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Three connected conductors A, B and C have a total charge of 48 $μ$ C . The ratio of their capacitances are 1:3:2. The charges on them individually.

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16 $μ$ C ,16 $μ$ C , 16 $μ$ C

24 $μ$ C ,12 $μ$ C , 12 $μ$ C

8 $μ$ C ,18 $μ$ C , 22 $μ$ C

8 $μ$ C ,24 $μ$ C , 16 $μ$ C

answer is C.

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Detailed Solution

$q_{1} = \frac{q}{c_{1} + c_{2} + c_{3}} c_{1} and so on$

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