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A charged particle of mass $m$ and charge $q$ is released from rest in an uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after $t$ seconds is
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Verified Answer
The correct answer is:
$\frac{E^2 q^2 t^2}{2 m}$
According to given situation, force on charge particle in uniform electric field, $F=q E$
Acceleration of charge particle,
$$
a=\frac{F}{m}=\frac{q E}{m}
$$
Velocity of charge particle after time $t$ is given as
$$
v=u+a t=0+\frac{q E}{m} t \Rightarrow v=\frac{q E t}{m}
$$
$\therefore$ Kinetic energy of the charged particle
$$
K=\frac{1}{2} m v^2=\frac{1}{2} m\left(\frac{q E t}{m}\right)^2=\frac{E^2 q^2 t^2}{2 m}
$$
Acceleration of charge particle,
$$
a=\frac{F}{m}=\frac{q E}{m}
$$
Velocity of charge particle after time $t$ is given as
$$
v=u+a t=0+\frac{q E}{m} t \Rightarrow v=\frac{q E t}{m}
$$
$\therefore$ Kinetic energy of the charged particle
$$
K=\frac{1}{2} m v^2=\frac{1}{2} m\left(\frac{q E t}{m}\right)^2=\frac{E^2 q^2 t^2}{2 m}
$$
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