Asked by arvind95 | 16th Dec, 2012, 01:44: PM
As the distance between the capacitor plates is increased, the capacitance decreased as it is inversely propotional to the distance between the plates.
Since the external circuit remains same, the potential difference between the plates remain same as it is maintained due to the external emf.
The electric field decreases because it is inversely proportional to the distance between the plates.
the charge on the plates is given as Q=CV and since V remains same but C decreases thus Q decreases.
Answered by | 22nd Jan, 2013, 09:11: AM
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