Question
The dimensions of $\frac{1}{2} \varepsilon_{0} E^{2}\left(\varepsilon_{0}=\right.$ permittivity of free space ; $E=$ electric field ) is(a) $M L T^{-1}$(b) $M L^{2} T^{-2}$(c) $M L^{-1} T^{-2}$(d) $M L^{2} T^{-1}$
Step 1
The energy density is given by the formula $\frac{1}{2} \varepsilon_{0} E^{2}$, where $\varepsilon_{0}$ is the permittivity of free space and $E$ is the electric field. Show more…
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The dimensions of $\frac{1}{2} \varepsilon_{0} E^{2}\left(\varepsilon_{0}=\right.$ permittivity of free space and $E=$ electric field are a. $\left[M L^{2} T^{-1}\right]$ b. $\left[M L^{-1} T^{-2}\right]$ c. $\left[M L^{2} T^{-2}\right]$ d. $\left[M L T^{-1}\right]$
The dimensions of permittivity $\varepsilon_{0}$ are $\quad[B V P 2006]$ (a) $\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~A}^{2} \mathrm{~T}^{4}\right]$ (b) $\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~A}^{-2} \mathrm{~T}^{-4}\right]$ (c) $\left[\mathrm{M}^{-1} \mathrm{~L}^{-1} \mathrm{~A}^{2} \mathrm{~T}^{2}\right]$ (d) $\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~A}^{2} \mathrm{~T}^{-4}\right]$
Units and Measurements
Round 2
If $X=\varepsilon_{0} L \frac{\Delta V}{\Delta t}$ Here, $\varepsilon_{0}=$ electric permittivity of free space $L=$ length $\Delta V=$ potential difference $\Delta t=$ time interval What are the dimensions of $X ?$ Guess the physical quantity: (a) Electric current, $\left[\mathrm{A}^{0} \mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$ (b) Electric potential, $\left[\mathrm{AM}^{0} \mathrm{~L}^{0} \mathrm{~T}^{8}\right]$ (c) Electric current, $\left[\mathrm{AM}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$ (d) None of the above
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