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?


1 Answer 1


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

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

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

  • $\begingroup$ Quite close, but the battery is not necessarily time-invariant. $\endgroup$
    – Tan En De
    Commented May 14, 2019 at 5:49
  • $\begingroup$ @TED The battery does not have to be there and it could be a capacitor discharging situation. $\endgroup$
    – Farcher
    Commented May 14, 2019 at 5:51
  • $\begingroup$ 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. $\endgroup$
    – Tan En De
    Commented May 14, 2019 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.