91Ó°ÊÓ

The initial charge of small metal sphere is q₀ .

The radius of small metal sphere is r .

The final charge of small metal sphere is q

The final charge of large metal sphere is Q

The radius of the large metal sphere is R .

The potential of each sphere become equal after joining both spheres by metal wire.

The charge on the small sphere after connection of both spheres is given as: q = (q₀- Q) ....................(1)

The final electric potential on small sphere is given as: $V_{sf}=\frac{k_q}{r}$

The final electric potential on large sphere is given as: $V_{if}=\frac{kQ}{R}$

At equilibrium final potential of small and large sphere becomes equal, so $V_{sf}=V_{if}$

Substitute all the values in the above equation.

$$\begin{gathered} \frac{kq}{r}=\frac{kQ}{R} \\ \frac{q}{r}=\frac{Q}{R} \\ \frac{q_0-Q}{r}=\frac{Q}{R} \\ Q=\frac{q_0}{\left(1+{\displaystyle \frac{r}{R}}\right)} \end{gathered}$$

Substitute this value in equation (1) to find final charge on small sphere.

$$\begin{gathered} q=\left(q_0-\frac{q_0}{\left(1+{\displaystyle \frac{r}{R}}\right)}\right) \\ q=\frac{q_0}{\left(1+{\displaystyle \frac{R}{r}}\right)} \end{gathered}$$

Therefore, the final charge on small and large sphere is $\frac{q_0}{1+{\displaystyle \frac{R}{r}}}$ and $\frac{q_0}{1+{\displaystyle \frac{r}{R}}}$.