Question

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

Answer 1

Step 1: Calculation of the total charge

The total charge of the metallic spheres is the sum of the charges.

$Q_{\mathrm{total}}=Q_1+Q_2$

Here, $Q_{\mathrm{total}}$ is the total charge, $Q_1$ and $Q_2$ are the charges.

$$\begin{gathered} Q_{\mathrm{total}}=-1\times {10}^{-2}\mathrm{C}+5\times {10}^{-2}\mathrm{C} \\ {\mathrm{Q}}_{\mathrm{total}}=4\times {10}^{-2}\mathrm{C} \end{gathered}$$

Step 2: Calculation of charge distribution when connected

The charge distribution is directly proportional to the radius.

Hence, the final charge in the bigger sphere can be calculated using the following formula

$Q_{2\left(\mathrm{final}\right)}=\frac{r_2}{r_1+r_2}\times Q_{\mathrm{total}}$

Here, $Q_{2\left(\mathrm{final}\right)}$ is the final charge in the bigger sphere, $r_1$ and $r_2$ are the radius of the smaller and the bigger sphere respectively.

$$\begin{gathered} Q_{2\left(\mathrm{final}\right)}=\frac{3\mathrm{cm}}{1\mathrm{cm}+3\mathrm{cm}}\times \left(4\times {10}^{-2}\mathrm{C}\right) \\ {Q}_{2\left(\mathrm{final}\right)}=3\times {10}^{-2}\mathrm{C} \end{gathered}$$

Therefore, the final charge in the bigger sphere is $3\times {10}^{-2}\mathrm{C}$.