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A particle of mass m and charge q is fastened to one end of a string of length l . The other end of the string is fixed to the point O . The whole system lies on a frictionless horizontal plane . Initially , the mass is at rest at A . A uniform electric field in the direction shown is the switched on . Then speed of the particle at B and tension in the string are

$(a)\;\sqrt{\large\frac{2qEL}{m}} , 2qE\qquad(b)\;\sqrt{\large\frac{qEL}{m}} ,2qE \qquad(c)\;\sqrt{\large\frac{qEL}{m}} ,qE \qquad(d)\;\sqrt{\large\frac{2qEL}{m}} ,3qE $

1 Answer

Answer : (b) $\;\sqrt{\large\frac{qEL}{m}} ,2qE $
Explanation :
From A to B
Work done by electric field = charge in kinetic energy of particle
$l^{|}\;$ = distance travelled by particle along the direction of electric field .
$l^{|}=l-l cos 60^{0}=\large\frac{l}{2}$
Work done by electric field = $\;qEl^{|}$
$=\large\frac{qEl}{2}$
$\large\frac{qEl}{2}=\large\frac{1}{2}\;mv^2$
$v=\sqrt{\large\frac{qEl}{m}}$
At B , F .B .O of particle is
$T - qE=\large\frac{mv^2}{l}$
$T=qE+qE$
$T=2qE\;.$
answered Feb 13, 2014 by yamini.v
edited Feb 13, 2014 by yamini.v
 

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