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A proton is released from rest , 10 cm from a charged sheet carrying charge density of $\;-2.21\times10^{-9}\;c/m^2\;.$ It will strike the sheet after time $\;(approximately)$

$(a)\;4\;\mu s\qquad(b)\;2\;\mu s\qquad(c)\;2 \sqrt{2}\;\mu s\qquad(d)\;4 \sqrt{2}\;\mu s$

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Answer : (a) $\;4\;\mu s$
Explanation : $\;F= \large\frac{\sigma\; q}{\in_{0}}$
$a= \large\frac{\sigma\; q}{\;m_{p}\in_{0}}$
$\large\frac{1}{2}\;at^2 = l\quad\;=>\;t\sqrt{\large\frac{2 l m_{p} \in_{0}}{\sigma\;q}}$
$t=\sqrt{\large\frac{2\times0.1\times1.675\times10^{-27}\times8.84\times10^{-12}}{2.21\times10^{-9}\times1.6\times10^{-19}}}$
$t\;\approx\;4\times10^{-6}\;s$
answered Jan 30, 2014 by yamini.v
 

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