If ∈0 is the permittivity of free space and E is the electric field, then ∈0 E^2 has the dimensions :

Question

If \(\varepsilon_{0}\) is the permittivity of free space and E is the electric field, then \(\varepsilon_{0} \mathrm{E}^{2}\) has the dimensions :

(1) \(\left[\mathrm{M}^{0} \mathrm{L}^{-2} \mathrm{T\ A}\right]\)

(2)\( \left[\mathrm{M} \ \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]\)

(3) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]\)

(4) \(\left[\mathrm{M} \ \mathrm{L}^{2} \mathrm{~T}^{-2}\right]\)   

Answer 1

Correct option is : (2) \(\left[\mathrm{M} \ \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]\) 

\(E=\frac{K Q}{R^{2}}\)

\(\mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{\mathrm{o}} \mathrm{R}^{2}}\)

\(\varepsilon_{0}=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^{2} \mathrm{E}}\)

Now, \(\varepsilon_{0} \mathrm{E}^{2}=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^{2} \mathrm{E}} \cdot \mathrm{E}^{2}=\frac{\mathrm{Q}}{4 \pi \mathrm{R}^{2}} \cdot \mathrm{E}\)  

\(\left[\varepsilon_{0} \mathrm{E}^{2}\right]=\left[\frac{\mathrm{QE}}{\mathrm{R}^{2}}\right]=\frac{[\mathrm{Q}][\mathrm{E}]}{\left[\mathrm{R}^{2}\right]}=\frac{[\mathrm{Q}]\ \ \ [W]}{\left[\mathrm{R}^{2}\right][\mathrm{Q}][\mathrm{R}]}\)

\(=\frac{[\mathrm{W}]}{\left[\mathrm{R}^{3}\right]}=\frac{\mathrm{ML}^{2} \mathrm{~T}^{-2}}{\mathrm{~L}^{3}}=\mathrm{ML}^{-1} \mathrm{T}^{-2}\)