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Question

Eight charged liquid drops each of radius 1mm and charge 0.1nC, merge together to form a bigger drop. The electric potential at the surface of single big drop is:

A
3.6kV
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B
7.2kV
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C
1.8kV
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D
36kV
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Solution

The correct option is B 3.6kV
Let, the density of liquid =S
given, radius =1×103m
charge =0.1nc
let mass =x
Now, as we know, S=massvolume

The radius of the bigger drop is given by R.

and for small drop, S=x(43πr3)r3=x43πS

For bigger drop, S=8x(43πR3)R3=8x43πS

So, R3r3=8R3=8x3R=2r

R=2×103m; charge on the bigger drop is given by, Q=8×0.1nc

=8×1010C

Potential on the surface is determined as,
=KQR=9×109×8×10102×103=3.6kV

Option A is the correct answer.

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