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A particle of mass $M$ and charge $q$ is released from rest in a region of uniform electric field 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 $T / S$
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The correct answer is:
remains constant with time t
Given, mass of the particle $=M$.
Charge on the particle $=q$
Electric field $=E$
Initial velocity, $u=0$
$\therefore$ Acceleration, $a=\frac{F}{M}=\frac{q E}{M}$
:.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, $\quad \frac{T}{S}=\frac{\frac{1}{2} M\left(\frac{q E t}{M}\right)^{2}}{\frac{1}{2}\left(\frac{q E}{M}\right) t^{2}}=q E$
$\Rightarrow$ Ratio of $\frac{T}{S}$ remains constant with time t.
Charge on the particle $=q$
Electric field $=E$
Initial velocity, $u=0$
$\therefore$ Acceleration, $a=\frac{F}{M}=\frac{q E}{M}$
:.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, $\quad \frac{T}{S}=\frac{\frac{1}{2} M\left(\frac{q E t}{M}\right)^{2}}{\frac{1}{2}\left(\frac{q E}{M}\right) t^{2}}=q E$
$\Rightarrow$ Ratio of $\frac{T}{S}$ remains constant with time t.
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