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Q. A particle of mass $M$ and charge $q$ is released from rest in a region of uniform electric field of magnitude $E$. After a time $t$, the distance travelled by the charge is $S$ and the kinetic energy attained by the particle is $T$. Then, the ratio $T/S$

WBJEEWBJEE 2013Electric Charges and Fields

Solution:

Given, mass of the particle $=M$
Charge on the particle $=q$
Electric field $=E$
nitial velocity, $u=0$
$\therefore $ Acceleration, $a=\frac{F}{M}=\frac{q E}{M}$
$\therefore $ Distance travelled in electric field,
$S=u t+\frac{1}{2} a t^{2} $
$ S=\frac{1}{2}\left(\frac{q E}{M}\right) t^{2}$
Also, kinetic energy $T=\frac{1}{2} M\left(\frac{q E t}{M}\right)^{2}$
So, $ \frac{T}{S}=\frac{\frac{1}{2} M\left(\frac{q E t}{M}\right)^{2}}{\frac{1}{2}\left(\frac{q E}{M}\right) t^{2}}=q E $
$\Rightarrow $ Ratio of $\frac{T}{6}$ remains constant with timet.