If n drops of equal size and same potential are converted into a bigger drop, then the potential of bigger drop will be
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This implies, the capacity of the bigger drop is n1/3 times the capacity of each small drop. Ptential of 1 big drop is given by V=n. So, the potential of the bigger drop is n2/3 times the potential of the smaller drop. Two charges 2μC and –2μC are placed at points A and B, 6 cm apart
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