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Imagine two poles of a battery to be one meter separated from each other. We connect the poles with a conducting wire after which a current flows. Can we stop the current from flowing by placing the battery between two large charged parallel plates? If so, how high must the voltage between the plates be compared to the voltage of the battery?
I know that a static electric field can't penetrate a metal. But what will be the case if a current flows through it?

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    $\begingroup$ Accept the correct answer. The pictures were pretty, but the analysis was flawed. $\endgroup$
    – user121330
    May 14 '21 at 23:26
  • $\begingroup$ @Deschele Schilder I have updated my answer. I am not saying that this answer is 100% correct and I assume ideal case. However after further investigation things seem more complicated and unless an analytical solution to the problem is given I regard this question a big question mark! $\endgroup$
    – Markoul11
    May 22 '21 at 11:12
  • $\begingroup$ This situation can be modeled using Bruce Sherwood's circuit/wire/field simulation. I think the program would need modification to include an external object (the capacitor) but I don't think that would be too difficult. $\endgroup$
    – garyp
    May 22 '21 at 11:39
  • $\begingroup$ @Deschele Schilder Try to tag the question with "physics" and "electromagnetism" as well. $\endgroup$
    – Markoul11
    May 23 '21 at 11:54
  • $\begingroup$ @Deschele Schilder I believe the answers here accepted as being the most valid are under serious dispute. physics.stackexchange.com/questions/437511/… $\endgroup$
    – Markoul11
    Jun 11 '21 at 10:39
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Can we stop the current from flowing by placing the battery between two large charged parallel plates?

No.

[I am assuming that the wire is not in electrical contact with the plates. Perhaps there are holes in the plates for the wires to pass through, or alternatively, the wire enters the cavity between the plates from the sides, then turns perpendicular to the plates, and then again, exits through the side. One final arrangement is that the entire circuit is in the cavity between the plates. The result is the same for all three configurations as long as the wire is not in electrical contact with the plates.]

As the voltage across the plates is increased, the wire will develop charges at the points of nearest approach to the plates. [That is, the electrons in the wire, will re-arrange themselves]. These charges will be opposite to the charges in the plate. They will create a $\vec{E}$ field in the wire equal and opposite to that created by the plates in the wire. The net effect will be that the emf applied by the battery will drive the same current through the wire with or without charges on the plates.

[On a side note, the electric field between two plates with nothing but a dielectric between them appears as parallel lines (with some fringing at the sides) as drawn in the original question. However, the electric field between two plates with a wire between them, even though not electrically connected, will be substantially different, whether or not there is current flowing through the wire.]

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  • $\begingroup$ This seems in contradiction with the first answer, but I can't see why. What if the entire battery-wire system is placed between the plates, so without holes? $\endgroup$ May 13 '21 at 12:14
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    $\begingroup$ @DescheleSchilder It is in contradiction to the first answer. The first answer is incorrect. If there are no holes, then the plates of the capacitor are in electrical contact with the wires, at least according to the diagram in the first answer. Then you have a system with two power supplies (one for the wire, one for the plates), and two resistors (the wire between the plates, and the wire outside of the plates). Kirchhoff's laws will determine the net current. $\endgroup$ May 13 '21 at 12:22
  • $\begingroup$ I haven't drawn a diagram. I only described the situation. The wire is only in contact with the two terminals of the battery. Without being in contact with the plates (Imagine a small 9V battery with a wire between the poles; this system we put between the plates, without touching the plates) $\endgroup$ May 13 '21 at 12:22
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    $\begingroup$ @Markoul11 "There is no difference between a field mediated by a medal medium with that mediarted by air." That is incorrect. If conductive material replaces dielectric material in an electric field, it will change the shape and strength of the field, because charges in the conductor will re-arrange themselves, and if the electric field was generated by charged conductive plates, the charges in those plates will re-arrange themselves as well. Conductors in an electric field do not act the same as dielectrics. $\endgroup$ May 14 '21 at 16:49
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    $\begingroup$ This situation can be modeled using Bruce Sherwood's circuit/wire/field simulation. I think the program would need modification to include an external object (the capacitor) but I don't think that would be too difficult. $\endgroup$
    – garyp
    May 22 '21 at 11:39
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enter image description here

Assuming a straight electrical wire with no dielectric insulation (i.e. dielectric constant of wire insulation would reduce the displacement Ec field of the capacitor), positioned perpendicular to the plates of the capacitor as shown in the figure and provided, also that the wire is powered by a different source than the capacitor and the capacitor is fully charged, then the short answer is ideally,

when Ec(inside wire)=-Ew (Eq.1)

where Ec(inside wire)≠Ec (i.e. Ec value in vacuum)

The two electric fields Ec(inside wire) of the capacitor and Ew of the voltage source Vw inside the wire will cancel out and no electric current Iw will flow through the wire.

