Two charges $-q$ each are separated by distance $2d$. A third charge $+ q$ is kept at mid point $O$. Find potential energy of $+ q$ as a function of small distance $x$ from $O$ due to $-q$ charges. Sketch $P.E.$ $v/s$ $x$ and convince yourself that the charge at $O$ is in an unstable equilibrium.
Two charges $-q$ each are separated by distance $2d$. A third charge $+ q$ is kept at mid point $O$. Find potential energy of $+ q$ as a function of small distance $x$ from $O$ due to $-q$ charges. Sketch $P.E.$ $v/s$ $x$ and convince yourself that the charge at $O$ is in an unstable equilibrium.
Suppose $+q$ charge is displaced towards $-q$ and hence total potential energy of system,
$\mathrm{U}=k\left[\frac{-q^{2}}{(d-x)}+\frac{-q^{2}}{(d+x)}\right]$ $=-k q^{2}\left[\frac{d+x+d-x}{d^{2}-x^{2}}\right]$
$=-k q^{2}\left[\frac{2 d}{d^{2}-x^{2}}\right]$
$\therefore$ By differentiating U w.r.t. $x$,
$\frac{d \mathrm{U}}{d x}=-k q^{2} \times 2 d\left(\frac{-2 x}{\left(d^{2}-x^{2}\right)^{2}}\right)$
If $\frac{d \mathrm{U}}{d x}=0$, then $\mathrm{F}=0$
$\therefore 0=\frac{4 k q^{2} d x}{\left(d^{2}-x^{2}\right)^{2}}, \quad \therefore x=0$
Hence $+q$ is in stable and unstable equilibrium
By differentiating again w.r.t. $x$,
$\frac{d^{2} \mathrm{U}}{d x^{2}}=\left[\frac{-2 d q^{2}}{4 \pi \epsilon_{0}}\right]\left[\frac{2}{\left(d^{2}-x^{2}\right)^{2}}-\frac{8 x^{2}}{\left(d^{2}-x^{2}\right)^{3}}\right]$
$=\left(\frac{-2 d q^{2}}{4 \pi \epsilon_{0}}\right) \frac{1}{\left(d^{2}-x^{2}\right)^{3}}\left[2\left(d^{2}-x^{2}\right)-8 x^{2}\right]$
system is in unstable equilibrium.
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