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# The electric potential varries in space according to the relation V= 4x+3y . A particle of mass 1 kg starts from rest from point $\;(8 , 3.9 m )\;$ under the influence of this field . Find the velocity of the particle when it crosses the y - axis . The charge on the particle ic $\;+1 \mu C$ .

$(a)\;10^{-3} m/s\qquad(b)\;10^{-2} m/s\qquad(c)\;10^{-1} m/s\qquad(d)\;10^{-4} m/s$

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## 1 Answer

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Answer : (b) $\;10^{-2} m/s$
Explanation :
$E_{x}=-\large\frac{\partial V }{\partial x} \hat {i}= -4 \hat{i}$
Particle is initially at rest therefore u=0
$a_{x}=\large\frac{q\;E_{x}}{m}$
$a_{x}=- \large\frac{4\times10^{-6}}{1} = - 4 \times 10^{-6}$
When the particle crosses the y-axis it's x coordinate is zero
Then $\;\bigtriangleup x= 0-8=-8$
$-8=ut+\large\frac{1}{2}\;a_{x} t^2$
$-8=0+\large\frac{1}{2}\times(-4\times10^{-6})\;t^2$
$t^2=4\times10^6$
$t=2 \times 10^3 s$
At $\;t=2\times10^3 s \quad \; v_{x}=u_{x}+a_{x}t$
$=0-4\times10^{-6}\times2\times10^{3}$
$v_{x}=-8\times10^{-3}$
$v_{y}=u_{y}+a_{y} t$
$=0-3\times10^{-6}\times2\times10^{3}$
$=-6\times10^{-3}$
$|v|=\sqrt{v_{x}^2+v_{y}^2}=10^{-3}\;\sqrt{64+36}$
$|v|=10^{-2} m/s\;.$
answered Feb 18, 2014 by

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