A parallel plate capacitor is charged to a potential V , and stores an amount of charge Q it is then disconnected from the battery. The distance between the plates is then halved. What happens to (a) the charge on plate? the electric field? (c) the energy stored in the electric field?
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Expert's answer
2012-11-22T07:53:50-0500
A parallel plate capacitor is charged to a potential V , and stores an amount of charge Q it is then disconnected from the battery. The distance between the plates is then halved. What happens to (a) the charge on plate? (b) the electric field? (c) the energy stored in the electric field? Answer (a) No Change.& We aren’t attached to a battery, so the charge is fixed.& (b) No Change.& The charge is constant so, in the planar geometry, the field doesn’t change. (c) Halves.& The volume in which we have field halves, so the energy does too.
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