Practically, this means that you must make sure that the polarity (+/-) of the source powering the wire is opposite to the polarity of the voltage source V powering the two plates of the capacitor and quantitatively that the displacement current between the capacitor plates is made equal and opposite sign of the d.c. current Iw flowing through the wire thus,

Ic(maximum peak)=-Iw (Eq.2)

The displacement current between the plates of a fully charged capacitor (i.e. when fully charged, current on the capacitor circuit lines is ideally zero Ic=0) is the maximum peak value measured of Ic current at the beginning of the capacitor charging period.

Meaning that you have to choose the correct value of source voltage V for your capacitor circuit that would cancel out the current on the wire Iw.

You could do a theoretical calculation but it would be complicated with a possible large error bar, knowing the exact resistivity and impedance values of the two circuits but if your question is related to a practical application and assuming your capacitor in the circuit is fast charging, you could just keep increasing V voltage value until you measure Iw=0.

Updated figure added 14 May 2021: Because it seems that there is a lot of confusion created about the previous figure also seen in this answer, I have added a second version of the circuit schematic that I hope will clarify things. The conventional flow of current is used. updatd version of figure

Answer update added 22 May 2021: The electric field Ew inside a current carrying wire connected to a voltage source is not zero. The E=0 case inside a conductor holds only for Electrostatics and not for a current carrying conductor where the electric field inside the conductor is not zero, E≠0. The whole point of this circuit is that the naked wire (marked as red) inside the field of the capacitor is connected to a voltage source therefore electrostatic equilibrium is continuously disturbed and never reached. physics.stackexchange.com/a/250651/183646 There is definitely Ew present opposite to Ec all the time after the capacitor is charged. We can assume the capacitor is in vacuum to avoid dielectric leakage and therefore an ideal case. The fields will interact and therefore affect the current on the wire. Of course because the huge difference of conductivity between the air and the wire a huge voltage Vc would be needed on the plates of the capacitor to zero even a small current on the wire Iw.

The system operation is not linear. Putting Vw at 0 results to electrostatic equilibrium and surface charge on the wire with the electric field inside the wire Ew=0. However, putting Vw to any non-zero value gives Ew≠0 inside the wire. Therefore you cannot superposition for Vw=0. Things are more complicated, https://link.springer.com/article/10.1023/A:1018874523513 , https://www.if.ufrj.br/~dore/Fis3/Assis_et_al.pdf .

I am not saying that my answer is 100% correct and it is assuming ideal case. However after further investigation things seem more complicated and unless an exact analytical solution to the problem is given or a related experiment is presented I regard this question a big question mark!

Answer update 28 May 2021: https://tinyurl.com/2ca5byv7 There is a experiment done by Oleg Jefimenko showing clearly the axial homogeneous electric field formed inside an d.c. electric current carrying wire. The question now is will the electric homogeneous vectors inside the wire of this field interact with the electrostatic aligned field of the capacitor?

Final answer update 29 May 2021:

Except the case of Vw=0 where there is electrostatic Equilibrium and Ew=0 inside the wire and therefore the operation of our system is not linear compared with all other values of Vw≠0 where there is an electric field created inside the wire due the current Ew≠0 by the Vw voltage source and there is no electrostatic equilibrium formed on the wire that can repel an external electrostatic field and thus the case Vm=0 cannot be used in the analysis, then,

"Electric fields are fully described by Maxwells eqs, which are linear in both E and B. Linearity implies superposition, which means that the fields do not interact with each other, but simply add vectorially. If they have equal and opposite amplitudes at any point, the resultant will be zero - i.e.exact cancellation at that point. That is true for both static and time/spatially varying fields." [https://physics.stackexchange.com/a/411049/183646]

Therefore my initial answer is correct .

