Question

Two metallic spheres of radii $2\text{cm}$ and $6\text{cm}$ are given charge $3\times {10}^{-2}\text{C}$ and $7\times {10}^{-2}\text{C}$, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is :

• A. $7.5\times {10}^{-2}\text{C}$
• B. $2.5\times {10}^{-2}\text{C}$
• C. $10\times {10}^{-2}\text{C}$
• D. $5\times {10}^{-2}\text{C}$

Answer 1

The correct option is A $7.5\times {10}^{-2}\text{C}$

Let the final charge on the bigger sphere be $Q_1$ and on the smalller sphere be $Q_2$.

We know that after redestribution of charges potential of both the spheres will be constant.
$\Rightarrow V_1=V_2$

$\Rightarrow \frac{k{Q}_1}{6}=\frac{k{Q}_2}{2}$

Or, $Q_1=3Q_2$ -------$\left(1\right)$

Also, sum of charges on both of the spheres will remain same.
i.e. $Q_1+Q_2=(3\times {10}^{-2}+7\times {10}^{-2})\text{C}$

$\Rightarrow Q_1+Q_2=10\times {10}^{-2}\text{C}$

$\Rightarrow 3Q_2+Q_2=10\times {10}^{-2}\text{C}$
[From $\left(1\right)$]

Or, $Q_2=2.5\times {10}^{-2}\text{C}$

$\Rightarrow Q_1=7.5\times {10}^{-2}\text{C}$