Question

Consider the situation of the previous problem. Define displacement resistance R_d = V/i_d of the space between the plates, where V is the potential difference between the plates and i_d is the displacement current. Show that R_d varies with time as $R_d=R(e^{t/\tau }-1)$.

Answer 1

Electric field strength for a parallel plate capacitor = $E=\frac{Q}{{\in }_{0}A}$
$$\begin{gathered} \mathrm{Electric}\mathrm{flux},\varphi \mathit{=}E\mathit{.}A\mathit{=}\frac{Q}{{\mathit{\in }}_{0}A}\mathit{.}A=\frac{Q}{{\in }_0} \\ \mathrm{Displacement}\mathrm{current},i_{\mathrm{d}}={\in }_0\frac{d{\varphi }_{\mathrm{E}}}{dt}={\in }_0\frac{d}{dt}\left(\frac{\mathrm{Q}}{{\in }_0}\right) \\ {i}_{\mathrm{d}}=\left(\frac{dQ}{dt}\right) \\ \mathrm{Also},Q\mathit{}\mathit{=}\mathit{}CV \\ {i}_{\mathrm{d}}=\frac{d}{dt}\left(E_{\mathit{0}}{C_e}^{\mathit{-}t\mathit{/}RC}\right) \\ {i}_{\mathrm{d}}=E_{0}C\left(\mathit{-}\frac{\mathit{1}}{RC}\right)e^{\mathit{-}t\mathit{/}RC} \\ \mathrm{Displacement}\mathrm{resistance},R_d=\frac{E_{\mathit{0}}}{i_{\mathrm{d}}}-R \\ \Rightarrow R_d=\frac{E_{\mathit{0}}}{\frac{E_{\mathit{0}}}{R}{\mathrm{e}}^{\mathit{-}t\mathit{/}RC}}-R \\ \Rightarrow R_d=R{\mathrm{e}}^{t\mathit{/}RC}-R \\ \mathit{\Rightarrow }{R}_{\mathit{d}}\mathit{=}R({\mathrm{e}}^{\mathit{t}\mathit{/}\mathit{R}\mathit{C}}-1) \end{gathered}$$