The dimensions of \(\frac{1}{2} \varepsilon_{0} E^{2}\left(\varepsilon_{0}=\right.\) permittivity of free…

The dimensions of \(\frac{1}{2} \varepsilon_{0} E^{2}\left(\varepsilon_{0}=\right.\) permittivity of free space and \(E=\) electric field) are
  1. \(\left.\mid M L^{2} T^{-1}\right]\)
  2. \(\left[M L^{-1} T^{-2}\right]\)
  3. \(\left[M L^{2} T^{- 2}\right]\)
  4. \(\left[M L T^{-1}\right]\)

Solution

\(\frac{1}{2} \varepsilon_{0} E^{2}\) is the expression for electrostatic energy density, i.e., the energy stored per unit volume in a parallel plate capacitor.
\(\begin{aligned}
\therefore \frac{1}{2} \varepsilon_{0} E^{2} &=\frac{\text { energy }}{\text { volume }} \\
&=\frac{M L^{2} T^{-2}}{L^{3}}=\left[M L^{-1} T^{-2}\right]
\end{aligned}\)
Alternatively, \(\varepsilon_{0}=\frac{1}{4 \pi F} \times \frac{q_{1} q_{2}}{r^{2}}\) and \(E=\frac{F}{q}\)
\(\therefore \frac{1}{2} \varepsilon_{0} E^{2}=\frac{1}{8 \pi F} \frac{q 1 q 2}{r^{2}} \times \frac{F^{2}}{q^{2}}=\frac{F}{r^{2}}=\frac{M L T^{-2}}{L^{2}}\) .

Asked in: JEE Mains - Units and Dimensions - Chapter Test

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