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I am using an electric clearing field of 15v/10cm (~150V/m) to remove ions from a test chamber, however I'm wondering to what extent the localisation of the field's reference voltage plays a role.

schematic

simulate this circuit – Schematic created using CircuitLab

If, regardless of what reference value the "ground" is at, the clearing field is at 15V, does the field always remove both positive ions and negative ions?

Or does a field of -7.5 to 7.5V behave differently to a field of 100 to 115V, say removing mostly negative ions?

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  • \$\begingroup\$ Calculate the force on an ion. The answer should then be apparent. \$\endgroup\$ – Andrew Morton Sep 29 '17 at 8:40
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The circuit ground is just a reference point for measuring voltages in the circuit.

The clearing field is generated by the difference of electric potential (i.e. voltage) across the plates of the "capacitor", therefore the field configuration inside the chamber is not influenced by the actual potential of the ground node.

Of course this assumes that the chamber is well shielded/insulated and there is no other path (other than the one through that generator, that is) that charges may follow to reach ground.

Note that when you are saying that your ground node is at some voltage other than 0V (ground is at 0V by definition) you are implying there is another reference in the system with respect to which your ground node has a different potential. You should examine your overall system and ask yourself if this other reference point (some other object, e.g. the earthing ground of your electrical system) might interact with your chamber and if there is a low impedance path from that "new" reference to your chamber that might allow charges to follow a different route.

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