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A 10.0 Kg block carrying a charge 100 $\;\mu C\;$ is connected to a spring for which $\;K=100 N/m\;$ . The block lies on a frictionless horizontal track , and the system is immersed in a uniform electric field $\;E=5\times10^{5}\;V/m\;$ , directed as shown in figure . If the block is released from rest when the spring is amount does the spring expand

$(a)\;0.5 m \qquad(b)\; 1 m\qquad(c)\;0.25 m\qquad(d)\;2 m$

1 Answer

Answer : (b) 1 m
Explanation :
Work done by electric field = Work done against spring
Let $\;W_{E}\;$ be the work done by electric field . Then
$W_{E}=Q\;E\;x$
Work done by spring = - $\large\frac{1}{2}\;k\;x^2$
Therefore $\;Q\;E\;x-\large\frac{1}{2}\;k\;x^2= \bigtriangleup K.E$
For maximum expansion $\;K.E_{final} =0$
But $\;K.E_{initial}=0$
Then $\;\bigtriangleup K. E=0$
Thus $\;Q\;E\;x-\large\frac{1}{2}\;k\;x^2=0$
$x=\large\frac{2QE}{K}$
$x=\large\frac{2\times100\times10^{-6}\times5\times10^{5}}{100}$
$x=1m$
answered Feb 14, 2014 by yamini.v
 

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