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A particle of mass m and charge +q is projected from the origin with a speed v at an angle $\;30^{0}\;$ with the horizontal into an electric field of intensity E directed downward . Equation of trajectory of particle is . Neglect gravity

$(a)\;y=\large\frac{x}{\sqrt{3}}+\large\frac{2}{3}\;\large\frac{qEx^2}{mv^2}\qquad(b)\;y=x \sqrt{3} +\large\frac{2}{3}\;\large\frac{qEx^2}{mv^2}\qquad(c)\;y=\large\frac{x}{\sqrt{3}}-\large\frac{2}{3}\;\large\frac{qEx^2}{mv^2}\qquad(d)\;y=x \sqrt{3} -\large\frac{3}{2}\;\large\frac{qEx^2}{mv^2}$

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Answer : (c) $y=\large\frac{x}{\sqrt{3}}-\large\frac{2}{3}\;\large\frac{qEx^2}{mv^2}$
Explanation :
There is no acceleration along x - axis Therefore
$x=(v cos 30^{0})\;t$
$x=\large\frac{v \sqrt{3} t}{2}$
Acceleration along y-axis $a_{y}=-\large\frac{qE}{m}$
Thus
$y=(v sin 30^{0})\;t+\large\frac{a_{y}}{2}\;t^2$
$y=(v sin 30^{0})\;t-\large\frac{qE}{2 m}\;t^2$
$y=\large\frac{x}{\sqrt{3}}-\large\frac{qE}{2 m}\;\large\frac{x^2}{v^2}\times\large\frac{4}{3}$
$y=\large\frac{x}{\sqrt{3}}-\large\frac{2}{3}\;\large\frac{qEx^2}{mv^2}\;.$

 

answered Feb 14, 2014 by yamini.v
edited Aug 16, 2014 by thagee.vedartham
 

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