Three point charges Q, 4Q and 16Q 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, 4Q, 16Q one placed on a straight line 9cm long. To get minimum potential energy larger charges must be kept at ends of wire.

Let distance of Q from 4Q is x, potential energy of the system is

$\begin{array}{l}{\mathrm{U}}_{\mathrm{system}}=\frac{4{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0\left(\mathrm{x}\right)}+\frac{16{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0(9-\mathrm{x})}+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(4\mathrm{Q}\right)16\mathrm{Q}}{9^2}\\ \left[\because \mathrm{Potential} \mathrm{Energy}: \mathrm{U}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{r}}\right]\end{array}$

We get minimum value of potential energy when  $\frac{{\mathrm{dU}}_{\mathrm{system}}}{\mathrm{dx}}=0$

$\Rightarrow \mathrm{x}=3 \mathrm{cm}$

distance of charge Q from 16Q = $9-x=6 \mathrm{cm}$ 

 $\mathrm{Electric} \mathrm{field}:\hspace{0.17em}\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

Electric field at P = ${\mathrm{E}}_{\mathrm{P}}=\frac{4{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0\left(9\right)}-\frac{16{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0\left(36\right)}=\frac{4{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0\left(9\right)}-\frac{4{\mathrm{Q}}^2}{4{\mathrm{\pi \epsilon }}_0\left(9\right)}=0$