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# A charged particle of mass 100g and charge $\;10^{-3} C\;$ is suspended through a thread . A horizontal electric field $\; E=\sqrt{3}\times10^{3} NC^{-1}\;$ exists in region . Find the angle made by thread with verticle .

$(a)\;60^{0}\qquad(b)\;30^{0}\qquad(c)\;45^{0}\qquad(d)\;None$

Answer : (a) $\;60^{0}$
Explanation :
$T cos \theta =mg$
$T sin \theta=F$
$Tan \theta =\large\frac{F}{mg}$
$Tan \theta = \large\frac{\sqrt{3}\times10^{3}\times10^{-3}}{10^{-2}\times10}$
$Tan \theta = \sqrt{3}$
$\theta= 60^{0}\;.$
edited Feb 6, 2014 by yamini.v