I've come across a tricky question and would appreciate some hints or explanations as to why the given solution is the way it is. The question reads as follows:

A coaxial cable consists of a wire with radius $a$ (the core of the cable), which is wrapped with insulating material with dielectric constant $\varepsilon$, until radius $b$ (called the insulator). Around the cable there is a layer of conducting material (radius $c$ from the center of the cable and is called the wrapper).

The wire's length is $d$ such that $d \gg a,b,c$. At one side of the cable, a voltage source $V_0$ with inner resistance $R_0$ is connected to both the wire and the wrapper, and at the other side, a resistor $R$ is connected instead of a voltage source.

enter image description here

It asks to find the magnetic and electric fields $B(r)$, $E(r)$, where $r$ is the distance from the center of the cable (from $z$-axis in the picture), when $t\rightarrow+\infty $.

In the solution, they said that when $t\rightarrow +\infty$, no current will pass through the cylindrical capacitor so: $I=\frac{V_0}{R_0+R}$ therefore $V\left(\text{final}\right)=V_0 \frac{R}{R_0+R}$.

I do not get this, how can one imagine how this circuit works? Is there an equivalent and more simple circuit? According to what they said, after infinite time, no current passes through the capacitor, but the wires are connected to the wrapper so how can there be current at all in the circuit? All I know is when an uncharged capacitor is charged, it will act as an open switch in the circuit after a long time.

Possible equivalent Circuit?: enter image description here

  • 1
    $\begingroup$ "According to what they said, after infinite time, no current passes through the capacitor" Right. Flowing through the capacitor would mean between the center and the wrapper. It doesn't mean that current isn't flowing along the center and along the wrapper. $\endgroup$ – BowlOfRed Jun 26 '15 at 16:02
  • $\begingroup$ Do you mean there is no current flowing through the dielectric? Suppose there was no dielectric at all, the capacitor still will be charged after an infinite time and will act as an open switch, so how will the current still flow in the circuit? Is the suggested circuit I've uploaded appropriate ? $\endgroup$ – Dylan132 Jun 26 '15 at 16:14
  • $\begingroup$ Capacitors only allow current when voltage is changing. When the voltage reaches steady state ($t=\infty$), it acts as an open switch. Current will still flow along the wrapper and along the center (along the outside of your box drawing). $\endgroup$ – BowlOfRed Jun 26 '15 at 16:18
  • $\begingroup$ But how did it flow before $t$ reached $infinity$? As I understand, at $t =inf$ it simply ignores the whole cylindrical wrapper and only passes through a bottom strip of it. I assume that at first the current spreads over the wrapper and charge is continuously stacked on the inner shell of the wrapper (at radius b), until the capacitor is fully charged, then no current passes through it, only at its bottom. Is this conclusion right? $\endgroup$ – Dylan132 Jun 26 '15 at 16:24
  • $\begingroup$ If you think about having $R$ connected to the cable and then connecting the battery/$R0$, the capacitor starts completely discharged. As the capacitor cannot change voltage instantly, the current will flow through $C$. With a time constant of $RC$ the current will shift to the resistor. As $t \to \infty$ the current in $C$ will go to zero and you will have a constant current in $R$. $\endgroup$ – Ross Millikan Jun 26 '15 at 20:00

Your "possible equivalent circuit" is correct and you have to just understand that the capacitor, in this case, happens to be particularly long, so that it happens to incorporate into its body both the wire up-top and the wire beneath.

As for "how can current flow?" the answer is "for long time scales a capacitor looks like a break in the circuit, that's why we draw it with this space in the middle. Eventually the capacitor gets charged up and no charge flows over the capacitor. However charge does still flow across the wire up-top and the wire on the bottom; it just doesn't flow by means of the capacitor: it follows the wires on the top and the bottom."


There is only one connection to the wrapper, at the power supply end. One can read the problem to imply that the resistor at the other end is connected to the wrapper, but you shouldn't. If there were a connection there, the resistor would be shorted out. As the wrapper has only one connection, at long $t$ it will have come to the proper equilibrium potential and there will be no current in the wrapper. Your basic circuit is then an ideal battery in series with two resistors, which is where the final current calculation comes from.

  • $\begingroup$ If there is only one connection, between the wrapper and the voltage source, then what is the resistance R connected to? Could you please upload a drawing or something? Or at least tell me what this circuit is equal to (in a simple form)? $\endgroup$ – Dylan132 Jun 26 '15 at 15:35
  • $\begingroup$ I had seen the picture as showing an external wire from $R_0$ to $R$. If the bottom line is the bottom edge of the wrapper and $R$ is connected from the wrapper to the center conductor, you have continuous current in the wrapper. $\endgroup$ – Ross Millikan Jun 26 '15 at 15:39
  • $\begingroup$ If there is an external wire from $R0$ to $R$ then how is the wrapper gonna be charged? I think the wires are connected to the bottom of the wrapper. I've drawn somewhat a circuit which actually fits the problem come to look at the mathematical equations. I'm not sure if the drawing and the assumptions are correct, I'll upload the drawing. $\endgroup$ – Dylan132 Jun 26 '15 at 16:10
  • $\begingroup$ As I reread it, I think you are correct that the wrapper is connected at both ends. Your circuit is correct as well. They say that at large time no current flows through the capacitor, which I misread as the wrapper. In your circuit, you have current flowing around the outside loop permanently in just the amount stated in the problem. Now C charges according to the voltage divider formed by the two resistors across the battery. $\endgroup$ – Ross Millikan Jun 26 '15 at 16:16
  • $\begingroup$ Does that mean that the capacitor is basically the inner shell of the wrapper (at radius b) and the outer shell of the wire (at radius a), where charge is continuously stacked? $\endgroup$ – Dylan132 Jun 26 '15 at 16:27

I think the solution is incorrect.

If the current becomes zero after an infinite time, the potential of the rod and the wrapper would become equal ( because they are connected by a wire of $R$ resistance, and if $I=0$, the $V$ across the wire $=0$. )

Also, your diagram seems incorrect. You assume that there are different wires connecting the capacitor and the resistance $R$ but, in the question, the wire IS a part of the capacitor.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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