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    $\begingroup$ There are a number of errors in your answer. I will only point out one in this comment. There is no displacement current in a capacitor except when it is charging or discharging. "The displacement current between the plates of a fully charged capacitor (i.e. when fully charged, current on the capacitor circuit lines is ideally zero Ic=0) is the maximum peak value measured of Ic current at the beginning of the capacitor charging period." No, the displacement current through the dielectric of a capacitor is always equal to the conduction current through the leads of that capacitor. $\endgroup$ May 13 '21 at 12:40
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    $\begingroup$ @Markoul11 "The value of the displacement current after the capacitor is charged ideally then we say equals the maximum value of the the conduction current during charge" That is just plain wrong. If the capacitor is not charging or discharging, there is NO displacement current. If you were taught otherwise, you were misinformed. $\endgroup$ May 14 '21 at 20:53
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    $\begingroup$ The question is indeed: will a constant electric field still be unable to enter a metal if the metal is carrying a current. I'm not so sure anymore it can't! $\endgroup$ May 22 '21 at 11:40
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    $\begingroup$ @Markoul11 it is clear that you do not understand either linearity or superposition. If you claim non-linearity then please write down exactly which law is nonlinear. Maxwell’s equations and Ohm’s law are linear, so what other law do you believe governs this system. $\endgroup$
    – Dale
    May 22 '21 at 17:13
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    $\begingroup$ @Markoul11 I do apologize, I should not use inflammatory language. I am just frustrated because you are not merely wrong but you are obviously wrong. I do not understand how you can not see it. Maxwell’s equations are linear (you have yet to produce a non-linear equation). Linearity means you can use the principle of superposition. That means you can find the effect of two sources together simply by finding the effect of each source alone and adding them. Your new quote confirms that. When you add the two sources together you get current. You “agree to disagree” with Maxwell and linearity $\endgroup$
    – Dale
    May 30 '21 at 11:20
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The answer by @Markoul11 is incorrect and the answer by @Math Keeps Me Busy is correct. The easiest way to analyze this problem is to use the principle of superposition as follows:

Both Maxwell’s equations and the equations of circuit theory are linear. This means that you can find the total current through the wire by simply finding the current due to each source individually and summing them.

To get rid of a voltage source we simply give it a voltage of 0. And in all cases we are considering the voltage source (a battery) to be operating at DC steady state.

Setting the external voltage source to 0 and the connected source to $V_{\text{connected}}$ can be analyzed with the usual circuit theory, and gives the standard Ohm’s law current. So in this case there is a current in the wire and a linearly changing voltage inside the wire, and a surface charge which produces the correct field distribution inside the wire. https://doi.org/10.1119/1.18112

Setting the connected source to zero and the external source to $V_{\text{external}}$ cannot be analyzed with circuit theory, but is easy to analyze with Maxwell’s equations. The $V_\text{connected}=0$ "short circuits" that connection, giving a closed conductive loop. Since it is DC steady state this is a simple conductor in an external electrostatic field. A surface charge is induced on the loop and the current is zero. Importantly, the surface charge opposes the external field so that there is no field inside the conductor. So in this case there is no current inside the wire and no voltage inside the wire, just an external surface charge canceling out the external field.

Once we have analyzed each source independently we can simply add the two results together to determine the current from both sources together. So when both $V_{\text{external}}$ and $V_{\text{connected}}$ are on we have the non-zero current from the connected source added to the zero current from the external source to produce a non-zero current. We also have the non-zero internal voltage gradient from the connected source added to the zero internal voltage from the external source to produce a non-zero internal voltage gradient. Finally, we have the non-zero surface charge distribution from the connected source and also the non-zero surface charges from the external source added to produce a non-zero total surface charge distribution. In summary, the external source produces 0 current and cannot stop the current from the connected source. Its effect is solely to alter the usual surface charge distribution on the wire.

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  • $\begingroup$ Wouldn't the external field cancel the "inner voltage" of the battery? $\endgroup$ May 14 '21 at 23:48
  • $\begingroup$ @Deschele Schilder no, the only effect of the external field is to induce a surface charge on the wire. That is the whole point of using the principle of superposition. $\endgroup$
    – Dale
    May 14 '21 at 23:51
  • $\begingroup$ I disagree. The whole point of this circuit is that the naked wire (marked as red) inside the field of the capacitor is connected to a voltage source therefore electrostatic equilibrium is continuously disturbed and never reached. physics.stackexchange.com/a/250651/183646 There is definitely Ew present opposite to Ec all the time after the capacitor is charged. We can assume the capacitor is in vacuum to avoid leakage. The fields will interact and therefore affect the current on the wire. $\endgroup$
    – Markoul11
    May 21 '21 at 19:20
  • $\begingroup$ @Markoul11 your argument is incorrect. Maxwell’s equations are linear, and your argument is only valid for a non-linear system. Please read up on the principle of superposition before posting any more EM answers. The fields don’t interact other than to simply add. We know the effect of each source individually, the total effect is just the sum. Your vague idea of “electrostatic equilibrium is continuously disturbed” is made precise with superposition. $\endgroup$
    – Dale
    May 21 '21 at 22:28
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    $\begingroup$ @Markoul11 said "You treat Vw=0 case in your analysis as an open loop circuit which does not apply in our present circuit which is for Vw=0 still a closed loop circuit" Setting the connected source to zero gives a short circuit, not an open circuit. So I agree that it is a closed loop circuit. That is well understood, but perhaps poorly communicated since this is one of the few points that we agree on. I have updated my answer accordingly to communicate more clearly on this point $\endgroup$
    – Dale
    May 30 '21 at 13:05

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