Two small identical metal balls are equally charged and placed at a fixed distance away from each other.…
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
$\frac{F}{2}$
F
$4 \mathrm{~F}$
$2 \mathrm{~F}$
Solution
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.
$\therefore \mathrm{F}_{\text {net }}=\mathrm{F}_1-\mathrm{F}_2=\frac{2 \mathrm{kQ}^2}{\mathrm{r}^2}-\frac{\mathrm{kQ}^2}{\mathrm{r}^2}=\mathrm{F}$