Question

Show that the dimensions of the displacement current ${\mathrm{\epsilon }}_0\frac{d{\phi }_E}{dt}$ are that of an electric current.

Answer 1

Displacement current,
$I_{\mathrm{D}}=\frac{{\in }_{0}d{\phi }_{\mathrm{e}}}{dt}$
Electric flux,
${\varphi }_{\mathrm{e}}=EA$
$$\begin{gathered} \left[{\varphi }_e\right]\mathit{=}\mathit{}\left[E\right]\left[A\right] \\ =\left[\frac{\mathit{1}}{\mathit{4}\pi {\mathit{\in }}_{\mathit{0}}}\frac{q}{r^{\mathit{2}}}\right]\left[A\right] \\ \mathrm{Also},\left[{\mathit{\in }}_0\right]=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{A}}^2\right] \\ \Rightarrow \left[{\varphi }_{\mathrm{e}}\right]=\left[{\mathrm{M}}^1{\mathrm{L}}^3{\mathrm{T}}^{-4}{\mathrm{A}}^{-2}\right]\left[\mathrm{AT}\right]\left[{\mathrm{L}}^{-2}\right]\left[{\mathrm{L}}^2\right] \\ =\left[{\mathrm{ML}}^3{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}\right] \end{gathered}$$

Displacement current,
$$\begin{gathered} \left[I_{\mathrm{D}}\right]=\left[{\in }_0\right]\left[{\varphi }_{\mathrm{e}}\right]\left[{\mathrm{T}}^{-1}\right] \\ \left[I_{\mathrm{D}}\right]=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{A}}^2\right]\left[{\mathrm{ML}}^3{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}\right]\left[{\mathrm{T}}^{-1}\right] \\ \left[I_{\mathrm{D}}\right]=\left[\mathrm{A}\right] \end{gathered}$$
[I_D]=[current]