Two small identical metal balls are equally charged and placed at a fixed distance away from each other. They experience the electrostatic force ' \(F\) ' A similar uncharged ball after touching one of them is placed at the middle point between the two balls. The force experienced by this ball is

The force between equally charged identical balls is:
\(\mathrm{F}=\frac{\mathrm{kQ}^2}{\mathrm{r}^2}\)
When a similar uncharged ball, touches one of them. This ball aquires a charge \(\frac{\mathrm{Q}}{2}\) and the ball being touched looses \(\frac{\mathrm{Q}}{2}\).
Now, the force experienced by the third ball is given by,
\(\mathrm{F}_{\text {net }}=\mathrm{F}_1-\mathrm{F}_2\)
where, \(\mathrm{F}_1=\frac{\mathrm{kQ}^2 / 2}{\left(\frac{\mathrm{r}}{2}\right)^2}=\frac{2 \mathrm{kQ}^2}{\mathrm{r}^2}\) is the repulsion from the first ball
and \(\mathrm{F}_2=\frac{\mathrm{k}\left(\frac{\mathrm{Q}}{2}\right)^2}{\left(\frac{\mathrm{r}}{2}\right)^2}=\frac{\mathrm{kQ}^2}{\mathrm{r}^2}\) is the repulsion from the second ball.
\(\therefore \mathrm{F}_{\text {net }}=\mathrm{F}_1-\mathrm{F}_2=\frac{2 \mathrm{kQ}^2}{\mathrm{r}^2}-\frac{\mathrm{kQ}^2}{\mathrm{r}^2}=\mathrm{F}\)