Three point charges Q, 4Q and 16Q are place | vedclass

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Question

For the system to be at minimum potential energy, the higher charged particles should be far apart.

and now potential energy

$\mathrm{U}=\frac{(

Vedclass · Accepted Answer

$B$ and $C$ both

Vedclass · Answer

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

$=\frac{k(4 Q)}{(3 c m)^{2}}-\frac{k(16 Q)}{(6 c m)^{2}}=0$

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

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