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The assumption is that, initially, there is a charge distribution on the outer surface of the inner conductor (highlighted blue). Why does this lead to flux being terminated on a negative charge on the outer conductor (highlighted red)?

Is the negative charge (on the inner surface of the outer cylinder) simply induced by the positive charge (on the outer surface of the inner cylinder)? If so, since the outer cylinder should remain electrically neutral (which is the initial state), shouldn't the positive charge again be induced on the outer surface of the outer cylinder? The book does not cover this explicitly so I am confused.

What exactly is happening in this situation?

Mathematics 22C
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2 Answers2

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Why does this lead to flux being terminated on a negative charge on the outer conductor (highlighted red)?

Let's assume there is a positive (as opposed to negative) charge on the inner conductor. Then there will be electric field lines radiating away from the inner conductor.

But the inner conductor is completely surrounded by the outer conductor, so the field lines must intersect the outer conductor.

The outer conductor is made of a good conductor material, so there is no electric field within its bulk. Therefore the electric field lines that intersect the inner surface of the outer conductor must terminate there.

And Gauss's law tells us that when a field line terminates there must be a charge for it to terminate on. It must be a negative charge because the field is pointing toward it rather than away from it.

If so, since the outer cylinder should remain electrically neutral (which is the initial state), shouldn't the positive charge again be induced on the outer surface of the outer cylinder?

Yes, the charge must come from somewhere. If the outer conductor is floating it will come from the outer surface and a positive charge will be left on the outer surface.

If the outer conductor is electrically connected to some other object (for example, if it's grounded, as it often is in the case of actual coaxial transmission lines) then the charge can come from that other object and it's possible no charge is left on the outer surface. In this case the outer conductor does not remain electrically neutral.

The book does not cover this explicitly so I am confused.

It doesn't cover it because the charge on the outer surface doesn't affect the behavior of the capacitor formed between the two conductors.

The Photon
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  • Thank you for the detailed answer, but I also want to know about the ‘thickness’ issue. As far as I understood this book is saying the outer cylinder has zero thickness. Is ‘inner’ and ‘outer’ surface even properly defined? – Mathematics 22C Dec 24 '23 at 01:34
  • @Mathematics22C, what thickness issue? There's nothing about it in your question post. The charge on a perfect conductor will be confined to a 2-dimensional (0 thickness) sheet on the surface of the conductor. Is that what you mean? – The Photon Dec 24 '23 at 04:29
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You buy a coax cable, both conductors are neutral. Then you connect it to a circuit: one end connects to a battery (initially neutral), the other end to some load (also initially neutral). The circuit as a whole is still neutral, and so is the coax, although there might be opposite charges on the inner and outer conductors.

The book is implicitly assuming that the coax is neutral. You can certainly concoct a scenario where this is not the case, as The Photon has pointed out. However, much like in a capacitor, this is not the typical use case and is usually unintended.

Puk
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