Question

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

Answer 1

Displacement current,

${\displaystyle I_D=\frac{{\in }_{0}d{\phi }_e}{dt}}$

Electric flux,

${\displaystyle {Ø}_e=EA}$

${\displaystyle \left[{Ø}_e\right]=\left[E\right]\left[A\right]}$

= ${\displaystyle \left[\frac{1}{4\pi {\in }_0}\frac{q}{r^2}\right]\left[A\right]}$

Also , ${\displaystyle \left[{\in }_0\right]=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]}$

⇒ ${\displaystyle \left[{Ø}_e\right]=\left[M^1{L}^3{T}^{-4}{A}^{-2}\right]\left[AT\right]\left[L^{-2}\right]\left[L^2\right]}$

= ${\displaystyle \left[M{L}^3{T}^{-3}{A}^{-1}\right]}$

Displacement current,

${\displaystyle \left[I_D\right]=\left[{\in }_0\right]\left[{Ø}_e\right]\left[T^{-1}\right]}$

${\displaystyle \left[I_D\right]=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]\left[M{L}^3{T}^{-3}{A}^{-1}\right]\left[T^{-1}\right]}$

${\displaystyle \left[I_D\right]=\left[A\right]}$

${\displaystyle \left[I_D\right]=\left[\text{Current}\right]}$