A particle of mass m and charge q is placed at rest in a uniform electric field E and then released.

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1. Electric Charges and Fields

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A particle of mass $m$ and charge $q$ is placed at rest in a uniform electric field $E$ and then released. The $KE$ attained by the particle after moving a distance $y$ is

A

$qEy^2$

B

$qE^2y$

C

$qEy$

D

$q^2Ey$

Solution

As $v^{2}=0^{2}+2 a y=2(F / m) y=2\left(\frac{q E}{m}\right) y$

$K.E.$ $=\frac{1}{2} m v^{2}$

$\therefore$ K.E. $=\frac{1}{2} m\left[2 \frac{(q E)}{m} y\right] \Rightarrow \mathrm{K.E.}=q E y$

Standard 12

Physics

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