# Conduction current for parallel-plate capacitor

I am reading Sadiku's Elements of Electromagnetics and here is the proof given to show that conduction current = displacement current for parallel-plate capacitor.

Displacement current: \begin{align*} E &= \frac{V}{d}\\ D &= \epsilon E = \epsilon \frac{V}{d}\\ J_d &= \frac{\partial D}{\partial t} = \frac{\epsilon}{d}\frac{dV}{dt}\\ I_d &= J_d \cdot S = \frac{\epsilon S}{d}\frac{dV}{dt} = C\frac{dV}{dt}\\ \end{align*}

Conduction current: \begin{align*} Q &=\rho_s S \\ \rho_s &= D\\ I_c &= \frac{dQ}{dt} = S \frac{d\rho_s}{dt} = S \frac{dD}{dt} = \epsilon S \frac{dE}{dt} = \frac{\epsilon S}{d}\frac{dV}{dt} \end{align*} the same with displacement current.

Contradictorily, it seems like this second method is also valid: \begin{align*} RC &= \frac{\epsilon}{\sigma}\\ R &= \frac{d}{\sigma S} \\ I_c &= \frac{V}{R} = \frac{\sigma S}{d} V \neq \frac{\epsilon S}{d}\frac{dV}{dt} \end{align*}

Which method is the correct one to find conduction current then?

Consider a series circuit consisting of a battery with emf $$\mathcal E$$, a capacitor $$C$$ and a resistor $$R$$.

$$\mathcal E = V_{\rm R} + \frac QC$$

Differentiating this equation with respect to time gives

$$0= \frac{dV_{\rm R}}{dt} + \frac IC$$

$$I = \frac {V_\rm R}{R} \Rightarrow 0= C\frac{dV_{\rm R}}{dt} +\frac1R V_R \Rightarrow \frac{\epsilon S}{d}\frac{dV_{\rm R}}{dt} = -\frac{\sigma S}{d} V_R$$

• Quite close, but the battery is not necessarily time-invariant. – TED May 14 at 5:49
• @TED The battery does not have to be there and it could be a capacitor discharging situation. – Farcher May 14 at 5:51
• I think I see what's wrong now. The $\displaystyle R=\frac{d}{\sigma S}$ I found in my question is the resistance between capacitor plates, and $V/R$ gives the current in between the plates and not the current in the circuit loop. – TED May 14 at 7:15