When one of the plates of an isolated capacitor is grounded, does the charge become zero on that plate or just the charge on the outer surface become zero?
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$\begingroup$ The charge on that plate becomes the same as the charge on Earth. $\endgroup$– Hot LicksCommented Jul 9, 2015 at 13:03
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$\begingroup$ No, it can get charged, see answer here: physics.stackexchange.com/q/191102 $\endgroup$– RolCommented Jul 9, 2015 at 18:04
1 Answer
The voltage becomes the same as Earth, but this doesn't mean that the charge goes to "zero". By "zero" here, I mean that the positive charges (nuclei) are perfectly balanced by the negative charges (electrons). You can call the voltage of Earth 0 Volts, but this is a relative measure. Charge, in the usage here, is not a relative measure because it is a material property of the conductor.
From the phrasing of the question, I predict that the other sheet in the capacitor has a charge and is electrically isolated. By problem specification, the voltage difference between the isolated plate and Earth is some voltage (say V) and this voltage difference is the same as between the plates (because the grounded plate is same voltage as Earth).
As a rule of thumb, a capacitor's plates have opposite and equal charges. This means that the grounded plate has the opposite charge of the isolated (charged) plate, even though it's voltage is zero. This charge, yes, will be mostly located on the surfaces or other edges.
It's the electric field from the isolated plate that does this. The presence of its charge pushes other charges in/out of the grounded plate. That charge goes into or out of the Earth because it is an infinite charge sink/source.
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$\begingroup$ Let's say I have a charged plate capacitor. There's a non-zero voltage across the plates. What if I connected only one plate to some object (possibly ground)? Can it change the voltage across the plates? I'm almost sure the answer is yes, it's sufficient if the object we connect the plate to is of a different potential than the plate. Then both the plate and the object will have equal potential, and the voltage across the plates will be different than it was before we connected the plate to this object. $\endgroup$ Commented Jul 5, 2016 at 18:30