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Question
One hundred and twenty five small liquid drops, each carrying a charge of 0.5 µC and of diameter 0.1 m form a bigger drop. Calculate the potential at the surface of the bigger drop.
Solution
Given: n = 125, q = 0.5 x 10-6 C, d = 0.1 m
To find: Electrical Potential (V) =?
Formula: `V = Q/(4piε_0P)`
The radius of each small drop, r = d/2 = 0.05 m
The volume of the larger drop being equal to the volume of the n smaller drops, the radius of the larger drop is
R = `root(3)("nr") = root(3)(125) (0.05) = 5 xx 0.05 = 0.25` m
The charge applied on the larger drop,
Q = nq = 125 × (0.5 × 10-6) C
∴ The electric potential of the surface of the larger drop,
V = `1/(4piε_0) "Q"/"R" = (9 xx 10^9) xx (125 xx (0.5 xx 10^-6))/0.25`
= 9 × 125 × 2 × 103 = 2.25 × 106 V
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