# A system of three charges are placed as shown in the figure :If $D> > d,$ the potential energy of the system is best given by : Option 1)   $\frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}-\frac{q\; Q\; d}{2D^{2}} \right ]$ Option 2)    $\frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}+\frac{2\; q\; Q\; d}{D^{2}} \right ]$ Option 3)   $\frac{1}{4\pi \epsilon _{0}}\left [ +\frac{q^{2}}{d}+\frac{q\; Q\; d}{D^{2}} \right ]$      Option 4) $\frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}-\frac{q\; Q\; d}{D^{2}} \right ]$

$u=\frac{-kq^{2}}{d}+\frac{k\: q\:Q}{D+\frac{d}{2}}-\frac{k\: q\: Q}{D-\frac{d}{2}}$

$u=\frac{-kq^{2}}{a}+\frac{kqQ\left ( -d \right )}{D^{2}-\frac{a^{2}}{4}}$

$u=k\left [ \frac{-q^{2}}{a}-\frac{q\; Q\; d}{D^{2}} \right ]$

$u=\; \frac{1}{4\pi \epsilon _{0}}\left [ \frac{-q^{2}}{d}-\frac{q\; Q\; d}{D^{2}} \right ]$

$D> > a$

Option 1)

$\frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}-\frac{q\; Q\; d}{2D^{2}} \right ]$

Option 2)

$\frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}+\frac{2\; q\; Q\; d}{D^{2}} \right ]$

Option 3)

$\frac{1}{4\pi \epsilon _{0}}\left [ +\frac{q^{2}}{d}+\frac{q\; Q\; d}{D^{2}} \right ]$

Option 4)

$\frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}-\frac{q\; Q\; d}{D^{2}} \right ]$

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