Example - 1 1000 equal drops of radius 1cm, and charge 1 × 10-6 C are fused to form one bigger drop. The ratio of potential of bigger drop to one smaller drop, and the electric field intensity on the surface of bigger drop will be respectively –
Solution : Let the potential of one smaller drop be B then potential of bigger drop, is V’ = n2/3 V
Þ (V’/V) = n2/3 = (1000)2/3 = 100
\ V’:V = 100:1
Also let the electric field on the surface of smaller drop be E then electric field on bigger drop is
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