JEE Main 2020 — Electrostatics Question with Solution

Acceleration produced by electric field,
$a=\frac{q{E}_0}{m}\hat{\mathrm{i}}$

After time $t$, velocity of the particle,
$v=u+at$

$\Rightarrow v=v_0\hat{\mathrm{j}}+\left(\frac{q{E}_{0}t}{m}\right)\hat{i} \left[\because u=v=v_0\hat{\mathrm{j}}\right]$

So, speed of the particle,
$\left|v\right|=\sqrt{v_0^2+{\left(\frac{q{E}_{0}t}{m}\right)}^2}$

Now, given $\left|v\right|=2v_0$after time $t$, so
$2v_0=\sqrt{v_0^2+{\left(\frac{q{E}_{0}t}{m}\right)}^2}$

$\Rightarrow 4v_0^2=v_0^2+{\left(\frac{q{E}_{0}t}{m}\right)}^2$

$\Rightarrow 3v_0^2={\left(\frac{q{E}_{0}t}{m}\right)}^2$

So, time interval, $t=\frac{\sqrt{3}m{v}_0}{q{E}_0}$

Here, we must note that no change in magnitude of velocity is caused by a perpendicular magnetic field. So, we are not taking effect of magnetic field while calculating change in speed.