The dimensional formula of frac{1}{2}epsilon_ | vedclass

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Question

(B) The expression $\frac{1}{2}\epsilon_0 E^2$ represents the energy density of an electric field,which is defined as energy per unit volume.Dimension

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$1$

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(B) The expression $\frac{1}{2}\epsilon_0 E^2$ represents the energy density of an electric field,which is defined as energy per unit volume.Dimensions of energy = $[ML^2 T^{-2}]$.Dimensions of volume = $[L^3]$.Therefore,dimensions of energy density = $\frac{[ML^2 T^{-2}]}{[L^3]} = [ML^{-1} T^{-2}]$.Comparing this with $[M^a L^b T^c]$,we get $a = 1$,$b = -1$,and $c = -2$.Now,calculating the value of $2a - b + c$:$2(1) - (-1) + (-2) = 2 + 1 - 2 = 1$.

The dimensional formula of $\frac{1}{2}\epsilon_0 E^2$ ($\epsilon_0 = $ permittivity of vacuum and $E = $ electric field) is $M^a L^b T^c$. The value of $2a - b + c$ is . . . . . . .

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