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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
$4Q$ and $16Q$ must be at the ends and $Q$ at a distance of $3$ $cm$ from the $16Q$.
$4Q$ and $16Q$ must be at the ends and $Q$ at a distance of $6 $ $cm$ from the $16Q.$
Electric field at the position of $Q$ is zero.
$B$ and $C$ both
Solution
For the system to be at minimum potential energy, the higher charged particles should be far apart.
and now potential energy
$\mathrm{U}=\frac{(\mathrm{k})\left(64 \mathrm{Q}^{2}\right)}{\mathrm{d}^{2}}+\frac{\mathrm{k}\left(16 \mathrm{Q}^{2}\right)}{\mathrm{d}-\mathrm{x}}+\frac{\mathrm{k} 4 \mathrm{Q}^{2}}{\mathrm{x}}$
$\frac{d U}{d x}=0 \Rightarrow \frac{+16}{(d-x)^{2}}=\frac{4}{x^{2}}$
$\Rightarrow \pm 2 x=d-x$
$\Rightarrow x \pm 2 x=d$
$\Rightarrow x=d / 3$ or $x=-d$
$\Rightarrow x=\frac{9}{3}=3 \mathrm{cm}$
Field at $Q$ is
$=\frac{k(4 Q)}{(3 c m)^{2}}-\frac{k(16 Q)}{(6 c m)^{2}}=0$
