Question

A parallel-plate capacitor of plate-area A and plate separation d is joined to a battery of emf ε and internal resistance R at t = 0. Consider a plane surface of area A/2, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time.

Answer 1

Electric field strength for a parallel plate capacitor,
$E=\frac{Q}{{\in }_{0}A}$
Electric flux linked with the area,
${\varphi }_E\mathit{=}EA\mathit{=}\frac{Q}{{\mathit{\in }}_{\mathit{0}}A}\mathit{\times }\frac{A}{\mathit{2}}=\frac{Q}{2{\in }_0}$
Displacement current,
$$\begin{gathered} I_{\mathrm{d}}={\in }_0\frac{d{\varphi }_{\mathrm{E}}}{dt}={\in }_0\frac{d}{dt}\left(\frac{Q}{{\mathit{\in }}_{\mathit{0}}\mathit{2}}\right) \\ {I}_{\mathrm{d}}=\frac{1}{2}\left(\frac{d\mathrm{Q}}{dt}\right)...\left(\mathrm{i}\right) \end{gathered}$$

Charge on the capacitor as a function of time during charging,
$Q=\epsilon C\left[1-e^{\mathit{-}t\mathit{/}RC}\right]$

Putting this in equation (i), we get:
$$\begin{gathered} I_{\mathrm{d}}=\frac{1}{2}\epsilon C\frac{d}{dt}\left(1\mathit{-}{e}^{\mathit{-}t\mathit{/}RC}\right) \\ {I}_{\mathrm{d}}=\frac{1}{2}\epsilon C\left(\mathit{-}{e}^{\mathit{-}t\mathit{/}RC}\right)\times \left(\mathit{-}\frac{1}{RC}\right) \\ \mathrm{C}=\frac{A{\mathit{\in }}_{\mathit{0}}}{d} \\ \Rightarrow I_{\mathrm{d}}=\frac{\epsilon }{2R}\times e^{\mathit{-}\frac{td}{{\epsilon }_{\mathit{0}}AR}} \end{gathered}$$