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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 point on the x-axis where electric field is zero .

$(a)\;\pm \sqrt{\large\frac{5}{2}}\qquad(b)\;\pm \sqrt{\large\frac{3}{2}}\qquad(c)\;\pm \sqrt{\large\frac{27}{2}}\qquad(d)\;\pm \sqrt{\large\frac{15}{2}}$

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Answer : (a) $\;\pm \sqrt{\large\frac{5}{2}}$
Explanation :
In the figure
$q=10 \mu C=10^{-6} C$
$q_{0}=0.1 \mu C = 10^{-7} C$
$m=6\times10^{-4} kg$
$Q=8 \mu C =8 \times 10^{-6} C$
Let P be any point at a distance x from origin then
$AP=CP$
$BP=DP$
Electric potential at P will be
$V=\large\frac{2kQ}{BP}-\large\frac{2kq}{AP}$
$V=2k\;[\large\frac{Q}{(\large\frac{27}{2}+x^2)^{\large\frac{1}{3}}}-\large\frac{q}{(\large\frac{3}{2}+x^2)^{\large\frac{1}{2}}}]$
$V=2\times9\times10^{9}\times10^{-6}\;[\large\frac{8}{(\large\frac{27}{2}+x^2)^{\large\frac{1}{2}}}-\large\frac{1}{(\large\frac{1}{2}+x^2)^{\large\frac{1}{2}}}]$
Electric field at P is
$E=-\large\frac{dV}{dx}=-1.8\times10^{4}\;[8\times(-\large\frac{1}{2})\times\large\frac{2x}{(\large\frac{27}{2}+x^2)^{\large\frac{3}{2}}}+\large\frac{1}{2}\;\large\frac{2x}{(\large\frac{3}{2}+x^2)^{\large\frac{3}{2}}}]$
$E=1.8\times 10^{4}\;[\large\frac{8x}{(\large\frac{27}{2}+x^2)^{\large\frac{3}{2}}}-\large\frac{x}{(\large\frac{3}{2}+x^2)^{\large\frac{3}{2}}}]$
$E=0 \;$ at
$\large\frac{8x}{(\large\frac{27}{2}+x^2)^{\large\frac{3}{2}}}=\large\frac{x}{(\large\frac{3}{2}+x^2)^{\large\frac{3}{2}}}$
$\large\frac{4}{(\large\frac{27}{2}+x^2)}=\large\frac{1}{(\large\frac{3}{2}+x^2)}$
$\large\frac{4\times3}{2}+4x^2=\large\frac{27}{2}+x^2$
$3x^2=\large\frac{27-12}{2}=\large\frac{15}{2}$
$x=\pm\sqrt{\large\frac{5}{2}}\;.$
answered Feb 19, 2014 by yamini.v
edited Feb 20, 2014 by yamini.v
 

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