Question

Derive dimensional formula of. 1/2.Epsilon not E^2

Answer 1

Dear Student,

$$\begin{gathered} Weknowthat \\ F=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_2}{r^2} \\ whereFisforce, \\ {q}_{1}and{q}_{2}arecharge \\ r=dis\tan ce \\ {\mathrm{\epsilon }}_0=\frac{1}{4\mathrm{\pi F}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2} \\ where\frac{1}{4\mathrm{\pi }}\mathrm{is}\mathrm{constant} \\ {\mathrm{\epsilon }}_0=\frac{1}{\left[{\mathrm{MLT}}^{-2}\right]}\times \frac{\left[\mathrm{AT}\right]\left[\mathrm{AT}\right]}{\left[{\mathrm{L}}^2\right]} \\ {\mathrm{\epsilon }}_0=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{A}}^2\right] \\ \mathrm{Also}, \\ \mathrm{F}=\mathrm{qE} \\ \mathrm{where}\mathrm{E}\mathrm{is}\mathrm{electric}\mathrm{field} \\ \mathrm{E}=\frac{\mathrm{F}}{\mathrm{q}} \\ \mathrm{E}=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[\mathrm{AT}\right]}=\left[{\mathrm{MLT}}^{-3}\mathrm{A}\right] \\ \mathrm{Now}\mathrm{dimensional}\mathrm{formula}\mathrm{of}\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}^2\mathrm{is} \\ \left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{A}}^2\right]\times {\left[{\mathrm{MLT}}^{-3}\mathrm{A}\right]}^2 \\ \mathrm{where}\frac{1}{2}\mathrm{is}\mathrm{constant} \\ \left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{A}}^2\right]\times \left[{\mathrm{M}}^2{\mathrm{L}}^2{\mathrm{T}}^{-6}{\mathrm{A}}^2\right] \\ \left[{\mathrm{M}}^1{\mathrm{L}}^{-1}{\mathrm{T}}^{-2}{\mathrm{A}}^4\right] \end{gathered}$$