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The kinetic energy of a charged particle decreases by 10 J as it moves from a point at potential 100 V to a point at potential 200 V. Find the charge on the particle.

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Here, decrease in K.E. = work done = 10 J.
$V_2 = 200 V, q=?$
As work done = charge × potential diff.
$10 =q(V_2-V_1) =q(200-100)$
$q= \large\frac{10}{100}$$=0.1 \;C$
Hence D is the correct answer.
answered Jun 11, 2014 by meena.p

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