Question 12 Marks

Show that the dimensions of the displacement current $\epsilon_0\frac{\text{d}\phi_\text{E}}{\text{dt}}$ are that of an electric current.

Answer

$\frac{\epsilon_0\text{d}\phi_\text{E}}{\text{dt}}=\frac{\epsilon_0\text{EA}}{\text{dt}4\pi\epsilon_0\text{r}^2}$$=\frac{\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2}{\text{M}^{-1}\text{L}^{-3}\text{A}^2}\times\frac{\text{A}^1\text{T}^1}{\text{L}^2}\times\frac{\text{L}^2}{\text{T}}=\text{A}^1$
$=\text{(Current)}$ Proved.

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