Two metal spheres of radii 0.01 m and 0.02 m are given a charge of 15mC and 45mC respectively. They are then connected by a wire. The final charge on the first sphere is _____ `xx10^(-3)C`

Options

A. 40
B. 30
C. 20
D. 10

Answer

C

Explanation

The situation is as shown in the figure.

When they are connected by a wire the charge flows and both the spheres attain the common potential which is given by
`V= (Q_(1)+Q_(2))/(C_(1) +C_(2)) = (Q_(1) + Q_(2))/(4pi in_(0) (R_(1) + R_(2)))` ....(i)
`:.` Final charge on the first sphere is
`Q_(1) = C_(1)V = (4pi in_(0)R_(1) (Q_(1) + Q_(2)))/(4pi in_(0) (R_(1) + R_(2)))` (Using (i))
`=(R_(1)(Q_(1) + Q_(2)))/(R_(1) +R_(2))`
Here, `R_(1) = 0.01 m, R_(2) = 0.02m`
`Q_(1) = 15mC= 15 xx 10^(-3)C`
`Q_(2) = 45mC = 45 xx 10^(-3)C`
`:. Q'_(1) = (0.01(15 xx 10^(-3) + 45 xx 10^(-3)))/(0.01 + 0.02)`
`=(0.01 xx 60 xx 10^(-3))/(0.03) = 20 xx 10^(-3)C`