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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. Find the least value of $\;V_{0}\;$ for which the particle will cross the origin .

$(a)\;2 \sqrt{2} m/s\qquad(b)\;3 \sqrt{3} m/s\qquad(c)\;2 m/s\qquad(d)\;3 m/s$

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Answer : (d) 3 m/s
Explanation :
The least value of kinetic energy of the particle at infinity should be enough to take the particle upto $x=+\sqrt{\large\frac{5}{2}}\;$ because
at $x=+\sqrt{\large\frac{5}{2}}\;,E=0\;$ Electrostatic force on $\;q_{0}\;$ is zero
for $\;x > \sqrt{\large\frac{5}{2}}\;,E\;$ is repulsive (towards +ve x-axis)
for $\;x < \sqrt{\large\frac{5}{2}}\;,E\;$ is attractive (towards _ve x-axis)
Now potential at $\;\sqrt{\large\frac{5}{2}} m\;$ is
$V=1.8\times10^{4}\;[\large\frac{8}{(\large\frac{27}{2}+\large\frac{5}{2})^{\large\frac{1}{2}}}-\large\frac{1}{(\large\frac{3}{2}+\large\frac{5}{2})^{\large\frac{1}{2}}}]$
$V=1.8\times1o^{4}\;[\large\frac{8}{4}-\large\frac{1}{2}]$
$V=1.8\times10^{4}\times\large\frac{3}{2}$
$V=2.7\times10^{4}V$
Applying energy conservation at $\;x=\infty\;and\;$ $\;x=\sqrt{\large\frac{5}{2}} m$
$\large\frac{1}{2}\;mv_{0}^{2}=q_{0}V$
$v_{0}=\sqrt{\large\frac{2q_{0}V}{m}}$
$v_{0}=\sqrt{\large\frac{2\times10^{-7}\times2.7\times10^{4}}{6\times10^{-4}}}$
$v_{0}=3 m/s$
Minimum value of $\;v_{0}\;$ is 3 m/s
answered Feb 19, 2014 by yamini.v
 

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