Question

The electric field between two plates of a parallel plate(area 8×10)capacitor is given by E=6.8
×10^6-8.4×10^5t.what is the magnitude of the displacement current between the plates

Answer 1

It is given the area of the plates = $8*10^{-2}m^2$ and the electric field $E=6.8*10^6-8.4*10^5t$.

The displacement current density is given by: $J_d=\epsilon_0\frac{\delta E}{\delta t}$; where $\epsilon_0$ is the permittivity of free space = $8.854*10^{-12}F/m$

Therefore, $J_d=8.854*10^{-12}*-8.4*10^5=7.4*10^{-6}A/m^2$

N.B: we ignore the negative sign in the electric field because we only care about the magnitude.

Current density is the magnitude of the current per unit area, therefore,

$I_d=J_d*A=7.4*10^{-6}*8*10^{-2}=5.9*10^{-7}Ampere$