Question

Two metallic spheres of radii $1\text{cm}$ and $3\text{cm}$ are given charges of $-1\times {10}^{-2}\text{C}$$\text{and}5\times {10}^{-2}\text{C}$, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
[NEET-2012 (Mains)]

• A. $2\times {10}^{-2}C$
• B. $3\times {10}^{-2}C$
• C. $1\times {10}^{-2}C$
• D. $4\times {10}^{-2}C$

Answer 1

The correct option is B $3\times {10}^{-2}C$

Charge flow from high potential to low potential. If these metallic spheres are connected by a conducting wire final potential of the spheres will be equal.

Let, $Q_1$ and $Q_2$ are the final charges on the spheres, then
Common potential, $V=\frac{Q_1+Q_2}{C_1+C_2}$

Charge on bigger sphere is $Q_2^1=C_{2}V$

$Q_2^1=\left(\frac{C_2}{C_1+C_2}\right)(Q_1+Q_2)$

$C_1=4\pi {ϵ}_o{R}_1$

$C_2=4\pi {ϵ}_o{R}_2$

$\Rightarrow Q_2^1=\left(\frac{R_2}{R_1+R_2}\right)(Q_1+Q_2)=\left(\frac{3}{3+1}\right)(5-1)\times {10}^{-2}=\frac{3}{4}\times 4\times {10}^{-2}=3\times {10}^{-2}C$

Hence, option $\left(C\right)$ is correct.