Eight drops of mercury, each of same radius and same charge combine to form a bigger drop. The ratio of capacitance of the bigger drop to that of each smaller drop is

Question

Accepted Answer

Correct answer: C. (c) Number of drops, n=8 Let, r is radius of smaller drop and R is the radius of Bigger drop. Volume of smaller drops & bigger drop remains same 8 43 r^3= 43 r^3 ⇒R=2r Capacitance of spherical conductor, C=4 ϵ_0R Ratio of capacitance of bigger drop to smaller drop is given as CC^ '= 4 ϵ_0R4 ϵ_0r= Rr= 2r4=2 :1 CC^ '= 21

$Eight drops of mercury, each of same radius and same charge combine to form a bigger drop. The ratio of capacitance of the bigger drop to that of each smaller drop is$

Solution

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