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Two point particles of same mass M are attached to two ends of a massless non-conducting rigid rod of length l. The charge on particle are +q and -q. The rod is held in a region of uniform electric field E making an angle $\theta $( very small) with E. Magnitude of torgue on rod will be equal to:


$(A)\;qEl \\ (B)\;qEl \cos \theta \\ (C)\;zero \\ (D)\;qEl \sin \theta $

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Force $F=qE$
The force acting at both the charged constitute a couple.
z= Force $\times$ perpendicular disatnce
=>$z= qE \times AB \sin \theta$
$\qquad= qEl \sin \theta$
Hence D is the correct answer.
answered Jan 15, 2014 by meena.p

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