$(a)\;3\times10^{-4} J \qquad(b)\;2.6\times10^{-4} J\qquad(c)\;2\times10^{-4} J\qquad(d)\;1.5\times10^{-4} J$

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Answer : (b) $\;2.6\times10^{-4} J$

Explanation :

Potential at origin is

$V_{0}=1.8\times10^{4}\;[\large\frac{8}{(\large\frac{27}{2})^{\large\frac{1}{2}}}-\large\frac{1}{(\large\frac{3}{2})^{\large\frac{1}{2}}}]$

$V_{0}=1.8\times10^{4}\;[\large\frac{8}{3 \sqrt{\large\frac{3}{2}}}-\large\frac{1}{\sqrt{\large\frac{3}{2}}}]$

$V_{0}=1.8\times10^{4}\times\sqrt{\large\frac{2}{3}}\times\large\frac{5}{3}$

$V_{0}=3\times10^{4} \sqrt{\large\frac{2}{3}}$

$V_{0}=\sqrt{6}\times10^{4}$

$V_{0} \approx 2.44\times10^{4}$

Let T be the kinetic energy of the particle at origin .

Applying energy conservation at x=0 and at $\;x=\infty$

$T+q_{0} V_{0}=\large\frac{1}{2} mV_{0}^{2}$

$T=\large\frac{1}{2}\;m V_{0}^{2}-q_{0}V_{0}$

$T=\large\frac{1}{2}\times6\times10^{-4}\times3^2-2.44\times10^{4}\times10^{-7}$

$T=27\times10^{-4}-2.44\times10^{-4}$

$T=2.6\times10^{-4} J\;.$

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