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Two metallic spheres of radii 1 cm and 3 cm are given charges of $-1\times10^{-2}$ C and $5\times10^{-2}$ C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
A
$1\times10^{-2}$ C
B
$2\times10^{-2}$ C
C
$3\times10^{-2}$ C
D
$4\times10^{-2}$ C
Detailed Solution
When connected, the spheres reach a common potential: $V = \frac{Q_1 + Q_2}{C_1 + C_2}$
Charge on the bigger sphere: $Q_2' = C_2V = \left(\frac{C_2}{C_1 + C_2}\right)(Q_1 + Q_2)$
For a sphere $C = 4\pi\varepsilon_0R$, so $C_1 = 4\pi\varepsilon_0R_1$ and $C_2 = 4\pi\varepsilon_0R_2$
$Q_2' = \left(\frac{R_2}{R_1 + R_2}\right)(Q_1 + Q_2) = \left(\frac{3}{3 + 1}\right)(5 - 1)\times10^{-2}$
$Q_2' = \frac{3}{4}\times4\times10^{-2} = 3\times10^{-2}$ C
Charge on the bigger sphere: $Q_2' = C_2V = \left(\frac{C_2}{C_1 + C_2}\right)(Q_1 + Q_2)$
For a sphere $C = 4\pi\varepsilon_0R$, so $C_1 = 4\pi\varepsilon_0R_1$ and $C_2 = 4\pi\varepsilon_0R_2$
$Q_2' = \left(\frac{R_2}{R_1 + R_2}\right)(Q_1 + Q_2) = \left(\frac{3}{3 + 1}\right)(5 - 1)\times10^{-2}$
$Q_2' = \frac{3}{4}\times4\times10^{-2} = 3\times10^{-2}$ C