Consider two metal plates (very large). The distance between the two plates is $2a$. For the first half, the space between plates is filled by a material with dielectric constant $\varepsilon_1$ and resistivity $\rho_1$. The second half is filled by a material with dielectric constant $\varepsilon_2$ and resistivity $\rho_2$. For simplicy, suppose that the object lies on the $x$ axis: the material $(\varepsilon_1, \rho_1)$ is located in $-a < x < 0$, the second one in $0 < x <a.$

Suppose to connect this system to a battery with voltage $V$ at time $t=0$. I was wondering to say something about the charge present in the system at $t=0$ and at $t=+\infty$. Specifically, the charge on the metal plates, on the interface between the two materials and the charge inside the two materials. If a volumetric distribution is present for the latter case, how can I find the expression as a function of $x$?

I think that each dielectric/resistivity couple can be represented as a resistance and a capacitance sharing the same voltage, as reported in the next figure:

enter image description here

(In the figure, $S$ stands for the area of each plate).

In this form, I'm able to solve the circuit. For example, the tension on the capacitors at $t=0$ are $V_{C1}(0^+)= \frac{C_2}{C_1+C_2}V$ and $V_{C2}(0^+)= \frac{C_1}{C_1+C_2}V$, respectively. In this case, I imagine the on each plate there is the same superficial charge $ Q = \frac{C_1C_2}{C_1+C_2}V$. This causes a non uniform electric field inside the the object, so that I guess there is charge elsewhere too.

Instead, for $t \to +\infty$ they are $V_{C1}(+\infty)= \frac{R_1}{R_1+R_2}V$ and $V_{C2}(+\infty) = \frac{R_2}{R_1+R_2}V$. Thanks to these expression, I can find the charge on each capacitors at $t=0$ and at $t = +\infty$. In this case, it seems that on each plate I have a different charge distribution. Again, what about the inside of the object?

As you noticed, the approach using an "equivalent circuit" does not teach me anything about what is happening to the charge internally. Which are the physical principle that lead me to a concrete answer to this question?

How can I approach this kind of problems?


1 Answer 1


Your equivalent circuit is fine.

As drawn with an ideal cell (no internal resistance) and connecting wire having no resistance the system will arrange itself so the the voltages are those of a resistive potential divider with the capacitors appropriately charged as you stated in your question.

$t \to +\infty$ they are $V_{C1}(+\infty)= \frac{R_1}{R_1+R_2}V$ and $V_{C2}(+\infty) = \frac{R_2}{R_1+R_2}V$.

This analysis would require a finite amount of charge to flow (to charge up the capacitors) in an instant of time which is not possible in the real world.

In the real world with the connecting wires and the cell will have resistance (and the whole circuit will have some “parasitic” self inductance).

The analysis would now have the capacitors uncharged at the instant the switch is closed.
As time progressed the capacitors would charge eventually reaching a steady value with the voltage across them dictated by the resistances of the resistors in the circuit and the emf of the cell.

  • $\begingroup$ Thanks. I'm aware that real wires have resistance. Anyway, what can we say about the charge inside the object when the wires are ideal? $\endgroup$ Sep 6, 2018 at 21:44
  • $\begingroup$ @the_candyman If you ignore the parasitic inductance and there is no resistance the circuit gets to the final state instantaneously. $\endgroup$
    – Farcher
    Sep 6, 2018 at 21:48
  • $\begingroup$ Do you mean that in the presented configuration it is not true that $V_{C1}(0^+)= \frac{C_2}{C_1+C_2}V$ and $V_{C2}(0^+)= \frac{C_1}{C_1+C_2}V$? $\endgroup$ Sep 7, 2018 at 12:44
  • $\begingroup$ @the_candyman If it were true then the capacitors are charged and the currents through the two resistors (capacitors) are unequal as it is unlikely that $\frac{C_2}{C_1+C_2}V$ is equal to $\frac{R_1}{R_1+R_2}V$ $\endgroup$
    – Farcher
    Sep 7, 2018 at 13:31
  • $\begingroup$ ok, I think I got your point. In each case, it seems that the electric field between plates is non-homogeneous along $x$. Does this mean that there is other charge inside the materials? How is this charge distributed? $\endgroup$ Sep 7, 2018 at 19:12

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