{- 2}$
${M L}^{2} T^{- 1}$
${M L T}^{- 1}$
${M L}^{2} T^{- 2}$

As we know that
Energy Density $= \frac{e n e r g y}{v o l u m e} = (\frac{1}{2}) ε_{0} E^{2}$

Hence ,
Dimesion of $(\frac{1}{2}) ε_{0} E^{2}$ is, = $\frac{[M L^{2} T^{- 2}]}{L^{3}} = M L^{- 1} T^{- 2}$

The dimension of (12)ε0E2 is(ε0: permittivity of free space, Electric field)ML−1T−2ML2T−1MLT−1ML2T−2

As we know that Energy Density=energyvolume=(12)ε0E2Hence , Dimesion of (12)ε0E2 is, =[ML2T−2]L3=ML−1T−2

The dimension of $(\frac{1}{2}) ε_{0} E^{2}$ is( $ε_{0}$ : permittivity of free space, Electric field)
${M L}^{- 1} T^{- 2}$
${M L}^{2} T^{- 1}$
${M L T}^{- 1}$
${M L}^{2} T^{- 2}$

As we know that
Energy Density $= \frac{e n e r g y}{v o l u m e} = (\frac{1}{2}) ε_{0} E^{2}$

Hence ,
Dimesion of $(\frac{1}{2}) ε_{0} E^{2}$ is, = $\frac{[M L^{2} T^{- 2}]}{L^{3}} = M L^{- 1} T^{- 2}$