Question

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

• A. $x=\frac{1}{2}\left(E_{0}s\right)t^3$
• B. $x=\frac{1}{6}\left(E_{0}s\right)t^3$
• C. $v=\left(E_{0}s\right)t^2$
• D. $v=\frac{1}{25}\left(E_{0}s\right)t^2$

Answer 1

The correct option is B $x=\frac{1}{6}\left(E_{0}s\right)t^3$


$E=E_{0}t$
Since $a=\frac{qE}{m}=\frac{q{E}_{0}t}{m}=E_{0}st$
$\Rightarrow \frac{dv}{dt}=E_{0}st\Rightarrow dv=E_{0}stdt$
$\Rightarrow v=E_{0}s{\int }_0^{t}dt=\frac{1}{2}{E}_{0}s{t}^2$
$\Rightarrow \frac{dx}{dt}=\frac{1}{2}{E}_{0}s{t}^2$
$\Rightarrow dx=\frac{1}{2}{E}_{0}s{t}^{2}dt$
$\Rightarrow {\int }_0^{x}dx=\frac{1}{2}{E}_{0}s{\int }_0^t{t}^{2}dt$
$x=\frac{E_{0}s{t}^3}{6}$