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Question
A metallic sphere A isolated from ground is charged to +50 μC. This sphere is brought in contact with other isolated metallics sphere B of half the radius of sphere A. The charge on the two-sphere will be now in the ratio
Options
1 : 2
2 : 1
4 : 1
1 : 1
Solution
2 : 1
Explanation:
Two metallic spheres brought together in contact with each other.
Radius of A = r1
Radius of B = r2
also r1 = 2 r2
A charge will flow from sphere A to sphere B till the potential becomes the same for both A and B.
∴ VA = VB
`∴ 1/(4piepsilon_0) "q"_"A"/"r"_1 = 1/(4piepsilon_0) "q"_"B"/"r"_2`
`∴ cancel(1/(4piepsilon_0)) "q"_"A"/"r"_1 = cancel(1/(4piepsilon_0)) "q"_"B"/"r"_2`
`∴ "q"_"A"/r_1 = q_"B"/r_2`
`∴ "q"_"A" = r_1/r_2 "q"_"B"`
`∴ "q"_"A" = (2 r_2)/r_2"q"_"B"`
`∴ "q"_"A" = 2" q"_"B"`
`∴ "q"_"A"/ " q"_"B" = 2/1 ~~ 2 : 1`
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