A particle having specific charge s starts from rest in a region where the electric field has constant direction but magnitude varying with time $t a s E = E_{0} t$ . In time $t$ , it is observed that the particle acquires a velocity $v$ after covering a distance $x$ , then

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$x = \frac{1}{24} (E_{0} s) t^{3}$

$v = \frac{1}{2} (E_{0} s) t^{2}$

$x = \frac{1}{5} (E_{0} s) t^{3}$

$x = \frac{1}{6} (E_{0} s) t^{3}$

answer is B, D.

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Detailed Solution

$E = E_{0} t$

Since $a = \frac{q E}{m} = \frac{q E_{0} t}{m} = E_{0} s t$

$\Rightarrow \frac{d v}{d t} = E_{0} s t \Rightarrow d v = E_{0} s t d t \Rightarrow v = E_{0} s \int_{0}^{t} t d t = \frac{1}{2} E_{0} s t^{2} \Rightarrow \frac{d x}{d t} = \frac{1}{2} E_{0} s t^{2} \Rightarrow d x = \frac{1}{2} E_{0} s t^{2} d t \Rightarrow \int_{0}^{x} d x = \frac{1}{2} E_{0} s \int_{0}^{t} t^{2} d t \Rightarrow x = \frac{1}{6} E_{0} s t^{3}$