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At work we faced the following problem:

enter image description here

We are working with some high voltage batteries which are located inside a casing which isolates the batteries. This means you cannot get any significant current by short circuting the batteries by using the casing.

The production teams, hoewever faces problems. When touching the front and the back of the casing they experience a small shock due to capacitive coupling. The electrical engineers reproduced the problem and measured exactly the voltage of the batteries between the front and the back.

They argued that this is completely expected and natural by this graphic (voltage source each 5 V):

enter image description here

On the left you see the 2 batteries, on the right the highohmian casing. The capacities show the coupling between capacities and casing.

I also understand that, but then I got thinking:

Why does the outside electric field of the batteries, which should be fairly low compared with the field inside the batteries generate the same voltage (U = $\int Eds$) as the two batteries do when connected directly.

Why does at all capacative coupling produce the same voltage?

In my opinion the electric field of the batteries should look like that:

enter image description here

(A very moderate estimation of the electric field produced by the batteries (stronger in the middle))

Edit: Improved field suggestion to make it look less confusing

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  • $\begingroup$ Are you suggesting a tangential discontinuity of the electric field at the case of the battery (some value inside, zero outside)? No discontinuity of the tangential component of $\mathbf{e}$ may occur, en.wikipedia.org/wiki/…. And even if you consider a thin layer, you'd need a time varying magnetic field $\endgroup$
    – basics
    Jan 21 at 12:35
  • $\begingroup$ I am saying that the potential energy (which voltage is) outside the batteries cannot possibly be the same as inside the batteries, since the electric field outside is much weaker. physics.stackexchange.com/q/398702 $\endgroup$
    – Niclas
    Jan 21 at 12:39

1 Answer 1

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In comments you clarified that the heart of your question is,

I am saying that the potential energy (which voltage is) outside the batteries cannot possibly be the same as inside the batteries, since the electric field outside is much weaker.

The potential energy change caused by moving a charged particle through a field depends on two things:

  1. How strong the field is

  2. How far you moved the particle.

The electrostatic field is a conservative field. That means, when you move the particle from the negative terminal of the battery to the positive terminal, it gains the same potential energy regardless of whether you move it by the short path through the body of the battery or the long path that goes to Chicago before it comes back to the other terminal.

Since the short path is much shorter than the long path that tells us that the field along the short path must be stronger, in order to achieve the same change in potential energy over a shorter distance.

Aside: About where you said "potential energy (which voltage is)". Voltage is not potential energy. Voltage is potential. Potential tells you how much potential energy would be required to put a unit of charge at a given location, but the voltage is defined even when the charge is not there, while the potential energy only exists when the charge is there.

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