A charge $\mathrm{Q}$ is to be divided between two objects. The values of the charges on the objects so that…
A charge $\mathrm{Q}$ is to be divided between two objects. The values of the charges on the objects so that the electrostatic force between them will be maximum is
$\frac{\mathrm{Q}}{2}, \frac{\mathrm{Q}}{2}$
$\frac{Q}{3}, \frac{2}{3} Q$
$\frac{\mathrm{Q}}{4}, \frac{3}{4} \mathrm{Q}$
$\frac{\mathrm{Q}}{5}, \frac{4}{5} \mathrm{Q}$
Solution
Let charge on one object be 'q', then on other object it is ' $\mathrm{Q}-\mathrm{q}^{\prime}$
So, $\mathrm{F}_{\mathrm{e}}=\frac{\mathrm{K} \mathrm{q}(\mathrm{Q}-\mathrm{q})}{\mathrm{r}^2}$.
For $\mathrm{F}_{\mathrm{e}}$ to be maximum, $\frac{\mathrm{dF}_{\mathrm{e}}}{\mathrm{dq}}=0$
Now, $\frac{\mathrm{dF}_{\mathrm{e}}}{\mathrm{dq}}=0 \Rightarrow \frac{\mathrm{d}}{\mathrm{dq}}\left[\mathrm{qQ}-\mathrm{q}^2\right]=0$
$\Rightarrow \mathrm{Q}-2 \mathrm{q}=0 \Rightarrow \mathrm{q}=\frac{\mathrm{Q}}{2}$
Then, $(\mathrm{Q}-\mathrm{q})=\mathrm{Q}-\frac{\mathrm{Q}}{2}=\frac{\mathrm{Q}}{2}$
So, force will be maximum when magnitude of charges are $\frac{\mathrm{Q}}{2}$ and $\frac{\mathrm{Q}}{2}$.