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Four point charges $\;+8 \mu C\;,-1 \mu C\;,-1 \mu C\;and\;+8 \mu C\;$ are fixed at the points $\;-\sqrt{\large\frac{27}{2}} m \;,- \sqrt{\large\frac{3}{2}} m \;,+ \sqrt{\large\frac{3}{2}} m \;and\;+\sqrt{\large\frac{27}{2}} m\;$ respectively on the Y-axis . A particle of mass $\;6\times10^{-4} kg\;$ and charge $\;0.1 \mu C \;$ moves along the - X direction . It's speed at $\;x=+ \infty\;$ is $\;V_{0}\;. \quad \;$ Answer the following question based on the passage. Kinetic energy of the particle at origin would be

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

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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 $\;x=\infty$
$T+q_{0} V_{0}=\large\frac{1}{2} mV_{0}^{2}$
$T=\large\frac{1}{2}\;m V_{0}^{2}-q_{0}V_{0}$
$T=\large\frac{1}{2}\times6\times10^{-4}\times3^2-2.44\times10^{4}\times10^{-7}$
$T=27\times10^{-4}-2.44\times10^{-4}$
$T=2.6\times10^{-4} J\;.$
answered Feb 19, 2014 by yamini.v
 

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