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I am uncomfortable with the notion of chassis ground but I can't pin down what exactly bugs me. So I'll start with an example and ask a couple of questions on the way. My background is this: I have studied quite a bit of advanced physics at college but somehow I dodged electronics and the practical aspects of electromagnetism almost completely.

Let's suppose we have a battery which is realized as a simple galvanic cell. Because of the difference in electronegativity, there are more electrons in the anode than there are in the kathode. Now if I connect the anode to the chassis, electrons flow from the anode to the chassis until both have the same charge density, right? And if I connect a load to the battery, electrons from the chassis and the anode are exchanged such that the charge density remains approximately the same? So could I say that after chassis grounding, I have the same kathode as before but a bigger anode which consists of the former anode and the chassis?

If the whole chassis gets charged, isn't there a problem that the charges may somehow escape to the environment? Do we have to pay attention to isolate the chassis from the environment?

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  • \$\begingroup\$ Rule number one: the sum of the voltage drops around any closed loop is zero. So, when you hook up the negative end of the battery to the cell, being that there is no closed loop, no voltage is dropped. No voltage difference means no current will flow. Rule number two: the sum of current from any node is zero. Basically, what goes in, must come out. By rule 1, nothing is coming out, so nothing goes in. This question seems to have you operating under the assumption that the chassis (a piece of metal) is charged. It is not. If it is, it is breaking the laws of thermodynamics \$\endgroup\$ – RYS Sep 3 '15 at 10:18
  • \$\begingroup\$ An enclosed piece of metal forms a Faraday cage, and we will find no net charge within it. Correspondingly, no charge from the outside will find its way in if you've made sure to handle all inputs and outputs to the case properly. This is a situation where using a chassis GND in a battery powered circuit would be useful. In higher powered grounds, chassis grounds are used for safety. \$\endgroup\$ – RYS Sep 3 '15 at 10:24
  • \$\begingroup\$ "If you were standing at arm’s length from someone and each of you had one percent more electrons than protons, the repelling force would be incredible. How great? Enough to lift the Empire State Building? No! To lift Mount Everest? No! The repulsion would be enough to lift a “weight” equal to that of the entire earth!"-Feynman \$\endgroup\$ – RYS Sep 3 '15 at 10:24
  • \$\begingroup\$ Thanks a lot for answering, RYS! However, I have problems to reconcile this with my naive picture of the situation. I have a neutral chassis and a negative anode. If I connect them with a wire, electrons can move between them. Diffusion then leads to an equilibirum where the charges of the anode are distributed equally over anode+chassis. Is this incorrect? \$\endgroup\$ – Fritz Sep 3 '15 at 10:56
  • \$\begingroup\$ no problem. I guess the point of me including the Feynman quote is to say that the metal is neutral. Charge density is just the charge per unit volume, but if the charge is balanced, then there is no potential. No potential means no current flow! Perhaps an experiment is in order. Hook the negative terminal of some cell up to a piece of metal and measure the current with the most sensitive ammeter you have available. You should see nothing. I'm an engineer, so I'll say 'just about nothing' :) \$\endgroup\$ – RYS Sep 3 '15 at 21:19
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I should start of by pointing out that when we talk about a "chassis ground" we usually mean we are connecting the chassis to some larger and better established ground potential like the earth itself. The reason to do this is for safety. If a high-voltage wire inside the equipment gets loose and contacts chassis, you'd rather the current be diverted to ground and a breaker gets thrown, than for the chassis to be sitting at high voltage waiting to electrocute the next user who touches it.

You are talking (sort of) about connecting the circuit ground to the chassis, which is not the usual meaning of "chassis ground".

Now if I connect the anode to the chassis, electrons flow from the anode to the chassis until both have the same charge density, right?

Charge will flow in a conductive structure until the two parts are at equal potentials.

Charge density will not be uniform throughout the structure. Charge will tend to accumulate near where there are external charges of opposite polarity, and will be depleted near where there are external charges of the same polarity.

And if I connect a load to the battery, electrons from the chassis and the anode are exchanged such that the charge density remains approximately the same?

Electrons are exchanged until the potential is equal between the battery's anode and the chassis.

So could I say that after chassis grounding, I have the same kathode as before but a bigger anode which consists of the former anode and the chassis?

Yes, that's reasonable.

If the whole chassis gets charged, isn't there a problem that the charges may somehow escape to the environment?

No. Kirchoff's current law says that current only flows in completed circuits. Charge may find a path back to the battery's cathode, and that would tend to deplete the battery. But it won't just flow away "into the environment" without the environment being connected back to the battery's cathode.

Do we have to pay attention to isolate the chassis from the environment?

Not particularly.

Edit:

how the potential can be equal if the charge density isn't.

See the electrostatic Poisson equation:

$$\nabla^2\varphi=-\frac{\rho}{\epsilon}$$

In fact, if the charge density is non-zero, then the potential will be nonuniform. Only when the net charge density is zero (which I guess is one example of a uniform charge density) will the potential be uniform.

In a condutor, the static behavior is for the free charge to move to the surfaces of the material, leaving zero net charge and a uniform potential within the bulk of the material.

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  • \$\begingroup\$ For further reading, this old question is slightly related: SPICE model for one terminal spherical capacitor \$\endgroup\$ – The Photon Sep 3 '15 at 15:57
  • \$\begingroup\$ Thanks a lot for answering and for correcting my terminology! What I still don't understand is how the potential can be equal if the charge density isn't. If after connecting the anode to the chassis, a surplus of electrons remains in the anode (due to the positively charged environment of the anode), how can the anode and the chassis have the same potential? Isn't the potential proportional to the amount of charge (i.e. something like the Coulomb potential)? \$\endgroup\$ – Fritz Sep 3 '15 at 16:45
  • \$\begingroup\$ On the other hand, it is clear that they must have the same potential because otherwise, charges would flow. I just can't reconcile this conflicting information and form a mental picture of the potential. I feel like I would understand what is going on if I could draw the analogy to a mass moving through the gravitational potential given by a landscape. But this probably isn't possible because in our case, the potential is generated by the mobile electrons themselves, in contrast to the landscape which constitutes a static background. \$\endgroup\$ – Fritz Sep 3 '15 at 17:01
  • \$\begingroup\$ @Fritz The electrons move so as to cancel out the external field. If you want to visualize a landscape, imagine pouring water over it: once it settles, the water is deeper (more charges) over the low spots, and flat (same potential) on top. \$\endgroup\$ – Kevin Reid Sep 3 '15 at 17:13
  • \$\begingroup\$ Thanks Kevin, that's an interesting idea! I'll have to think a bit about it in order to see whether it makes me comfortable with the situation. \$\endgroup\$ – Fritz Sep 3 '15 at 17:17

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