Questions 14 to 20 are all about the same problem: Imagine that a parallel plate
ID: 3161954 • Letter: Q
Question
Questions 14 to 20 are all about the same problem: Imagine that a parallel plate capacitor of capacitance C is charged so that the potential difference between its plates is V. The capacitor is then disconnected so that charges cannot get on the plates and cannot leave the plates. We physically pull the plates apart so that their separation doubles. This is done without adding charges to or subtracting charges from the plates. We neglect fringing effects. After the plate separation has been doubled: Has the charge on the plates increased, decreased, or stayed the same? Defend your answer. If it did change, by what factor? Has the electric field between the plates increased, decreased, or stayed the same? Defend your answer. If it did change, by what factor?Explanation / Answer
14)
the charge on the plates will stay the same.
as there is no battery or circuit , the charge can't move from the plates and it will stay the same.
It will not change.
15)
electric field between the plates will be same.
electric field between the plates is given as
Electric field = Q/(Area * epsilon)
as both area and Q are same
electric field will be same as well
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