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A pendulum bob of mass m carying a charge 'q' is at rest with its string making an angle $\theta$ with the vertical in a uniform horizontal electric field E. The tension in the string

$a)\;\frac{mg}{\sin \theta} \; and \;\frac{qE}{\cos \theta}\\ b)\;\frac{mg}{\cos \theta} \; and \;\frac{qE}{\sin \theta} \\ c)\; \frac{mg}{Eq}\\ d)\; \frac{qE}{mg} $

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1 Answer

At equillibrium
$T \cos \theta=mg$
$T \sin \theta=qE$
$T= \large\frac{mg}{\cos \theta}$
or $ T=\large \frac{qE}{\sin \theta}$
Hence b is the correct answer.
answered Aug 9, 2013 by meena.p
edited Jun 13, 2014 by lmohan717
 

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