Question

what is the dimension for permitivity of free space?

Answer 1

Dear Student,

$$\begin{gathered} {\mathrm{\epsilon }}_0=\frac{{\mathrm{q}}^2}{{\mathrm{Fd}}^2} \\ \mathrm{where},{\mathrm{\epsilon }}_0=\mathrm{permittivity} \\ \mathrm{q}=\mathrm{charge} \\ \mathrm{F}=\mathrm{force} \\ \mathrm{d}=\mathrm{distance} \\ \mathrm{According}\mathrm{to}\mathrm{their}\mathrm{dimensions}: \\ {\mathrm{\epsilon }}_0=\frac{{\left[{\mathrm{M}}^0{\mathrm{L}}^0\mathrm{TA}\right]}^2}{\left[{\mathrm{M}}^1{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]{\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]}^2} \\ \mathrm{where},\mathrm{A}=\mathrm{dimensional}\mathrm{formula}\mathrm{of}\mathrm{electric}\mathrm{charge} \\ {\mathrm{\epsilon }}_0=\frac{{\left[{\mathrm{M}}^0{\mathrm{L}}^0\mathrm{TA}\right]}^2}{\left[{\mathrm{M}}^1{\mathrm{L}}^{1+2}{\mathrm{T}}^{-2}\right]}=\frac{\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^2{\mathrm{A}}^2\right]}{\left[{\mathrm{M}}^1{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]} \\ {\mathrm{\epsilon }}_0=\left[{\mathrm{M}}^{0-1}{\mathrm{L}}^{0-3}{\mathrm{T}}^{2+2}{\mathrm{A}}^2\right] \\ {\mathrm{\epsilon }}_0=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{A}}^2\right] \end{gathered}$$