Three point charges Q, 2Q and 8Q are placed on a straight line 9 cm long. Charges are placed in such a way that the system has minimum potential energy. Then
 

Given charges Q, 2Q,8Q one placed on a straight line 9cm long. To get minimum potential energy large charges must be kept at ends of wire.
Let distance from 2Q is x, potential energy at ‘p’ is $\mathrm{P}=\frac{2{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0\left(\mathrm{x}\right)}+\frac{8{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0(9-\mathrm{x})}+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(\mathrm{Q}\right)16\mathrm{Q}}{9}[\mathrm{potential}\text{ energy}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1^2}{\mathrm{r}}]$
For minimum value, $\frac{\mathrm{dp}}{\mathrm{dx}}=0$

We get x=3, distance from 8Q
$9-\mathrm{x}=6\mathrm{cm}\mathrm{ }[\mathrm{Electric}\mathrm{ }\mathrm{field}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}]$
Electric field at $\mathrm{P}=\frac{2{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0\left(9\right)}-\frac{8{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0\left(36\right)}=\frac{4{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0\left(9\right)}-\frac{4{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0\left(9\right)}=0$