Question

A small metal sphere of radius r initially has a charge q₀ . Then a long copper wire is connected from this small sphere to a distant, large, uncharged metal sphere of radius R. Calculate the final charge q on the small sphere and the final charge on the large sphere. You may neglect the small amount of charge on the wire. What other approximations did you make? (Think about potential…)

Short answer

The final charge on small and large sphere is $\frac{q_0}{\left(1+{\displaystyle \frac{R}{r}}\right)}$ and $\frac{q_0}{\left(1+{\displaystyle \frac{r}{R}}\right)}$.

Step-by-step solution

Identification of given data

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 .

Conceptual Explanation

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

Determination of final charge on small and large sphere

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 $\frac{q_0}{1+{\displaystyle \frac{r}{R}}}$.